knowledge-route Class 9 Maths

S.No.	Topics	Page No.
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1.	Number System	$1-35$
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2.	Polynomials	$36-56$
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3.	Coordinate Geometry	$57-62$
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4.	Linear Equation in two Variable	$63-71$
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5.	Introduction of Euclid’s Geometry	$72-77$
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6.	Lines and Angles	$78-88$
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7.	Triangles	$89-98$
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8.	Quadrilateral	$99-114$
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9.	Area of parallelograms and triangle	$115-128$
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10.	Circle	$129-155$
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11.	Constructions	$156-164$
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12.	Heron’s Formula	$165-172$
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13.	Surface Area \And Volume	$173-181$
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14.	Statistics	182 - 197
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15.	Probability	$198-204$
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16.	Proof In Mathematics	$205-211$
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17.	Mathematical Modeling	$212-220$

》 ) NUMBER SYSTEM $\ll«$

CLASSIFICATION OF NUMBERS

(I) Natural numbers:

Set of all non-fractional number from 1 to $+\infty, N=\lbrace 1,2,3,4, \ldots \rbrace $

(II) Whole numbers :

Set of numbers from 0 to $+\infty, W=\lbrace 0,1,2,3,4, \ldots \rbrace $.

(III) Integers :

Set of all-non fractional numbers from $\infty$ to $+\infty$, I or $Z=\lbrace \ldots,-3,-2,-1,0,1,2,3, \ldots \rbrace $

(IV) Rational numbers :

These are real numbers which can be expressed in the form of $p / q$, where $\mathbf{p}$ and $\mathbf{q}$ are integers and $q \neq 0$.

e.g. $2 / 3,37 / 15,-17 / 19$.

All natural numbers, whole numbers and integers are rational.
Rational numbers include all Integers (without any decimal part to it), terminating fractions (fractions in which the decimal parts terminating e.g. 0.75, - 0.02 etc.) and also non-terminating but recurring decimals e.g. $0.666 \ldots .,-2.333 \ldots . .$, etc.

Fractions :

(a) Common fraction : Fractions whose denominator is not 10 .

(b) Decimal fraction $\quad$ : Fractions whose denominator is 10 or any power of 10.

(d) Improper fraction : Numerator $>$ Denominator i.e. $\frac{5}{3}$.

(e) Mixed fraction $\quad:$ Consists of integral as well as fractional part i.e. $3 \frac{2}{7}$.

(f) Compound fraction : Fraction whose numerator and denominator themselves are fractions. i.e. $\frac{2 / 3}{5 / 7}$. Improper fraction can be written in the form of mixed fractions.

(v) Irrational Numbers :

All real number which are not rational are irrational numbers. These are non-recurring as well as nonterminating type of decimal numbers e.g. $\sqrt{2}, \sqrt[3]{4}, 2+\sqrt{3}, \sqrt{2+\sqrt{3}}, \sqrt[4]{\sqrt[7]{3}}$ etc.

(vi) Real numbers : Number which can represent actual physical quantities in a meaningful way are known asreal numbers. These can be represented on the number line. Number line in geometrical straight line with arbitrarily defined zero (origin).

(vii) Prime number : All natural numbers that have one and itself only as their factors are calledprime numbers i.e. prime numbers are exactly divisible by 1 and themselves. e.g. 2,3,5,7,11,13,17,19,23 ….etc. If $P$ is the set of prime number then $\mathbf{P}={2,3,5,7 \ldots}$.

(viii) Composite numbers : All natural number, which are not prime are composite numbers. If $C$ is the set of composite number then $C={4,6,8,9,10,12, \ldots }$ 1 is neither prime nor composite number.

(ix) Co-prime numbers : If the H.C.F. of the given numbers (not necessarily prime) is 1 then they are known as co-prime numbers. e.g. 4,9 , are co-prime as H.C.F. of $(4,9)=1$. Any two consecutive numbers will always be co-prime.

(x) Even Numbers : All integers which are divisible by 2 are called even numbers. Even numbers are denoted by the expression $2 n$, where $n$ is any integer. So, if $E$ is a set even numbers, then $E={\ldots .,-4,-2,0,2, 4, \ldots }$

(xi) Odd Numbers: All integers which are not divisible by 2 are called odd numbers. Odd numbers are denoted by the general expression $2 n-1$ where $n$ is any integer. If $O$ is a set of odd numbers, then $O={\ldots ,5,-3,-1,1,3,5, \ldots } .$

(xii) Imaginary Numbers: All the numbers whose square is negative are called imaginary numbers. e.g. 3i, 4 i, i, \ldots where $i=\sqrt{-1}$.

(xiii) Complex Numbers : The combined form of real and imaginary numbers is known as complex numbers. It is denoted by $Z=A+i B$ where $A$ is real part and $B$ is imaginary part of $Z$ and $A, B \in R$.

The set of complex number is the super set of all the sets of numbers.

IDENTIFICATION PRIME NUMBER

Step 1 : Find approximate square root of given number.

Step 2 : Divide the given number by prime numbers less than approximate square root of number. If given number is not divisible by any of this prime number then the number is prime otherwise not.

Ex. 1 571, is it a prime?

Sol. Approximate square root of $571=24$.

Prime number $<24$ are $2,3,5,7,11,13,17,19, \And 23$. But 571 is not divisible by any of these prime numbers so 571 is a prime number.

Ex. 2 Is 1 prime or composite number ?

Sol. 1 is neither prime nor composite number.

REPRESENTATIO FO RATIONAL NUMBER OF A REAL NUMBER LINE

(i) 3/7 Divide a unit into 7 equal parts.

(ii) $\frac{13}{7}$

(iii) $-\frac{4}{9}$

(a) Decimal Number (Terminating) :

(i) 2.5

(ii) 2.65 (process of magnification)

Ex. $3 \quad$ Visualize the representation of $5.3 \overline{7}$ on the number line upto 5 decimal place. i.e. 5.37777 .

(b) Find Rational Numbers Between Two Integral Numbers :

Ex. 4 Find 4 rational numbers between 2 and 3.

Sol.

STEPS :

(i) Write 2 and 3 multipling in $N^{r}$ and $D^{r}$ with (4+1).

(ii) i.e. $2 \frac{2 \times (4+1)}{(4+1)} = \frac{1}{5} \And 3 = \frac{3 \times (4+1)}{(4+1)} = \frac{15}{5}$

(iii) So, the four required numbers are $\frac{11}{5}, \frac{12}{5}, \frac{13}{5}, \frac{14}{5}$.

Ex. 5 Find three rational no’s between $a$ and $b(a<b)$.

Sol. $\quad a<b$

$\Rightarrow a+a<b+a$

$\Rightarrow 2 a<a+b$

$\Rightarrow a<\frac{a+b}{2}$

Again, $a<b$

$\Rightarrow a+b<b+b$.

$\Rightarrow a+b<2 b$

$\Rightarrow \frac{a+b}{2}<b$.

$\therefore \quad a<\frac{a+b}{2}<b$.

i.e. $\frac{a+b}{2}$ lies between $a$ and $b$.

Hence 1st rational number between $a$ and $b$ is $\frac{a+b}{2}$.

For next rational number

$$ \frac{a+\frac{a+b}{2}}{2}=\frac{\frac{2 a+a+b}{2}}{2}=\frac{3 a+b}{4} \quad \therefore \quad a<\frac{3 a+b}{4}<\frac{a+b}{2}<b . $$

Next, $\quad \frac{\frac{a+b}{2}+b}{2}=\frac{a+b+2 b}{2 \times 2}=\frac{a+3 b}{4}$

$\therefore \quad a<\frac{3 a+b}{4}<\frac{a+b}{2}<\frac{a+3 b}{4}<b$, and continues like this.

Ex. 6 Find 3 rational numbers between $\frac{1}{3} \And \frac{1}{2}$.

Sol. 1st Method $\frac{\frac{1}{3}+\frac{1}{2}}{2}=\frac{\frac{2+3}{6}}{2}=\frac{5}{12} \quad \therefore \quad \frac{1}{3}, \frac{5}{12}, \frac{1}{2}$

$=\frac{\frac{1}{3}+\frac{5}{12}}{2}=\frac{\frac{4+5}{12}}{2}=\frac{9}{24}$ $\therefore \quad \frac{1}{3}, \frac{9}{24}, \frac{5}{12}, \frac{1}{2}$

$=\frac{\frac{5}{12}+\frac{1}{2}}{2}=\frac{\frac{5}{12}+\frac{6}{12}}{2}=\frac{11}{24}$

$\therefore \quad \frac{1}{3}, \frac{9}{24}, \frac{5}{12}, \frac{11}{24}, \frac{1}{2}$.

Verify : $\frac{8}{24}<\frac{9}{24}<\frac{10}{24}<\frac{11}{24}<\frac{12}{24} \Big( as\frac{8}{24}=\frac{1}{3} \And \frac{1}{2}\Big)$

$2^{\text {nd }}$ Method : Find $n$ rational numbers between $a$ and $b(a<b)$.

(i) Find $d=\frac{b-a}{n+1}$.

(ii) 1st rational number will be $a+d$.

2nd rational number will be a $+2 d$.

3 rd rational number will be $a+3 d$ and so on….

$n$th rational number is a + nd.

Ex. 7 Find 5 rational number between $\frac{3}{5}$ and $\frac{4}{5}$

Here, $a=\frac{3}{5}, b=\frac{4}{5} d=\frac{b-a}{n+1}=\frac{\frac{4}{5}-\frac{3}{5}}{5+1}=\frac{1}{5} \times \frac{1}{6}=\frac{1}{30}$.

$1^{\text {st }}=a+b=\frac{3}{5}+\frac{1}{30}=\frac{19}{20}, \quad 2^{\text {nd }}=a+2 d=\frac{3}{5}+\frac{2}{30}$,

$3^{rd}=a+3 d=\frac{3}{5}+\frac{3}{30}=\frac{21}{30}, \quad 4^{\text {th }}=a+4 d=\frac{3}{5}+\frac{4}{30}=\frac{22}{30} ，$

$5^{\text {th }}=a+5 d=\frac{3}{5}+\frac{5}{30}=\frac{23}{30}$.

RATIONAL NUMBER IN DECIMAL REPRESENTATION

(a) Terminating Decimal :

In this a finite number of digit occurs after decimal i.e. $\frac{1}{2}=0.5,0.6875,0.15$ etc.

(b) Non-Terminating and Repeating (Recurring Decimal) :

In this a set of digits or a digit is repeated continuously.

Ex.8 $\frac{2}{3}=0.6666——-=0 . \overline{6}$.

Ex. 9 $\frac{5}{11}=0.454545——=0 . \overline{45}$.

PROPERTIES OF RATIONAL NUMBER

If $a, b, c$ are three rational numbers.

(i) Commutative property of addition. $a+b=b+a$ (ii) Associative property of addition $(a+b)+c=a+(b+c)$

(iii) Additive inverse $a+(-a)=00$ is identity element, - a is called additive inverse of $a$.

(iv) Commutative property of multiplications a.b. = b.a. (v) Associative property of multiplication (a.b).c = a.(b.c)

(vi) Multiplicative inverse $(a) \times(\frac{1}{a})=1$

1 is called multiplicative identity and $\frac{1}{a}$ is called multiplicative inverse of a or reciprocal of a.

(vii) Distributive property $a .(b+c)=a . b+a . c$

$\therefore \quad$ L.H.S. $\neq$ R.H.S.

Hence in contradicts our assumption that $2+\sqrt{3}$ rational.

$\therefore \quad 2+\sqrt{3}$ is irrational.

Ex. 9 Prove that $\sqrt{3}-\sqrt{2}$ is an irrational number

Sol. Let $\sqrt{3}-\sqrt{2}=r$ where $r$ be a rational number

Squaring both sides

$\Rightarrow(\sqrt{3}-\sqrt{2})^{2}=r^{2}$

$\Rightarrow 3+2-2 \sqrt{6}=r^{2}$

$\Rightarrow \quad 5-2 \sqrt{6}=r^{2}$

Here, $\quad 5-2 \sqrt{6}$ is an irrational number but $r^{2}$ is a rational number

$\therefore \quad$ L.H.S. $\neq$ R.H.S.

Hence it contradicts our assumption that $\sqrt{3}-\sqrt{2}$ is a rational number.

(b) Irrational Number in Decimal Form :

$\sqrt{2}=1.414213 \ldots .$. i.e. it is not-recurring as well as non-terminating.

$\sqrt{3}=1.732050807$….. i.e. it is non-recurring as well as non-terminating.

Ex. 10 $\quad$ Insert an irrational number between 2 and 3 .

Sol. $\quad \sqrt{2 \times 3}=\sqrt{6}$

Ex. 11 Find two irrational number between 2 and 2.5

Sol. 1st Method : $\sqrt{2 \times 2.5}=\sqrt{5}$

Since there is no rational number whose square is 5 . So $\sqrt{5}$ is irrational..

Also $\sqrt{2 \times \sqrt{5}}$ is a irrational number.

2nd Method : 2.101001000100001…. is between 2 and 5 and it is non-recurring as well as non-terminating. Also, 2.201001000100001……… and so on.

Ex. 12 Find two irrational number between $\sqrt{2}$ and $\sqrt{3}$.

Sol. 1st Method : $\sqrt{\sqrt{2} \times \sqrt{3}}=\sqrt{\sqrt{6}}=\sqrt[4]{6} \quad$ Irrational number between $\sqrt{2}$ and $\sqrt[4]{6}$

$\sqrt{\sqrt{2} \times \sqrt[4]{6}}=\sqrt[4]{2} \times \sqrt[8]{6}$

2nd Method: As $\sqrt{2}=1.414213562 \ldots \ldots$ and $\sqrt{3}=1.732050808 \ldots \ldots$.

As, $\sqrt{3}>\sqrt{2}$ and $\sqrt{2}$ has 4 in the 1 st place of decimal while $\sqrt{3}$ has 7 is the 1 st place of decimal.

$\therefore \quad 1.501001000100001 \ldots . . . ., 1.601001000100001 \ldots . .$. etc. are in between $\sqrt{2}$ and $\sqrt{3}$

Ex. 13 Find two irrational number between 0.12 and 0.13 Sol. $\quad 0.1201001000100001 \ldots . . ., 0.12101001000100001$…….etc.

Ex. 14 Find two irrational number between 0.3030030003 ….. and 0.3010010001 …….

Sol. $\quad 0.302020020002 \ldots . . .0 .302030030003 \ldots .$. etc.

Ex. 15 Find two rational number between 0.2323323332 _…. and 0.252552555255552…….

Sol. 1st place is same 2.

2nd place is $3 \And 5$.

3rd place is 2 in both.

4 th place is $3 \And 5$.

Let a number $=0.25$, it falls between the two irrational number.

Also a number $=0.2525$ an so on.

(c) Irrational Number on a Number Line :

Ex. 16 Plot $\sqrt{2}, \sqrt{3}, \sqrt{5} \sqrt{6}$ on a number line.

Sol.

Another Method for :

(i) Plot $\sqrt{2}, \sqrt{3}$

So, $OC=\sqrt{2}$ and $OD=\sqrt{3}$

(ii) Plot $\sqrt{5}, \sqrt{6}, \sqrt{7} \sqrt{8}$

$OC=\sqrt{5}$

$OD=\sqrt{6}$

$OH=\sqrt{7} \ldots \ldots$

(d) Properties of Irrational Number :

(i) Negative of an irrational number is an irrational number e.g. $-\sqrt{3}-\sqrt[4]{5}$ are irrational.

(ii) Sum and difference of a rational and an irrational number is always an irrational number.

(iii) Sum and difference of two irrational numbers is either rational or irrational number.

(iv) Product of a non-zero rational number with an irrational number is either rational or irrationals

(v) Product of an irrational with a irrational is not always irrational.

Ex. 17 Two number’s are 2 and $\sqrt{3}$, then

Sum $=2+\sqrt{3}$, is an irrational number.

Difference $=2-\sqrt{3}$, is an irrational number.

Also $\sqrt{3}-2$ is an irrational number.

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Ex. 18 Two number’s are 4 and $\sqrt[3]{3}$, then

Sum $=4+\sqrt[3]{3}$, is an irrational number.

Difference $=4-\sqrt[3]{3}$, is an irrational number.

Ex. 19 Two irrational numbers are $\sqrt{3},-\sqrt{3}$, then

Sum $=\sqrt{3}+(-\sqrt{3})=0$ which is rational.

Difference $=\sqrt{3}-(-\sqrt{3})=2 \sqrt{3}$, which is irrational.

Ex. 20 Two irrational numbers are $2+\sqrt{3}$ and $2-\sqrt{3}$, then

Sum $=(2+\sqrt{3})+(2-\sqrt{3})=4$, a rational number

Two irrational numbers are $\sqrt{3}+3 m \sqrt{3}-3$

Difference $=\sqrt{3}+3-\sqrt{3}+3=6$, a rational number

Ex. 21 Two irrational numbers are $\sqrt{3}-\sqrt{2}, \sqrt{3}+\sqrt{2}$, then

Sum $=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}=2 \sqrt{3}$, an irrational

Ex. 22 2 is a rational number and $\sqrt{3}$ is an irrational.

$2 \times \sqrt{3}=2 \sqrt{3}$, an irrational.

Ex. 23 $\quad 0$ a rational and $\sqrt{3}$ an irrational.

$0 \times \sqrt{3}=0$, a rational.

Ex. 24 $\frac{4}{3} \times \sqrt{3}=\frac{4}{3} \sqrt{3}=\frac{4}{\sqrt{3}}$ is an irrational.

Ex. 25 $\sqrt{3} \times \sqrt{3}=\sqrt{3 \times 3}=\sqrt{9}=3$ a rational number.

Ex.26 2 $\sqrt{3} \times 3 \sqrt{2}=2 \times 3 \sqrt{3 \times 2}=6 \sqrt{6}$ and irrational number.

Ex. 27 $\quad \sqrt[3]{3} \times \sqrt[3]{3^{2}}=\sqrt[3]{3 \times 3^{2}}=\sqrt[3]{3^{3}}=3$ a rational number.

Ex. 28 $(2+\sqrt{3})(2-\sqrt{3})=(2)^{2}-(\sqrt{3})^{2}=4-3=1$ a rational number.

Ex. 29 $(2+\sqrt{3})(2+\sqrt{3})=(2+\sqrt{3})^{2}$

$ \begin{aligned} & =(2)^{2}+(\sqrt{3})^{2}+2(2) \times(\sqrt{3}) \\ & =4+3+4 \sqrt{3} \\ & =7+4 \sqrt{3} \text { an irrational number } \end{aligned} $

NOTE

(i) $\sqrt{-2} \neq-\sqrt{2}$, it is not a irrational number.

(ii) $\sqrt{-2} \times \sqrt{-3} \neq(\sqrt{-2 \times-3}=\sqrt{6})$

Instead $\sqrt{-2}, \sqrt{-3}$ are called Imaginary numbers.

$\sqrt{-2}=i \sqrt{2}$, where $i(=$ iota $)=\sqrt{-1}$

$\therefore \quad(.$ A) $i^{2}=-1$

(B) $i^{3}=i^{2} \times i=(-1) \times i=-i$

(iii) Numbers of the type $(a+i b)$ are called complex numbers where $(a, b) \in$ R.e.g. $2+3 i,-2+4 i,-3 i, 11-4 i$, are complex numbers.

GEOMETRICAL REPRESENTATION OF REAL NUMBERS

To represent any real number of number line we follows the followingSTEPS :

STEP I : Obtain the positive real number $x$ (say).

STEP II : Draw a line and mark a point A on it.

STEP III : Mark a point $B$ on the line such that $A B=x$ units.

STEP IV : From point B mark a distance of 1 unit and mark the new point as C.

STEP $V$ : Find the mid - point of $AC$ and mark the point as $O$.

STEP VI : Draw a circle with centre $O$ and radius OC.

STEP VII : Draw a line perpendicular to AC passing through B and intersecting the semi circle at D.

Length $BD$ is equal to $\sqrt{x}$.

~~ 4. Examine whether the following numbers are rational or irrational :

(i) $(2-\sqrt{3})^{2}$

(ii) $(\sqrt{2}+\sqrt{3})^{2}$

(iii) $(3+\sqrt{2})(3-\sqrt{2})$

(iv) $\frac{\sqrt{3}-1}{\sqrt{3}+1}$

~~ 5. Represent $\sqrt{8.3}$ on the number line.

~~ 6. Represent $(2+\sqrt{3})$ on the number line.

~~ 7. Prove that $(\sqrt{2}+\sqrt{5})$ is an irrational number.

~~ 8. Prove that $\sqrt{7}$ is not a rational number.

~~ 9. Prove that $(2+\sqrt{2})$ is an irrational number.

~~ 10. Multiply $\sqrt{27 a^{3} b^{2} c^{4}} \times \sqrt[3]{128 a^{7} b^{9} c^{2}} \times \sqrt[6]{729 a b^{12} c^{2}}$.

~~ 11. Express the following in the form of $p / q$.

(i) $0 . \overline{3}$

(ii) $0 . \overline{37}$

(iii) $0 . \overline{54}$

(iv) $0 . \overline{05}$

(v) $1 . \overline{3}$

(vi) $0 . \overline{621}$

~~ 12. Simplify : $0 . \overline{4}+.01 \overline{8}$

> > >
NUMBER SYSTEM < < <

SURDS

Any irrational number of the form $\sqrt[n]{a}$ is given a special name surd. Where ’ $a$ ’ is called radicand, it should always be a rational number. Also the symbol $\sqrt[n]{ }$ is called the radical sign and the index $n$ is called order of the surd.

$\sqrt[n]{a}$ is read as ’ $n^{\text {th }}$ root $a^{\prime}$ and can also be written as $a^{\frac{1}{n}}$.

(a) Some Identical Surds :

(i) $\sqrt[3]{4}$ is a surd as radicand is a rational number.

Similar examples $\sqrt[3]{5}, \sqrt[4]{12}, \sqrt[5]{7}, \sqrt{12}$

(i) $2 \sqrt{3}$ is a surd (as surd + rational number will give a surd)

Similar examples $\sqrt{3}+1, \sqrt[3]{3}+1, \ldots$.

(iii) $\sqrt{7-4 \sqrt{3}}$ is a surd as $7-4 \sqrt{3}$ is a perfect square of $(2-\sqrt{3})$

Similar examples $\sqrt{7+4 \sqrt{3}}, \sqrt{9-4 \sqrt{5}}, \sqrt{9+4 \sqrt{5}}$

(i) $\sqrt[3]{\sqrt{3}}$ is a surd as $\sqrt[3]{\sqrt{3}}=(3^{\frac{1}{2}})^{\frac{1}{3}}=3^{\frac{1}{6}}=\sqrt[6]{3}$

Similar examples $\sqrt[3]{\sqrt[3]{5}}, \sqrt[4]{\sqrt[5]{6}}, \ldots \ldots \ldots$

(b) Some Expression are not Surds :

(i) $\sqrt[3]{8}$ because $\sqrt[3]{8}=\sqrt[3]{2^{3}}=2$, which is a rational number.

(ii) $\sqrt{2+\sqrt{3}}$ because $2+\sqrt{3}$ is not a perfect square.

(iii) $\sqrt[3]{1+\sqrt{3}}$ because radicand is an irrational number.

LAWS OF SURDS

(i) $(\sqrt[n]{a})^{n}=\sqrt[n]{a^{n}}=a$

e.g. (A) $\sqrt[3]{8}=\sqrt[3]{2^{3}}=2 \quad$ (B) $\sqrt[4]{81}=\sqrt[4]{3^{4}}=3$

(ii) $\sqrt[n]{a} \times \sqrt[n]{b}=\sqrt[n]{ab} \quad \text{[Here order should be same]}$

e.g. (A) $\sqrt[3]{2} \times \sqrt[3]{6}=\sqrt[3]{2 \times 6}=\sqrt[3]{12}$

but, $\sqrt[3]{3} \times \sqrt[4]{6} \neq \sqrt{3 \times 6} \quad \text{[Because order is not same]}$

1 st make their order same and then you can multiply.

(iii) $\sqrt[n]{a}+\sqrt[n]{b}=\sqrt[n]{\frac{a}{b}}$

(iv) $\sqrt[n]{m} \sqrt{m}=\sqrt[n m]{a}=\sqrt[m]{\sqrt[n]{a}} \quad$ e.g. $=\sqrt{\sqrt{\sqrt{2}}}=\sqrt[8]{8}$

(v) $\sqrt[n]{a}=\sqrt[n \times p]{a^{p}} \quad \text{[Important for changing order of surds]}$

or, $\sqrt[n]{a^{m}}=\sqrt[n \times p]{a^{m \times p}}$

e.g. $\sqrt[3]{6^{2}}$ make its order 6 , then $\sqrt[3]{6^{2}}=\sqrt[3 \times 2]{6^{2 \times 2}}=\sqrt[6]{6^{4}}$.

e.g. $\sqrt[3]{6}$ make its order 15 , then $\sqrt[3]{6}=\sqrt[3 \times 5]{6^{1 \times 5}}=\sqrt[15]{6^{5}}$.

OPERATION OF SURDS

(a) Addition and Subtraction of Surds :

Addition and subtraction of surds are possible only when order and radicand are same i.e. only for surds.

Ex. 1 Simplify

(i) $\sqrt{6}-\sqrt{216}+\sqrt{96}=15 \sqrt{6}-\sqrt{6^{2}} \times 6+\sqrt{16 \times 6} \quad \text{[Bring surd in simples form]}$

$ \begin{aligned} & =(15-6+4) \sqrt{6} \\ & =13 \sqrt{6} \end{aligned} $

$ =15 \sqrt{6}-6 \sqrt{6}+4 \sqrt{6} $

Ans.

(ii) $5 \sqrt[3]{250}+7 \sqrt[3]{16}-14 \sqrt[3]{54}=5 \sqrt[3]{125 \times 2}+7 \sqrt[3]{8 \times 2}-14 \sqrt[3]{27 \times 2}$

$ =5 \times 5 \sqrt[3]{2}+7 \times 2 \sqrt[3]{2}-14 \times 3 \times \sqrt[3]{2} $

$ \begin{aligned} & =(25+14-42) \sqrt[3]{2} \\ & =-3 \sqrt[3]{2} \quad \text{Ans.} \end{aligned} $

(ii) $5 \sqrt[3]{250}+7 \sqrt[3]{16}-14 \sqrt[3]{54}=5 \sqrt[3]{125 \times 2}+7 \sqrt[3]{8 \times 2}-14 \sqrt[3]{27 \times 2}$

$ =5 \times 5 \sqrt[3]{2}+7 \times 2 \sqrt[3]{2}-14 \times 3 \times \sqrt[3]{2} $

$ \begin{aligned} & =(25+14-42) \sqrt[3]{2} \\ & =-3 \sqrt[3]{2} \end{aligned} $

Ans.

(iii) $4 \sqrt{3}+3 \sqrt{48}-\frac{5}{2} \sqrt{\frac{1}{3}}=4 \sqrt{3}+3 \sqrt{16 \times 3}-\frac{5}{2} \sqrt{\frac{1 \times 3}{3 \times 3}}$

$ \begin{aligned} & =4 \sqrt{3}+3 \times 4 \sqrt{3}-\frac{5}{2} \times \frac{1}{3} \sqrt{3} \\ & =4 \sqrt{3}+12 \sqrt{3}-\frac{5}{6} \sqrt{3} \\ & =(4+12-\frac{5}{6}) \sqrt{3} \\ & =\frac{91}{6} \sqrt{3} \quad \text{Ans.} \end{aligned} $

(b) Multiplication and Division of Surds :

Ex. 2 (i) $\sqrt[3]{4} \times \sqrt[3]{22}=\sqrt[3]{4 \times 22}=\sqrt[3]{2^{3} \times 11}=2 \sqrt[3]{11}$

(i) $\sqrt[3]{2} \times \sqrt[4]{3}=\sqrt[12]{2^{4}} \times \sqrt[12]{3^{3}}=\sqrt[12]{2^{4} \times 3^{3}}=\sqrt[12]{16 \times 27}=\sqrt[12]{432}$

Ex. 3 Simplify $\sqrt{8 a^{5} b} \times \sqrt[3]{4 a^{2} b^{2}}$

Hint : $\sqrt[6]{8^{3} a^{15} b^{3}} \times \sqrt[6]{4^{2} a^{4} b^{4}}=\sqrt[6]{2^{13} a^{19} b^{7}}=\sqrt[6]{2 a b}$.

Ans.

Ex. 4 Divide $\sqrt{24} \div \sqrt[3]{200}=\frac{\sqrt{24}}{\sqrt[3]{200}}=\frac{\sqrt[6]{(24)^{3}}}{\sqrt[6]{(200)^{2}}}=\sqrt[6]{\frac{216}{625}}$

Ans. .

(c) Comparison of Surds :

It is clear that if $x>y>0$ and $n>1$ is a positive integer then $\sqrt[n]{x}>\sqrt[n]{y}$.

Ex. $5 \sqrt[3]{16}>\sqrt[3]{12}, \sqrt[5]{35}>\sqrt[5]{25}$ and so on.

Ex. 6 Which is greater is each of the following :

(i) $\sqrt[3]{16}$ and $\sqrt[5]{8}$

(ii) $\sqrt{\frac{1}{2}}$ and $\sqrt[3]{\frac{1}{3}}$

L.C.M. of 3 and 515.

L.C.M. of 2 and 3 is 6.

$ \begin{aligned} & \sqrt[3]{6}=\sqrt[3 \times 5]{6^{5}}=\sqrt[15]{7776} \\ \\ & \sqrt[6]{\Big(\frac{1}{2}\Big)^{3}} \text { and } \sqrt[3]{\Big(\frac{1}{3}\Big)^{2}} \\ \\ & \sqrt[5]{8}=\sqrt[3 \times 5]{8^{5}}=\sqrt[15]{512}\\ \\ & \sqrt[6]{\frac{1}{8}} \text { and } \sqrt[6]{\frac{1}{9}} \quad \Big[\text { As } 8<9 \therefore \frac{1}{8}>\frac{1}{9}\Big] \\ \\ & \therefore \quad \sqrt[75]{7776}>\sqrt[15]{512} \\ \\ & \text { so, } \quad \sqrt[6]{\frac{1}{8}}>\sqrt[6]{\frac{1}{9}} \\ \\ & \Rightarrow \sqrt[3]{6}>\sqrt[5]{8} \\ \\ & \Rightarrow \quad \sqrt{\frac{1}{2}}>\sqrt[3]{\frac{1}{3}} \end{aligned} $

Ex. 7 Arrange $\sqrt{2}, \sqrt[3]{3}$ and $\sqrt[4]{5}$ is ascending order.

Sol. L.C.M. of 2, 3, 4 is 12. $ \begin{aligned} \therefore \quad \sqrt{2} & =\sqrt[2 \times 6]{2^{6}}=\sqrt[12]{64} \\ \sqrt[3]{3} & =\sqrt[3 \times 4]{3^{4}}=\sqrt[12]{81} \\ \sqrt[4]{5} & =\sqrt[4 \times 3]{5^{3}}=\sqrt[12]{125} \end{aligned} $

As, $64<81<125$.

$\therefore \quad \sqrt[12]{64}<\sqrt[12]{81}<\sqrt[12]{125}$

$ \Rightarrow \sqrt{2}<\sqrt[3]{3}<\sqrt[4]{5} $

Ex. 8 Which is greater $\sqrt{7}-\sqrt{3}$ or $\sqrt{5}-1$ ?

Sol. $\quad \sqrt{7}-\sqrt{3}=\frac{(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})}{(\sqrt{7}+\sqrt{3})}=\frac{7-3}{\sqrt{7}+\sqrt{3}}=\frac{4}{\sqrt{7}+\sqrt{3}}$

And, $\quad \sqrt{5}-1=\frac{(\sqrt{5}-1)(\sqrt{5}+1)}{(\sqrt{5}+1)}=\frac{5-1}{\sqrt{5}+1}=\frac{4}{\sqrt{5}+1}$

Now, we know that $\sqrt{7}>\sqrt{5}$ and $\sqrt{3}>1$, add

So, $\sqrt{7}+\sqrt{3}>\sqrt{5}+1$

$\Rightarrow \quad \frac{1}{\sqrt{7}+\sqrt{3}}<\frac{1}{\sqrt{5}+1}$

$\Rightarrow \quad \frac{4}{\sqrt{7}+\sqrt{3}}<\frac{4}{\sqrt{5}+1} \quad \Rightarrow \quad \sqrt{7}-\sqrt{3}<\sqrt{5}-1$

So, $\sqrt{5}-1>\sqrt{7}-\sqrt{3}$

RATIONALIZATION OF SURDS

Rationalizing factor product of two surds is a rational number then each of them is called the rationalizing factor (R.F.) of the other. The process of converting a surd to a rational number by using an appropriate multiplier is known asrationalization.

Some examples :

(i) R.F. of $\sqrt{a}$ is $\sqrt{a} \quad(\therefore \sqrt{a} \times \sqrt{a}=a)$.

(ii) R.F. of $\sqrt[3]{a}$ is $\sqrt[3]{a^{2}}(\therefore \sqrt[3]{a} \times \sqrt[3]{a^{2}}=\sqrt[3]{a^{3}}=a)$.

(iii) R.F. of $\sqrt{a}+\sqrt{b}$ is $\sqrt{a}-\sqrt{b}$ $ \And \text{vice versa} $ $\lfloor \therefore(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b\rfloor$.

(iv) R.F. of $a+\sqrt{b}$ is $a-\sqrt{b} \And \text{vice versa}$ $[\therefore(a+\sqrt{b})(a-\sqrt{b})=a^{2}-b]$

(v) R.F. of $\sqrt[3]{a}+\sqrt[3]{b}$ is $(\sqrt[3]{a^{2}}-\sqrt[3]{a b}+\sqrt[3]{b^{2}})[\therefore(\sqrt[3]{a}+\sqrt[3]{b})(\sqrt[3]{a^{2}}-\sqrt[3]{a b}+\sqrt[3]{b^{2}})]$ $[\therefore(\sqrt[3]{a})^{3}+(\sqrt[3]{b})^{3}=a+b]$ which is rational.

(vi) R.F. of $(\sqrt{a}+\sqrt{b}+\sqrt{c})$ is $(\sqrt{a}+\sqrt{b}-\sqrt{c}) nd(a+b-c+2 \sqrt{a b})$.

Ex. 9 Find the R.G. (rationalizing factor) of the following : (i) $\sqrt{10}$ (ii) $\sqrt{12}$ (iii) $\sqrt{162}$ (iv) $\sqrt[3]{4}(v) \sqrt[3]{16}$ (vi) $\sqrt[4]{162}$ (vii) $2+\sqrt{3}$

(viii) $7-4 \sqrt{3}$

(ix) $3 \sqrt{3}+2 \sqrt{2}$

(x) $\sqrt[3]{3}+\sqrt[3]{2}$

(xi) $1+\sqrt{2}+\sqrt{3}$

(i) $\sqrt{10}$

Sol. $[\therefore \sqrt{10} \times \sqrt{10}=\sqrt{10 \times 10}=10]$ as 10 is rational number.

$\therefore \quad$ R.F. of $\sqrt{10}$ is $\sqrt{10}$Ans.

(ii). $\quad \sqrt{12}$Sol. First write it’s simplest from i.e. $2 \sqrt{3}$.

Now find R.F. (i.e. R.F. of $\sqrt{3}$ is $\sqrt{3}$ )

$\therefore$ R.F. of $\sqrt{12}$ is $\sqrt{3}$Ans.

(iii) $\quad \sqrt{162}$Sol. Simplest from of $\sqrt{162}$ is $9 \sqrt{2}$. R.F. of $\sqrt{2}$ is $\sqrt{2}$.

$\therefore$ R.F. of $\sqrt{162}$ is $\sqrt{2}$Ans.

(iv) $\quad \sqrt[3]{4}$Sol. $\sqrt[3]{4} \times \sqrt[3]{4^{2}}=\sqrt[3]{4^{3}}$

$\therefore \quad$ R.F. of $\sqrt[3]{4}$ is $\sqrt[3]{4^{2}} \quad$Ans.

(v). $\quad \sqrt[3]{16}$

Sol. Simplest from of $\sqrt[3]{16}$ is $2 \sqrt[3]{2} \quad$ Now R.F. of $\sqrt[3]{2}$ is $\sqrt[3]{2^{2}}$

$\therefore \quad$ R.F. of $\sqrt[3]{16}$ is $\sqrt[3]{2^{2}} \quad$Ans.

(vi) $\sqrt[4]{162}$

Sol. Simplest form of $\sqrt[4]{162}$ is $3 \sqrt[4]{2}$

Now R.F. of $\sqrt[4]{2}$ is $\sqrt[4]{2^{3}}$

$ \text { R.F. of }(\sqrt[4]{162}) \text { is } \sqrt[4]{2^{3}} $

Ans. (vii) $2+\sqrt{3}$

Sol. As $(2+\sqrt{3})(2-\sqrt{3})=(2)^{2}-(\sqrt{3})^{2}=4-3=1$, which is rational.

$\therefore$ R.F. of $(2+\sqrt{3})$ is $(2-\sqrt{3})$Ans.

(viii) $7-4 \sqrt{3}$

Sol. As $(7-4 \sqrt{3})(7+4 \sqrt{3})=(7)^{2}-(4-\sqrt{3})^{2}=49-48=1$, which is rational

$\therefore \quad$ R.F. of $(7-4 \sqrt{3})$ is $(7+4 \sqrt{3})$

Ans. (ix). $\quad 3 \sqrt{3}+2 \sqrt{2}$

Sol. As $(3 \sqrt{3}+2 \sqrt{2})(3 \sqrt{3}-2 \sqrt{2})=(3 \sqrt{3})^{2}-(2 \sqrt{2})^{2}=27-8=19$, which is rational.

$\therefore \quad$ R.F. of $(3 \sqrt{3}+2 \sqrt{2})$ is $(3 \sqrt{3}-2 \sqrt{2})$Ans.

(x) $\sqrt[3]{3}+\sqrt[3]{2}$

Sol. As $(\sqrt[3]{3}+\sqrt[3]{2})(\sqrt[3]{3^{2}}-\sqrt[3]{3} \times \sqrt[3]{2}+\sqrt[3]{2^{2}})=(\sqrt[3]{3^{3}}+\sqrt[3]{2^{3}})=3+2=5$, which is rational.

$\therefore \quad$ R.F. of $(\sqrt[3]{3}+\sqrt[3]{2})$ is $(\sqrt[3]{3^{2}}-\sqrt[3]{3} \times \sqrt[3]{2}+\sqrt[3]{2^{2}})$Ans.

(xi) $1+\sqrt{2}+\sqrt{3}$

Sol. $\quad(1+\sqrt{2}+\sqrt{3})(1+\sqrt{2}-\sqrt{3})=(1+\sqrt{2})^{2}-(\sqrt{3})^{2}$

$ \begin{aligned} & =1)^{2}+(\sqrt{2})^{2}+2(1)(\sqrt{2})-3 \\ & =1+2+2 \sqrt{2}-3 \\ & =3+2 \sqrt{2}-3 \\ & =2 \sqrt{2} \\ 2 \sqrt{2} \times \sqrt{2} & =2 \times 2=4 \end{aligned} $

$\therefore \quad$ R.F. of $1+\sqrt{2}+\sqrt{3}$ is $(1+\sqrt{2}-\sqrt{3})$ and $\sqrt{2}$.** Ans.**

NOTE : R.F. of $\sqrt{a}+\sqrt{b}$ or $\sqrt{a}-\sqrt{b}$ type surds are also called conjugate surds \And vice versa.

Ex. 10

(i) $2-\sqrt{3}$ is conjugate of $2+\sqrt{3}$

(ii) $\sqrt{5}+1$ is conjugate of $\sqrt{5}-1$

NOTE : Sometimes conjugate surds and reciprocals are same.

Ex. 11 (i) $2+\sqrt{3}$, it’s conjugate is $2-\sqrt{3}$, its reciprocal is $2-\sqrt{3}$ \And vice versa.

(ii) $5-2 \sqrt{6}$, it’s conjugate is $5+2 \sqrt{6}$, its reciprocal is $5-2 \sqrt{6}$ \And vice versa.

(iii) $6-\sqrt{35}, 6+\sqrt{35}$

(iv) $7-4 \sqrt{3}, 7+4 \sqrt{3}$

(v) $8+3 \sqrt{7}, 8-3 \sqrt{7}$ $\ldots$and so on.

Ex. 12 Express the following surd with a rational denominator.

Sol. $.\frac{8}{\sqrt{15}+1-\sqrt{5}-\sqrt{3}}=\frac{8}{[(\sqrt{15}+1)-(\sqrt{15}+\sqrt{3})}] \times[\frac{(\sqrt{15}+1)+(\sqrt{5}+\sqrt{3})}{(\sqrt{15}+1)+(\sqrt{5}+\sqrt{3})}]$

$ \begin{aligned} & =\frac{8(\sqrt{15}+1+\sqrt{5}+\sqrt{3})}{(\sqrt{15}+1)^{2}-(\sqrt{5}+\sqrt{3})^{2}} \\ & =\frac{8(\sqrt{15}+1+\sqrt{5}+\sqrt{3})}{15+1+2 \sqrt{15}-(5+3+2 \sqrt{15})} \\ & =\frac{8(\sqrt{15}+1+\sqrt{5}+\sqrt{3})}{8} \end{aligned} $

$ =(\sqrt{15}+1+\sqrt{5}+\sqrt{3}) $

Ans.

Ex. 13 Rationalize the denominator of $\frac{a^{2}}{\sqrt{a^{2}+b^{2}+b}}$

Sol. $\frac{a^{2}}{\sqrt{a^{2}+b^{2}}+b}=\frac{a^{2}}{\sqrt{a^{2}+b^{2}+b}} \times \frac{\sqrt{a^{2}+b^{2}}-b}{a^{2}+b^{2}-b}$

$ \begin{aligned} & =\frac{a^{2}(\sqrt{a^{2}+b^{2}}-b)}{(\sqrt{a^{2}+b^{2}})^{2}-(b)^{2}} \\ & =\frac{a^{2}(\sqrt{a^{2}+b^{2}}-b)}{a^{2}+b^{2}-b^{2}}=(\sqrt{a^{2}+b^{2}}-b) \end{aligned} $

Ans.

Ex. 14 If $\frac{3+2 \sqrt{2}}{3-\sqrt{2}}=a+b \sqrt{2}$, where $a$ and $b$ are rational then find the values of $a$ and $b$.

Sol. L.H.S. $\frac{3+2 \sqrt{2}}{3-\sqrt{2}}=\frac{(3+2 \sqrt{2})(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$

$ \begin{aligned} & =\frac{9+3 \sqrt{2}+6 \sqrt{2}+4}{9-2} \\ & =\frac{13+9 \sqrt{2}}{7} \\ & =\frac{13}{7}+\frac{9}{7} \sqrt{2} \\ \therefore \quad \frac{13}{7} & +\frac{9}{7} \sqrt{2}=a+b \sqrt{2} \end{aligned} $

Equating the rational and irrational parts

We get $a=\frac{13}{7}, b=\frac{9}{7}$

Ans.

Ex. 15 If $\sqrt{3}=1.732$, find the value of $\frac{1}{\sqrt{3}-1}$

Sol. $\frac{1}{\sqrt{3}-1}=\frac{1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$ \begin{aligned} & =\frac{\sqrt{3}+1}{3-1} \\ & =\frac{\sqrt{3}+1}{2} \\ & =\frac{1.732+1}{2} \\ & =\frac{2.732}{2} \\ & =1.366 \end{aligned} $

Ans.

Ex. 16 If $\sqrt{5}=2.236$ and $\sqrt{2}=1.414$, then

Evaluate : $\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{4}{\sqrt{5}-\sqrt{2}}$

Sol. $\frac{3}{\sqrt{5}+\sqrt{2}}+\frac{4}{\sqrt{5}-\sqrt{2}}=\frac{3 \sqrt{5}-\sqrt{2}+4(\sqrt{5}+\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}$

$ \begin{aligned} & =\frac{3 \sqrt{5}-3 \sqrt{2}+4 \sqrt{5}+4 \sqrt{2}}{5-2} \\ & =\frac{7 \sqrt{5}+\sqrt{2}}{5-2} \end{aligned} $

$ \begin{aligned} & =\frac{7 \sqrt{5}+\sqrt{2}}{3} \\ & =\frac{7 \times 2.236+1.414}{3} \\ & =\frac{15.652+1.414}{3} \\ & =\frac{17.066}{3} \\ & =5.689 \text { (approximate) } \end{aligned} $

Ex. 17 If $=\frac{1}{2+\sqrt{3}}$ find the value of $x^{3}-x^{2}-11 x+3$.

Sol. As, $x=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}$

$ \begin{aligned} & \Rightarrow \quad x-2=-\sqrt{3} \\ & \Rightarrow \quad(x-2)^{2}=(-\sqrt{3})^{2} \quad \text { [By squaring both sides] } \\ & \Rightarrow \quad x^{2}+4-4 x=3 \\ & \Rightarrow \quad x^{2}-4 x+1=0 \end{aligned} $

Now, $x^{3}-x^{2}-11 x+3=x^{3}-4 x^{2}+x+3 x^{2}-12 x+3$

$$ =x(x^{2}-4 x+1)+3(x^{2}-4 x+1) $$

$$ =x(0)+3(0) $$

$$ =0+0=0 \quad$$

Ex. 18 If $x=3-\sqrt{8}$, find the value of $x^{3}+\frac{1}{x^{3}}$.

Sol. $x=3-\sqrt{8}$

$ \begin{aligned} & \therefore \quad \frac{1}{x}=\frac{1}{3-\sqrt{8}} \\ & \Rightarrow \quad \frac{1}{x}=3+\sqrt{8} \end{aligned} $

Now, $x+\frac{1}{x}=3-\sqrt{8}+3+\sqrt{8}=6$

$\Rightarrow x^{3}+\frac{1}{x^{3}}=\Big(x+\frac{1}{x})^{3}-3 x \frac{1}{x}(x+\frac{1}{x}\Big)$

$\Rightarrow \quad x^{3}+\frac{1}{x^{3}}=(6)^{3}-3(6)$

$\Rightarrow x^{3}+\frac{1}{x^{3}}=216-18$

$\Rightarrow x^{3}+\frac{1}{x^{3}}=198$

Ans.

Ex. 19

If $x=1+2^{1 / 3}+2^{2 / 3}$, show that $x^{3}-3 x^{2}-3 x-1=0$

Sol.

$\begin{aligned} & x=1+2^{1 / 3}+2^{2 / 3} \\ \Rightarrow \quad & x-1(2^{1 / 3}+2^{2 / 3})\end{aligned}$

$\Rightarrow \quad(x-1)^{3}=(2^{1 / 3}+2^{2 / 3})^{3}$

$\Rightarrow \quad(x-1)^{3}=(2^{1 / 3})+(2^{2 / 3})^{3}+3 \cdot 2^{1 / 3} \cdot 2^{2 / 3}(2^{1 / 3}+2^{1 / 3})$

$\Rightarrow(x-1)^{3}=2+2^{2}+3.2^{1}(x-1)$

$\Rightarrow \quad(x-1)^{3}=6+6(x-1)$

$\Rightarrow \quad x^{3}-3 x^{2}+3 x-1=6 x$

$\Rightarrow x^{3}-3 x^{2}-3 x-1=0$

Ans.

Ex. 20 Solve : $\sqrt{x+3}+\sqrt{x-2}=5$.

Sol. $\sqrt{x+3}=5-\sqrt{x-2}$

$\Rightarrow(\sqrt{x+3})^{2}=(5-\sqrt{x-2})^{2} \quad$ [By squaring both sides]

$\Rightarrow x+3=25+(x-2)-10 \sqrt{x-2}$

$\Rightarrow x+3=25+x-2-10 \sqrt{x-2}$

$\Rightarrow 3-23=-10 \sqrt{x-2}$

$\Rightarrow-20=-10 \sqrt{x-2}$

$\Rightarrow 2=\sqrt{x-2}$

$\Rightarrow \quad x-2=4 \quad$ [By squaring both sides]

$\Rightarrow x=6 \quad$Ans.

Ex. 21 If $x=1+\sqrt{2}+\sqrt{3}$, prove that $x^{4}-4 x^{3}-4 x^{2}+16-8=0$.

Hint : $x=1+\sqrt{2}+\sqrt{3}$

$\Rightarrow x-1=\sqrt{2}+\sqrt{3}$

$\Rightarrow \quad(x-1)^{2}=(\sqrt{2}+\sqrt{3})^{2} \quad$ [By squaring both sides]

$\Rightarrow x^{2}+1-2 x=2+3+2 \sqrt{6}$

$\Rightarrow x^{2}-2 x-4=2 \sqrt{6}$

$\Rightarrow(x^{2}-2 x-4)^{2}=(2 \sqrt{6})^{2}$

$\Rightarrow x^{4}+4 x^{2}+16-4 x^{3}+16 x-8 x^{2}=24$

$\Rightarrow x^{4}-4 x^{3}-4 x^{2}+16 x+16-24=0$

$\Rightarrow x^{4}-4 x^{3}-4 x^{2}+16 x-8=0$

Ans.

EXPONENTS OF REAL NUMBER

(a) Positive Integral Power :

For any real number a and a positive integer ’ $n$ ’ we define $a^{n}$ as :

$a^{n}=a \times a \times a \times \ldots \ldots \ldots \ldots . . . . \times a(n$ times)

$a^{n}$ is called then $n^{\text {th }}$ power of $a$. The real number ’ $a$ ’ is called the base and ’ $n$ ’ is called the exponent of the $n^{\text {th }}$ power of $a$.

e.g. $2^{3}=2 \times 2 \times 2=8$

NOTE : For any non-zero real number ’ $a$ ’ we define $a^{0}=1$.

e.g. thus, $3^{0}=1,5^{0},\Big(\frac{3}{4}\Big)^{0}=1$ and so on.

(b) Negative Integral Power :

For any non-zero real number ’ $a$ ’ and a positive integer ’ $n$ ’ we define $a^{-n}=\frac{1}{a^{n}}$

Thus we have defined $a^{n}$ find all integral values of $n$, positive, zero or negative. $a^{n}$ is called the $n^{\text {th }}$ power of a.

RATIONAL EXPONENTS OR A REAL NUMBER

(a) Principal of $\mathbf{n}^{\text {th }}$ Root of a Positive Real Numbers :

If ’ $a$ ’ is a positive real number and ’ $n$ ’ is a positive integer, then the principal $n^{\text {th }}$ root of a is the unique positive real number $x$ such that $x^{n}=a$.

The principal $n^{\text {th }}$ root of a positive real number a is denoted by $a^{1 / n}$ or $\sqrt[n]{a}$.

(b) Principal of $n^{\text {th }}$ Root of a Negative Real Numbers :

If ’ $a$ ’ is a negative real number and ’ $n$ ’ is an odd positive integer, then the principle $n^{\text {th }}$ root of a is define as $|a|^{1 / n}$ i.e. the principal $n^{\text {th }}$ root of $-a$ is negative of the principal $n^{\text {th }}$ root of $|a|$.

Remark :

It ’ $a$ ’ is negative real number and ’ $n$ ’ is an even positive integer, then the principle $n^{\text {th }}$ root of a is not defined, because an even power of real number is always positive. Therefore $(-9)^{1 / 2}$ is a meaningless quantity, if we confine ourselves to the set of real number, only.

(c) Rational Power (Exponents) :

For any positive real number ’ $a$ ’ and a rational number $\frac{p}{q} \neq$ where $q \neq 0$, we define $a^{p / q}=(a^{p})^{1 / q}$ i.e. ap/q is the principle $q^{\text {th }}$ root of ap.

LAWS OF RATIONAL EXPONETNS

The following laws hold the rational exponents

(i) $a^{m} \times a^{n}=a^{m+1}$

(ii) $a^{m} \div a^{n}=a^{m-n}$

(iii) $(a^{m})^{n}=a^{m n}$

(iv) $a^{-n}=\frac{1}{a^{n}}$

(v) $a^{m / n}=(a^{m})^{1 / n}=(a^{1 / n})^{m}$ i.e. $a^{m / n}=\sqrt[n]{a^{m}}=(\sqrt[n]{a})^{m}$

(vi) $(a b)^{m}=a^{m b} b^{m}$

(vii) $(\frac{a}{b})^{m}=\frac{a^{m}}{b^{m}}$

(viii) $a^{b n}=a^{b+b+b \ldots n}$ tmes

Where $a, b$ are positive real number and $m, n$ are relational numbers.

ILLUSTRATIONS :

Ex. 22 Evaluate each of the following:

(i) $5^{2} \times 5^{4}$ (ii) $5^{8} \div 5^{3}$ (iii) $(3^{2})^{2}$ (iv) $(\frac{11}{12})^{3}$ (v) $(\frac{3}{4})^{-3}$

Sol. Using the laws of indices, we have

(i) $5^{2} .5^{4}=5^{2+4}=5^{6}=15625 \quad \because a^{m} \times a^{n}=a^{m+n}$

(ii) $5^{8} \div 5^{3}=\frac{5^{8}}{5^{3}}=5^{8-3}=5^{5}=3125 \quad \because a^{m}+a^{n}=a^{m-n}$

(iii) $(3^{2})^{3}=3^{2 \times 3}=3^{6}=729 \quad \because(a^{m})^{n}=a^{m \times n}$

(iv) $\Big(\frac{11}{12}\Big)^{3}=\frac{11^{3}}{12^{3}}=\frac{1331}{1728} \quad \because \Big(\frac{a}{b}\Big)^{m}=\frac{a^{m}}{b^{m}}$

(v) $\Big(\frac{3}{4}\Big)^{-3}=\frac{1}{\Big(\frac{3}{4}\Big)^{3}}=\frac{1}{\frac{3^{3}}{4^{3}}}=\frac{1}{\frac{27}{64}}=\frac{64}{27} \quad \because a^{-n}=\frac{1}{a^{n}}$

Ex. 23****Evaluate each of the following :

(i) $\Big(\frac{2}{11}\Big)^{4} \times\Big(\frac{11}{3}\Big)^{2} \times\Big(\frac{3}{2}\Big)^{3}$ (ii) $\Big(\frac{1}{2}\Big)^{5} \times\Big(\frac{-2}{3}\Big)^{4} \times\Big(\frac{3}{5}\Big)^{-1}$ (iii) $2^{55} \times 2^{60}-2^{97} \times 2^{18}$ (iv) $\Big(\frac{2}{3}\Big)^{3} \times\Big(\frac{2}{5}\Big)^{-3} \times\Big(\frac{3}{5}\Big)^{2}$

Sol. We have.

(i) $\Big(\frac{2}{11}\Big)^{4} \times\Big(\frac{11}{3}\Big)^{2} \times\Big(\frac{3}{2}\Big)^{3}=\frac{2^{4}}{11^{4}} \times \frac{11^{2}}{3^{2}} \times \frac{3^{3}}{2^{3}}$

$ \begin{aligned} & =\frac{2 \times 3}{11^{2}} \\ & =\frac{6}{121} \end{aligned} $

Ans.

(ii) We have,

$ \begin{aligned} \Big(\frac{1}{2}\Big)^{5} \times\Big(\frac{-2}{3}\Big)^{4} & \times\Big(\frac{3}{5}\Big)^{-1}=\Big(\frac{1}{2}\Big)^{5} \times\Big(\frac{-2}{3}\Big)^{4} \times\Big(\frac{1}{3}\Big) \\ \\ & =\frac{1^{5}}{2^{5}} \times \frac{(-2)^{2}}{3^{4}} \times \frac{5}{3} \\ \\ & =\frac{1 \times 16 \times 5}{32 \times 81 \times 3}\\ \\ & =\frac{5}{2 \times 81 \times 3} \\ \\ & =\frac{5}{486} \end{aligned} $

Ans. (iii) We have,

$ \begin{aligned} 2^{55} \times 2^{60}-2^{97} \times 2^{18} & ={ }^{55+60}-2^{97+18} \\ & =2^{15}-2^{115} \\ & =0 \end{aligned} $

Ans.

(iv) We have,

$ \begin{gathered} \Big(\frac{2}{3}\Big)^{3} \times\Big(\frac{2}{5}\Big)^{-3} \times\Big(\frac{3}{5}\Big)^{-2}=\frac{2^{3}}{3^{3}} \times \frac{1}{2 / 5} \times \frac{3^{2}}{5^{2}} \\ \\ =\frac{2^{3}}{3^{3}} \times \frac{1}{2^{3} / 5^{3}} \times \frac{3^{2}}{5^{2}} \\ \\ =\frac{2^{3} \times 5^{3} \times 3^{2}}{3^{3} \times 2^{3} \times 5^{2}} \\ \\ =\frac{5}{3} \end{gathered} $

Ans.

Ex. 24 Simplify :

(i) $\frac{(25)^{3 / 2} \times(243)^{3 / 5}}{(16)^{5 / 4} \times(8)^{4 / 3}}$ (ii) $\frac{16 \times 2^{n+1}-4 \times 2^{n}}{16 \times 2^{n+2}-2 \times 2^{n+2}}$

Sol. We have,

(i) $\frac{(25)^{3 / 2} \times(243)^{3 / 5}}{(16)^{5 / 4} \times(8)^{4 / 3}}=\frac{(5^{2})^{3 / 2} \times(3^{5})^{3 / 5}}{(2^{4})^{5 / 4} \times(2^{3})^{4 / 3}}$

$ =\frac{5^{2 \times 3 / 2} \times 3^{5 \times 3 / 5}}{2^{4 \times 5 / 4} \times 2^{3 \times 4 / 3}} $

$=\frac{5^{3} \times 3^{3}}{2^{5} \times 2^{4}}$

$=\frac{125 \times 27}{32 \times 16}$

$=\frac{3375}{512}$

Ans.

(ii) $\frac{16 \times 2^{n+1}-4 \times 2^{n}}{16 \times 2^{n+2}-2 \times 2^{n+2}}=\frac{2^{4} \times 2^{n+1}-2^{2} \times 2^{n}}{2^{4} \times 2^{n+2}-2 \times 2^{n+2}}$

$ \begin{aligned} & =\frac{2^{n+5}-2^{n+2}}{2^{n+6}-2^{n+3}} \\ & =\frac{2^{n+5}-2^{n+2}}{2.2^{n+5}-2.2^{n+2}} \\ & =\frac{2^{n+5}-2^{n+2}}{2(2^{n+5}-2^{n+2})}=\frac{1}{2} \end{aligned} $

Ans.

Ex. 25 Simplify $\Big(\frac{81}{16}\Big)^{-3 / 4} \times\Big[\Big(\frac{25}{9}\Big)^{-3 / 2} \div\Big(\frac{5}{2}\Big)^{-3}\Big]$

Sol. We have

$$ \Big(\frac{81}{16}\Big)^{-3 / 4} \times \Big[\Big(\frac{25}{9}\Big)^{-3 / 2} \div\Big(\frac{5}{2}\Big)^{-3}\Big]=\Big(\frac{3^{4}}{2^{4}}\Big)^{-3 / 4} \times\Big[\Big(\frac{5^{2}}{3^{2}}\Big)^{-3 / 2} \div\Big(\frac{5}{2}\Big)^{-3}\Big] $$

$$ \quad=\Big[\Big(\frac{3}{2}\Big)^{4}\Big]^{-3 / 4} \times\Big[\Big(\frac{5}{3}\Big)^{2}\Big]^{-3 / 2} \div\Big[\Big(\frac{5}{2}\Big)^{-3}\Big] $$

$$ =\Big(\frac{3}{2}\Big)^{4 \times-3 / 4} \times\Big[\Big(\frac{5}{3}\Big)^{2 \times-3 / 2} \div\Big(\frac{5}{2})^{-3}] $$

$$ =\Big(\frac{3}{2}\Big)^{-3} \times\Big[(\frac{5}{3}\Big)^{-3} \div(\frac{5}{2}\Big)^{-3}\Big] $$

$$ =\Big(\frac{2}{3}\Big)^{3} \times\Big[\Big(\frac{5}{3})^{-3} \times \Big(\frac{5}{2}\Big)^{-3}\Big] $$

$$=\frac{2^{3}}{3^{3}} \times\Big[\frac{3^{3}}{5^{3}} \div \frac{2^{3}}{5^{3}}\Big] $$

$$ =\frac{2^{3}}{3^{3}} \times\Big[\frac{3^{3}}{5^{3}} \times \frac{5^{3}}{2^{3}}\Big] $$

$ =1 \quad $ {Ans. }

$$EXERCISE$$

OBJECTIVE DPP - 3.1

~~ 1. If $x=3+\sqrt{8}$ and $y=3-\sqrt{8}$ then $\frac{1}{x^{2}}+\frac{1}{y^{2}}=$

(A) -34

(B) 34

(D) $-12 \sqrt{8}$

~~ 2. If $\frac{3+\sqrt{7}}{3-\sqrt{7}}=a+b \sqrt{7}$ then $(a, b)=$

(A) $(8,-3)$

(B) $(-8,-3)$

(D) $(8,3)$

~~ 3. $\frac{\sqrt{5}-2}{\sqrt{5}+2}-\frac{\sqrt{5}+2}{\sqrt{5}-2}=$

(A) $8 \sqrt{5}$

(B) $-8 \sqrt{5}$

(D) $-6 \sqrt{5}$

~~ 4. If $x=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $y=1$, the value of $\frac{x-y}{x-3 y}$ is :

(A) $\frac{5}{\sqrt{5}-4}$

(B) $\frac{5}{\sqrt{6}+4}$

(D) $\frac{\sqrt{6}+4}{5}$

~~ 5. Which one is greatest in the following :

(A) $\sqrt{2}$

(B) $\sqrt[3]{3}$

(D) $\sqrt[3]{2}$

~~ 6. The value of $\sqrt[5]{(32)^{-3}}$ is :

(A) $1 / 8$

(B) $1 / 16$

(D) None

~~ 7. If $x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3+\sqrt{2}}}$ and $y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ the value of $x^{2}+xy+y^{2}$ is :

(A) 99

(B) 100

(D) 0

~~ 8. Simplify : $\frac{2}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{3}{\sqrt{5}+\sqrt{2}}$

(A) 1

(B) 0

(D) 100

~~ 9. Which of the following is smallest ?

(A) $\sqrt[4]{5}$

(B) $\sqrt[5]{4}$

(D) $\sqrt{3}$

~~ 10. The product of $\sqrt{3}$ and $\sqrt[3]{5}$ is :

(A) $\sqrt[6]{375}$

(B) $\sqrt[6]{675}$

(D) $\sqrt[6]{475}$

~~ 11. The exponential from of $\sqrt{\sqrt{2} \times \sqrt{2} \times \sqrt{2}}$ is :

(A) $2^{1 / 16}$

(B) $83 / 4$

(D) $81 / 2$

~~ 12. The value of $x$, if $5^{x-3} \cdot 3^{2 x-8}=225$, is :

(A) 1

(B) 2

(D) 5

~~ 13. If $2^{5 x} \div 2^{x}=\sqrt[5]{2^{20}}$ then $x=$

(A) 0

(B) -1

(D) 1

~~ 14. $\sqrt[3]{(729)^{2.5}}=$

(A) $\frac{1}{81}$

(B) 81

(D) 729

~~ 15. $\sqrt[4]{\sqrt[3]{x^{2}}}=$

(A) $x$

(B) $x^{\frac{1}{2}}$

(D) $x^{\frac{1}{6}}$

SUBJECTIVE DPP - 3.2

~~ 1. Arrange the following surds in ascending order of magnitude :

(i) $4 \sqrt{10}, 3 \sqrt{6}, \sqrt{3}$

(ii) $3 \sqrt{4}, 4 \sqrt{5}, \sqrt{3}$

~~ 2. Whish is greater :

$\sqrt{17}-\sqrt{12}$ or $\sqrt{11}-\sqrt{6}$.

~~ 3. Simplify : $\frac{8}{\sqrt{15}+1-\sqrt{5}-\sqrt{3}}$.

~~ 4. If $p$ and $q$ are rational number and $p-\sqrt{q}=\frac{4+\sqrt{2}}{3+\sqrt{2}}$ find $p$ and $q$.

~~ 5. Find the simplest R.F. of :

(i) $\sqrt[3]{32}$

(ii) $\sqrt[3]{36}$

(iii) $2^{3 / 5}$

~~ 6. Retionalise the denominator :

(i) $\frac{3}{\sqrt{5}}$

(ii) $\frac{\sqrt{2}+\sqrt{5}}{\sqrt{3}}$

~~ 7. Retionalise the denominator and simplify :

(i) $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

(ii) $\frac{1+\sqrt{2}}{3+2 \sqrt{2}}$

(iii) $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$

~~ 8. Simplify :

(i) $\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5-\sqrt{3}}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

(ii) $\frac{7+3 \sqrt{5}}{3+\sqrt{5}}-\frac{7-3 \sqrt{5}}{3-\sqrt{5}}$

~~ 9. Find the value of $a$ and $b$

(i) $\frac{\sqrt{11}-\sqrt{7}}{\sqrt{11}+\sqrt{7}}=a-b \sqrt{77}$

(ii) $\frac{5+\sqrt{6}}{5-\sqrt{6}}=a+b \sqrt{6}$ ~~ 10. If $x=\frac{\sqrt{3}+1}{2}$ find the value of $4 x^{3}+2 x^{2}-8 x+7$.

~~ 11. If $x=\frac{5-\sqrt{21}}{2}$ show that $(x^{3}+\frac{1}{x^{3}})-5(x^{2}+\frac{1}{x^{2}})+(x+\frac{1}{x})=0$.

~~ 12. Show that $a=x+1 / x$, where $x=\frac{\sqrt{a+2}+\sqrt{a-2}}{\sqrt{a+2}-\sqrt{a-2}}$.

~~ 13. Prove that : $\frac{1}{3-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2}=5$.

~~ 14. If $x=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}$ and $y=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}$ find the value of $3 x^{2}+4 x y-3 y^{2}$.

~~ 15. Evaluate:

$\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2 \sqrt{2}}$.

~~ 16. If $x=3, b=4$ then find the values of :

(i) $a^{b}+b^{a}$

(ii) $a^{a}+b^{b}$

(iii) $a^{b}-b^{a}$

~~ 17. Simplify :

$(\sqrt{x})^{-2 / 3} \sqrt{y^{4}} \div \sqrt{x y^{-1 / 2}}$.

~~ 18. Simplify :

(i) $[16^{-1 / 5}]^{5 / 2}$

(ii) $[0.001]^{\frac{1}{3}}$

~~ 19. If $\frac{9^{n} \times 3^{2} \times[3-{ }^{-n / 2}]^{-2}-(27)^{n}}{3^{3 m} \times 2^{3}}=\frac{1}{27}$, then prove than $m-n=1$.

~~ 20. Find the value of $x$, if $5^{x-3(2 x-3)}=625$.

$$ANSWER KEY$$

(Objective DPP # 1.1)

Qus.	1	2	3	4	5	6
Ans.	A	B	C	B	D	C

(Subjective DPP # 1.2)

~~ 1. (i) Non-terminating and repeating

(ii) Non-terminating and non-repeating

(iii) Non-terminating and repeating

(iv) Terminating

~~ 2. $\frac{-7}{6}, \frac{-4}{3}, \frac{-3}{2}, \frac{-5}{3}, \frac{-11}{6}$

~~ 3. $-4,-3,-2,-1$

~~ 4. $\frac{-5}{24} \quad$

~~ 5. $\frac{22}{7}, \frac{23}{7}, \frac{24}{7}, \frac{25}{7}, \frac{26}{7}, \frac{27}{7}$

~~ 5. $\frac{9}{24}, \frac{10}{24}, \frac{11}{24}$

~~ 6. $\frac{-5}{14}, \frac{-4}{14}, \frac{-3}{14}$

(Objective DPP # 2.1)

Qus.	1	2	3	4	5	6	7	8
Ans.	$A$	$B$	$A$	$A$	$C$	$D$	$C$	$C$

(Subjective DPP # 2.2)

~~ 3. 0.110101001000100001

~~ **4. **(i) irrational (ii) irrational (iii) rational (iv) irrational

~~ 10. $\quad 36 A^{4} B^{6} C^{3} \sqrt[6]{108}$

~~ 11.

(i) $1 / 3$ (ii) $\frac{37}{99}$ (iii) $\frac{6}{11}$ (iv) $\frac{5}{99}$ (v) $\frac{4}{3}$ (vi) $\frac{4}{3}$ (v) $\frac{23}{37}$

~~ 12. $\frac{19}{30}$

(Objective DPP # 3.1)

Qus.	1	2	3	4	5	6	7	8	9	10
Ans.	$B$	$D$	$B$	$D$	$C$	$A$	$A$	$B$	$B$	$B$
Qus.	11	12	13	14	15
Ans.	$C$	$D$	$D$	$C$	$D$
(Subjective DPP # 3.2)

~~ 1. (i) $4 \sqrt{10}>3 \sqrt{6}>\sqrt{3}$

(ii) $4 \sqrt{5}>3 \sqrt{4}>\sqrt{3}$

~~ 2. $\sqrt{11}-\sqrt{6}$

~~ 3. $\sqrt{16}+1+\sqrt{5}+\sqrt{5}$

~~ 4. $P=\frac{10}{7}, Q=\frac{2}{49}$

~~ 5. (i) $\sqrt[3]{2}$

(ii) $\sqrt[3]{6}$

(iii) $2^{2 / 5}$

~~ 6. (i) $\frac{3}{2} \sqrt{5}$

(ii) $\frac{\sqrt{6}+\sqrt{15}}{3}$

~~ 7. (i) $5-2 \sqrt{6}$

(ii) $7+5 \sqrt{2}$

(iii) $\frac{9+4 \sqrt{6}}{15}$

~~ 8. (i) 8 (ii) $\sqrt{5}$

~~ 9. (i) $a=9 / 2, b=1 / 2$

(ii) $a=\frac{31}{19}, b=\frac{10}{19}$

~~ 10. 10

~~ 11. $\frac{12+56 \sqrt{10}}{3}$

~~ 12. 1

~~ 13. (i) 145

(ii) 283

(iii) 17

~~ 17. $\frac{y^{9 / 4}}{x^{5 / 6}}$

~~ 18. (i) $1 / 4$

(ii) 0.1

20.1

> > > POLYNOMIALS < < <

POLYNOMIALS

An algebraic expression $(f(x).$ of the form $f(x)=a_0+a_1 x+a_2 x^{2}+$ $+a^{n} x^{n}$, where $a_0 a_1, a_2 \ldots \ldots ., a_n$ are real numbers and all the index of ’ $x$ ’ are non-negative integers is called a polynomials in $x$.

(a) Degree of the Polynomial :

Highest Index of $x$ in algebraic expression is called the degree of the polynomial, here $a_{0^{\prime}} a_1 x, a_2 x^{2} \ldots . . a_n x^{n}$, are called the terms o the polynomial and $z_0 \cdot a_1, a_2 \ldots \ldots, a_n$ are called various coefficients of the polynomial $f(x)$.

NOTE : A polynomial in $x$ is said to be in standard form when the terms are written either in increasing order or decreasing order of the indices of $x$ in various terms.

(b) Different Types of Polynomials :

Generally, we divide the polynomials in the following categories.

(i) Based on degrees :

There are four types of polynomials based on degrees. These are listed below :

(A) Linear Polynomials : A polynomials of degree one is called a linear polynomial. The general formula of linear polynomial is $ax+b$, where $a$ and $b$ are any real constant and $a \neq 0$.

(B) Quadratic Polynomials : A polynomial of degree two is called a quadratic polynomial. The general form of a quadratic polynomial is $ax^{2}+b+c$, where $a \neq 0$.

(C) Cubic Polynomials : A polynomial of degree three is called a cubic polynomial. The general form of a cubic polynomial is $a x^{3}+b x^{2}+c x+d$, where $a \neq 0$.

(D) Biquadratic (or quadric) Polynomials : A polynomial of degree four is called a biquadratic (quadratic) polynomial. The general form of a biquadratic polynomial is $a x^{4}+b x^{3}+c x^{2}+d x+e$, where $a \neq 0$.

NOTE : A polynomial of degree five or more than five does not have any particular name. Such a polynomial usually called a polynomial of degree five or six or ….etc.

(ii) Based on number of terms

There are three types of polynomials based on number of terms. These are as follows :

(A) Monomial : A polynomial is said to be monomial if it has only one term. e.g. $x, 9 x^{2}, 5 x^{3}$ all are monomials.

(B) Binomial : A polynomial is said to be binomial if it contains two terms e.g. $2 x^{2}+3 x, \sqrt{3} x+5 x^{3},-8 x^{3}+$ 3 , all are binomials.

(C) Trinomials : A polynomial is said to be a trinomial it if contains three terms. e.g. $3 x^{3}-8+\frac{5}{2}$, $\sqrt{7} x^{10} 8 x^{4}-3 x^{2}, 5-7 x+8 x^{9}$, are all trinomials.

NOTE : A polynomial having four or more than four terms does not have particular Name. These are simply called polynomials.

(iii) Zero degree polynomial : Any non-zero number (constant) is regarded as polynomial of degree zero or zero degree polynomial. i.e. $f(x)=a$, where $a \neq 0$ is a zero degree polynomial, since we can write $f(x)=a$ as $f(x)=ax$.

(iv) Zero polynomial : A polynomial whose all coefficients are zeros is called as zero polynomial i.e. $f(x)=$ 0 , we cannot determine the degree of zero polynomial.

ALGEBRAIC IDENTITY

An identity is an equality which is true for all values of the variables

Some important identities are

(i) $(a+b)^{2}=a^{2}+2 a b+b^{2}$

(ii) $(a-b)^{2}=a^{2}-2 a b+b^{2}$

(iii) $a^{2}-b^{2}=(a+b)(a-b)$

(iv) $a^{3}+b^{3}=(a+b)(a^{2}-a b+b^{2})$

(v) $a^{3}-b^{3}=(a-b)(a^{2}+a b+b^{2})$

(vi) $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$

(vii) $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$

(viii) $a^{4}+a^{2} b^{2}+b^{4}=(a^{2}+a b+b^{2})(a^{2}-a b+b^{2})$

(ix) $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)(a^{2}+b^{2}+c^{2}-a b-b c-a c)$

Special case : if $a+b+c=0$ then $a^{3}+b^{3}+c^{3}=3 a b c$.

(a) Value Form :

(i) $a^{2}+b^{2}=(a+b)^{2}-2 ab, \quad$ if $a+b$ and $ab$ are given

(ii) $a^{2}+b^{2}=(a-b)^{2}+2 a b \quad$ if $a-b$ and $a b$ are given

(iii) $a+b=\sqrt{(a-b)^{2}+4 a b} \quad$ if $a-b$ and $a b$ are given

(iv) $a-b=\sqrt{(a+b)^{2}-4 a b} \quad$ if $a+b$ and $a b$ are given

(v) $a^{2}+\frac{1}{a^{2}}=\Big(a+\frac{1}{a}\Big)^{2}-2 \quad$ if $a+\frac{1}{a}$ is given

(vi) $a^{2}+\frac{1}{a^{2}}=\Big(a+\frac{1}{a}\Big)^{2}+2 \quad$ if $a-\frac{1}{a}$ is given

(vii) $a^{3}+b^{3}=(a+b)^{3}-3 a b(a+b) \quad$ if $(a+b)$ and $a b$ are given

(viii) $a^{3}-b^{3}=(a-b)^{3}+3 a b(a-b) \quad$ if $(a-b)$ and $a b$ are given

(ix) $x^{3}+\frac{1}{a^{3}}=\Big(a+\frac{1}{a}\Big)^{3}-3\Big(a+\frac{1}{a}\Big) \quad$ if $a+\frac{1}{a}$ is given

(x) $a^{3}-\frac{1}{a^{3}}=\Big(a-\frac{1}{a}\Big)^{3}+3\Big(a-1 \frac{1}{a}\Big), \quad$ if $\Big(a-\frac{1}{a}\Big)$ is given

(xi) $a^{4}+b^{4}=(a^{2}+b^{2})^{2}-2 a^{2} b^{2}=[(a+b)^{2}-2 a b]^{2}-2 a^{2} b^{2}$, if $(a+b)$ and $a b$ are given

(xii) $a^4 - b^4 = (a^2 + b^2)(a^2 - b^2) = (a+b)^2 -2ab(a+b)(a-b)$

(xiii) $a^{5}+b^{5}=(a^{3}+b^{3})(a^{2}+b^{2})-a^{2} b^{2}(a+b)$

ILLUSTRATION

Ex. 1 Find the value of :

(i) $36 x^{2}+49 y^{2}+84 x y$, when $x=3, y=6$

(ii) $25 x^{2}+16 y^{2}-40 x y$, when $x=6, y=7$

Sol. (i) $36 x^{2}+49 y^{2}+84 x y=(6 x)^{2}+(7 y)^{2}+2 \times(6 x) \times(7 y)$

$$ =(6 x+7 y)^{2} $$

$$ =(6 \times 3+7 \times 6)^{2}[\text { When } x=3, y=6] $$

$$ =(18+42)^{2} $$

$$ =(60)^{2} $$

$$ =3600 . \quad \text {Ans} $$

(ii) $25 x^{2}+16 y^{2}-40 x y=(5 x)^{2}+(4 y)^{2}-2 \times(5 x) \times(4 y)$

$ \begin{aligned} & =(5 x-4 y)^{2} \\ & =(5 \times 6-4 \times 7)^{2}[\text { When } x=6, y=7] \\ & =(30-28)^{2} \\ & =2^{2} \\ & =4 \end{aligned} $

Ans.

Ex. 2 If $x^{2}+\frac{1}{x^{2}}=23$, find the value of $(x+\frac{1}{x})$.

Sol. $\quad x^{2}+\frac{1}{x^{2}}=23$

$\Rightarrow x^{2}+\frac{1}{x^{2}}+2=25$

[Adding 2 on both sides of (i)]

$\Rightarrow(x^{2})+(\frac{1}{x})^{2}+2 \cdot x \cdot \frac{1}{x}=25$

$\Rightarrow(x+\frac{1}{x})^{2}=(5)^{2}$

$\Rightarrow \quad x+\frac{1}{x}=5$Ans.

Ex. 3 Prove that $a^{2}+b^{2}+c^{2}-ab-bc-ca=\frac{1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$.

Sol. Here, L.H.S. $=a^{2}+b^{2}+c^{2}-a b-+b c-c a$

$ =\frac{1}{2}[2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a] $

$ \begin{aligned} & =\frac{1}{2}[(a^{2}-2 a b+b^{2})+(b^{2}-2 b c+c^{2})+(c^{2}-2 c a+a^{2})] \\ & =\frac{1}{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}] \end{aligned} $

$=$ RHS

Hence Proved.

Ex. 4 Evaluate : (i) $(107)^{2}$ (ii) $(94)^{2}$ (iii) $(0.99)^{2}$

Sol. (i)

$ (107)^{2} =(100+7)^{2} $

$ =(100)^{2}+(7)^{2}+2 \times 100 \times 7 $

$ =10000+49+1400 $

$ =11449 \quad \text {Ans.} $

$ (94)^{2} =(100-6)^{2} \quad \text {Ans} $

$ =(100)^{2}+(6)^{2}-2 \times 100 \times 6 $

$ =10000+36-1200 $

$ =8836 \quad \text {Ans.} $

$ (0.99)^{2} =(1-0.01)^{2} \quad $

$ =(1)^{2}+(0.01)^{2}-2 \times 1 \times 0.01 $

$ =+0.0001-0.02 $

$ =0.9801 $

(ii) $\quad(94)^{2}=(100-6)^{2}$

(iii) $\quad(0.99)^{2}=(1-0.01)^{2}$

NOTE : We may extend the formula for squaring a binomial to the squaring of a trinomial as given below.

$ \begin{aligned} (a+b+c)^{2} & =[a+(b+c)]^{2} & & \\ & =a^{2}+(b+c)^{2}+2 \times a \times(b+c) & & \text { [Using the identity for the square of binomial] } \\ & =a^{2}+b^{2}+c^{2}+2 b c+2(b+c) & & {[\text { Using }(b+c)^{2}=b^{2}+c^{2}+2 b c] } \\ & =a^{2}+b^{2}+c^{2}+2 b c+2 a b+2 a c & & \text { [Using the distributive law] } \\ & =a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c & & \\ \therefore(a+b+c)^{2} & =a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 a c & & \end{aligned} $

Ex. 5 Simplify : $(3 x+4)^{3}-(3 x-4)^{3}$.

Sol. We have,

$ (3 x+4)^{3}-(3 x-4)^{3} \quad=[(3 x)^{3}+(4)^{3}+3 \times 3 x \times 4 \times(3 x+4)]-[(3 x)^{3}-(4)^{3}-3 \times 3 x \times 4 \times(3 x-4)] $

$=[27^{3}+64+36 x(3 x+4)]-[27^{3}-64-36 x(3 x-4)] $

$ =[27 x^{3}+64+108 x^{2}+144 x]-[27 x^{3}-64-108 x^{2}+144 x] $

$ \quad=27 x^{3}+64+108 x^{2}+144 x-27 x^{3}+64+108 x^{2}-144 x $

$\quad=128+216 x^{2} $

$ \because \quad(3 x+4)^{3}-(3 x-4)^{3}=128+216 x^{2} \quad \text {Ans.}$

Ex. 6 Evaluate :

(i) $(1005)^{3}$ (ii) $(997)^{3}$

Sol. (i) $\quad(1005)^{3}=(1000+5)^{3}$

$ =(1000)^{3}+(5)^{3}+3 \times 1000 \times 5 \times(1000+5) $

$ =1000000000+125+15000+(1000+5) $

$ =1000000000+125+15000000+75000 $

$ =1015075125 \quad \text{Ans.} $

(ii) $(997)^{3} =(1000-3)^{3} $

$ =(1000)^{3}-(3)^{3}-3 \times 1000 \times 3 \times(1000-3) $

$ =1000000000-27-9000 \times(1000-3) $

$ =1000000000-27-900000+27000 $

$ =991026973 \quad \text{Ans.}$

Ex. 7 If $x-\frac{1}{x}=5$, find the value of $x^{3}-\frac{1}{x^{3}}$

Sol. We have, $x-\frac{1}{x}=5 \quad \ldots \text{(i)}$

$ \Rightarrow \quad \Big(x-\frac{1^{3}}{x}=(5)^{3}\Big) \quad \text { [Cubing both sides of (i)] } $

$ \Rightarrow \quad x^{3}-\frac{1}{x^{3}}-3 x \cdot \frac{1}{x} \cdot \Big(x-\frac{1}{x}\Big)=125 $

$ \Rightarrow \quad x^{3}-\frac{1}{x^{3}}-3\Big(x-\frac{1}{x}\Big)=125 $

$ .\Rightarrow \quad x^{3}-\frac{1}{x^{3}}-3 \times 5=125 \quad \text { [Substituting }\Big(x-\frac{1}{x}\Big)=5] $

$ \Rightarrow \quad x^{3}-\frac{1}{x^{3}}-15=125 $

$ \Rightarrow \quad x^{3}-\frac{1}{x^{3}}=(125+15)=140 \quad \text{Ans.}$

Ex. 8 Find the following products of the following expression : (i) $(4 x+3 y)(16 x^{2}-12 x y+9 y^{2})$ (ii) $(5 x-2 y)(25 x^{2}+10 x y+4 y^{2})$

Sol. (i) $(4 x+3 y)(16 x^{2}-12 x y+9 y^{2})$

$ \begin{matrix} =(4 x+3 y)[(4 x)^{2}-(4 x) \times(3 y)+(3 y)^{2}] & \\ \\ =(x+b)(x^{2}-a b+b^{2}) & \\ \\ =a^{3}+b^{3} & \\ \\ =(4 x)^{3}+(3 y)^{3}=64 x^{3}+27 y^{3} & \text { [Where } a=4 x, b=3 y] \end{matrix} $

(ii) $\quad(5 x-2 y)(25 x^{2}+10 x y+4 y^{2})$

$=(5 x-2 y)[(5 x^{2}+(5 x) \times(2 y)+(2 y)^{2}].$

$=(a-b)(a^{2}+a b+b^{2}) \quad$ [Where $a=5 x, b=2 y$ ]

$=a^{3}-b^{3}$

$=(5 x)^{3}-(2 y)^{3}$

$=125 x^{3}-8 y^{3} \quad$Ans.

Ex. 9 Simplify: $\frac{(a^{2}-b^{2})^{3}+(v^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}}{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}$.

Sol. Here $(a^{2}-b^{2})+(b^{2}-c^{2})^{3}+(c^{2}-a^{2})=0$

$\therefore \quad(a^{2}-b^{2})^{3}+(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}=3(a^{2}-b^{2})(b^{2}-c^{2})(c^{2}-a^{2})$

Also, $\quad(a-b)+(b-c)+(c-a)=0$

$\therefore \quad(a-b)^{3}+(b-c)^{3}+(c-a)^{3}=3(a-b)(b-c)(c-a)$

$\therefore$ Given expression $\quad=\frac{3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)}{3(a-b)(b-c)(c-a)}$

$=\frac{3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)}{3(a-b)(b-c)(c-a)}$

$=(a+b)(b+c)(c+a) \quad$Ans.

Ex. 10 Prove that: $(x-y)^{3}+(y-z)^{3}+(z-x)^{3}=3(x-y)(y-z)(z-x)$.

Sol. Let $(x-y)=a,(y-z)=b$ and $(z-x)=c$.

Then, $\quad a+b+c=(x-y)+(y-z)+(z-x)=0$

$\therefore a^{3}+b^{3}+c^{3}=3 abc$

Or $(x-y)^{3}+(y-z)^{3}+(z-x)^{3}=3(x-y)(y-z)(z-x) \quad$Ans.

Ex. 11 Find the value of $(28)^{3}-(78)^{3}+(50)^{3}$.

Sol. Let $a=28, b=-78, c=50$

Then, $\quad a+b+c=28-78+50=0$

$\therefore \quad a^{3}+b^{3}+c^{3}=3 abc$.

So, $(28)^{3}+(-78)^{3}+(50)^{3}=3 \times 28 \times(-78) \times 50$

Ex. 12 If $a+b+c=9 a$ and $a b+b c+a c=26$, find the value of $a^{3}+b^{3}+c^{3}-3 a b c$.

Sol. We have $a+b+c=9$

$\Rightarrow(a+b+c)^{2}=81 \quad \text{[On squaring both sides of (i)]}$

$\Rightarrow a^{2}+b^{2}+c^{2}+2(ab+bc+ac)=81$

$\Rightarrow a^{2}+b^{2}+c^{2}+2 \times 26=81 \quad[\because ab+bc+ac=26]$

$\Rightarrow a^{2}+b^{2}+c^{2}=(81-52)$

$\Rightarrow a^{2}+b^{2}+{ }^{2}=29$.

Now, we have

$ a^{3}+b^{3}+c^{3}-3 a b c=(a+b +c)(a^{2}+b^{2}+c^{2}-a b-b c-a c) $

$ =(a+b+c)[(a^{2}+b^{2}+c^{2})-(a b+b c+a c)] $

$ =9 \times[(29-26)] $

$ =(9 \times 3) $

$ =27 \quad \text {Ans.} $

(b) A Special Product :

We have $(x+a)(x+b)=x(x+b)+a(x+b)$

$=x^{2}+x b+a x+a b$

$=x^{2}+b x+a x+a b \quad[\because x b=b x]$

$=x^{2}+a x+b x+a b$

$=x^{2}+(a+b) x+a b$

Thus, we have the following identity

$ (x+a)(x+b)=x^{2}+(a+b) x+a b $

Ex. 13 Find the following products : (i) $(x+2)(x+3)$ (ii) $(x+7)(x-2)$ (ii) $(y-4)(y+3)$ (iv) $(y-7)(y+3)$ (v) $(2 x-3)(2 x-5)$ (vi) $(3 x+4)(3 x-5)$

Sol. Using the identity : $(x+a)(x+b)=x^{2}+(a+b) x+a b$, we have

(i) $(x+2)(x+3)=x^{2}+(2+3) x+2 \times 3$ $=x^{2}+5 x+6$.

Ans.

(ii) $(x+7)(x-2)=(x+7)(x+(-2))$

$=x^{2}+7 x+(-2) x+7 \times(-2)$

$=x^{2}+5 x-14$.

Ans.

(iii) $(y-4)(y-3)={y+(-4)}{y+(-3)}$

$=y^{2}+{(-4)+(-3)} y+(-4) \times(-3)$

$=y^{2}-7 y+12$

Ans.

(iv) $(y-7)(y+3)={y+(-7)}(y+3)$

$=y^{2}+{(-7)+3}+(-7) \times 3$

$=y^{2}-4 y-21$.

Ans.

(v) $(2 x-3)(2 x+5)=(y-3)(y+5)$, where $y=2 x$

$={y+(-3)}(y+5)$

$=y^{2}+{(-3)+5} y+(-3) \times 5$

$=y^{2}+2 y-15$

$=(2 x)^{2}+2 \times 2 x-15$

$=4 x^{2}+4 x-15$.** Ans.**

(vi) $(3 x+4)(3 x-5)=(y+4)(y-5)$, where $y=3 x$

$=(y+4){y+(-5)}$

$=y^{2}+{4+(-5)}+4 \times(-5)$

$=y^{2}-y-20$

$=(3 x)^{2}-3 x-20$

$=9 x^{2}-3 x-20$.

Ans.

Ex. 14 Evaluate : (i) $35 \times 37$

(ii) $103 \times 96$

Sol. (i) $35 \times 37=(40-5)(40-3)$

$=(40+(-5))(40+(-3))$

$=40^{2}+(-5-3) \times 40+(-5 \times-3)$

$=1600-320+15$

$=1615$ - 320

$=1295$

Ans.

(ii) $ 103 \times 96 $

$ =(100+3)[100+(-4)] $

$ =100^{2}+(3+(-4)) \times 100+(3 \times-4) $

$ =10000-100-12 $

$ =9888 \quad \text{Ans.} $

FACTORS OF A POLYNOMIAL

If a polynomial $f(x)$ can be written as a product of two or more other polynomial $f_1(x), f_2(x), f_3(x)$,…. then each of the polynomials $f_1(x), f_2(x)$,….. is called a factor of polynomial $f(x)$. The method of finding the factors of a polynomials is called factorisations.

(a) Factorisation by Making a Trinomial a Perfect Square :

Ex. 15 $\quad 81 a^{2} b^{2} c^{2}+64 a^{6} b^{2}-144 a^{4} b^{2} c$

Sol. $\quad 81 a^{2} b^{2} b c^{2}+64 a^{6} b^{2}-144 a^{4} b^{2} c$

$=[9 a b c]^{2}-2[9 a b c][8 a^{3} b]+[8 a^{3} b]^{2}$

$=[9 a b c-8 a^{3} b]^{2}=a^{2} b^{2}[9 c-8 a^{2}]^{2} \quad$Ans.

Ex. 16 $\quad\Big(3 a-\frac{1}{b}\Big)^{2}-\Big(3 a-\frac{1}{b}\Big)+9+\Big(c+\frac{1}{b}-2 a\Big)\Big(3 a-\frac{1}{b}-3\Big)$

Sol. $\quad\Big(3 a-\frac{1}{b}\Big)^{2}-6\Big(3 a-\frac{1}{b}\Big)+9+\Big(c+\frac{1}{b}-2 a\Big)\Big(3 a-\frac{1}{b}-3\Big)$

$ =\Big(3 a-\frac{1}{b}\Big)^{2}-2.3\Big(3 a-\frac{1}{b}\Big)+\Big(3\Big)^{2}+\Big(c+\frac{1}{b}-2 a\Big)\Big(3 a-\frac{1}{b}-3\Big) $

$ =\Big(3 a-\frac{1}{b}-3\Big)^{2}+\Big(c+\frac{1}{b}-2 a\Big)\Big(3 a-\frac{1}{b}-3\Big) $

$ =\Big(3 a-\frac{1}{b}-3\Big)\Big[3 a-\frac{1}{b}+3+\frac{1}{b}-2 a\Big] $

$ =\Big(3 a-\frac{1}{b}-3\Big)[a+c-3] \quad \text { Ans.} $

(b) Factorisation by Using the Formula for the Difference of Two Squares :

$ a^{2}-b^{2}=(+b)(a-b) $

Ex. 17 Factorise : $\quad 4(2 a+3 b-4 c)^{2}-(a-4 b+5 c) .2$

Sol.

$ =4(2 a+3 b-4 c)^{2}-(a-4 b+5 c)^{2} $

$ =[2(2 a+3 b-4 c)]^{2}-(a-4 b+5 c)^{2} $

$ =[4 a+6 b-8 c+a-4 b+5 c][4 a+6 b-8 c-a+4 b-5 c] $

$ =[5 a+2 b-3 c][3 a+10 b-13 c] \quad \text {Ans.}$

Ex. 18 Factorise : $4 x^{2}+\frac{1}{4 x^{2}}+2-9 y^{2}$.

Sol. $\quad 4 x^{2}+\frac{1}{4 x^{2}}+2-9 y^{2}$

$ \begin{aligned} & =(2 x)^{2}+2 \cdot(2 x) \cdot(\frac{1}{2 x})+(\frac{1}{2 x})^{2}-(3 y)^{2} \\ & =(2 x+\frac{1}{2 x})^{2}-(3 y)^{2} \\ & =(2 x+\frac{1}{2 x}+3 y)(2 x+\frac{1}{2 x}-3 y) \end{aligned} $

Ans.

Ex. 19 Factorise : $x^{4}+\frac{1}{a^{4}}-3$.

Sol. $\quad(a^{2})^{2}+(\frac{1}{a^{2}})^{2}-2 \cdot(a^{2})(\frac{1}{a^{2}})-1$

$ \begin{aligned} & =(a^{2}-\frac{1}{a^{2}})^{2}-(1)^{2} \\ & =(a^{2}-\frac{1}{a^{2}}+1)(a^{2}-\frac{1}{a^{2}}-1) \end{aligned} $

Ans.

Ex. 20 Factorise: $x^{4}+x^{2} y^{2}+y^{4}$.

Sol. $\quad x^{2}+x^{2} y^{2}+y^{4}=(x^{2})^{2}+2 \cdot x^{2} \cdot y^{2}+(y^{2})^{2}-x^{2} y^{2}$

$ =x^{2}+y^{2})^{2}-(x y)^{2} $

$ =(x^{2}+y^{2}+x y((x^{2}+y^{2}-x y) \quad. \text {Ans.}$

(c) Factorisation by Using Formula of $a^{3}+b^{3}$ and $a^{3}-b^{3}$ :

Ex. 21 Factorise : $64 a^{13} b+343 a b^{13}$.

Sol. $\quad 64 a^{13} b+343 a b^{13}=a b[64 a^{12}+343 b^{12}]$

$ =a b[(4 a^{4})^{3}+(7 b^{4})^{3}] $

$ =a b[4 a^{4}+7 b^{4}][(4 a^{4})^{2}-(4 a^{4})(7 b^{4})+(7 b^{4})^{2}] $

$ =a b[4 a^{4}+7 b^{4}][16 a^{8}-28 a^{4} b^{4}+49 b^{8}] \quad \text {Ans.}$

Ex. 22 Factorise : $p^{3} q^{2} x^{4}+3 p^{2} q x^{3}+3 p x^{2}+\frac{x}{q}-q^{2} r^{3} x$

Sol. In above question, If we take common then it may become in the form of ${ }^{3}+b^{3}$.

$ \therefore \quad p^{3} q^{2} x^{4}+3 p^{2} q x^{3}+3 p x^{2}+\frac{x}{q}-q^{2} r^{3} x $

$ \begin{aligned} & =\frac{x}{q}[p^{3} q^{3} x^{3}+3 p^{2} q^{2} x^{2}+3 p q x+1-q^{3} r^{3}] \\ & =\frac{x}{q}[(p q x)^{3}+3(p q x)^{2} \cdot 1+3 p q x \cdot(1)^{2}+(1)^{3}-q^{3} r^{3}] \end{aligned} $

Let $pqx=A \And 1=B$

$=\frac{x}{q}[A^{3}+3 A^{2} B+3 A B^{2}+B^{3}-q^{3} r^{3}]$

$=\frac{x}{q}[(p q x+1)^{3}-(q r)^{3}]=\frac{x}{q}[p q x+1-q r][(p q x+1)^{2}+(p q x+1) q r+(q r)^{2}]$

$=\frac{x}{q}[p q x+1-q r][p^{2} q^{2} x^{2}+1+2 p q x+p q^{2} x r+q r+q^{2} r^{2}]$Ans.

Ex. 23 Factorise : $x^{3}-6 x^{2}+32$

Sol. : $\quad x^{3}+32-6 x^{2}$

$=x^{3}+8+24-6 x^{2}$

$=[(x)^{3}+(2)^{3}]+6[4-x^{2}]$

$=(x+2)[x^{2}-2 x+4]+6[2+x][2-x]$

$=(x+2)[x^{2}-2 x+4+6(2-x)]$

$=(x+2)[x^{2}-2 x+4+12-6 x]$

$=(x+2)[x^{2}-8 x+16]$

$=(x+2)(x-4)^{2}$Ans.

$$EXERCISE$$

OBJECTIVE DPP - 4.1

~~ 1. The product of $(x+a)(x+b)$ is :

(A) $x^{2}+(a+b) x+a b$ (B) $x^{2}-(a-b) x+a b$ (C) $a^{2}+(a-b) x+a b$ (D) $x^{2}+(a-b) x-a b$.

~~ 2. The value of $150 \times 98$ is : (A) 10047 (B) 14800 (C) 14700 (D) 10470

~~ 3. The expansion of $(x+y-z)^{2}$ is :

(A) $x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x$ (B) $x^{2}+y^{2}-z^{2}-2 x y+y z+2 z x$ (C) $x^{2}+y^{2}+z^{2}+2 x y-2 y z-2 z x$ (D) $x^{2}+y^{2}-z^{2}+2 z y-2 y z-2 z x$

~~ 4. The value of $(x+2 y+2 z)^{2}+(x-2 y-2 z)^{2}$ is:

(A) $2 x^{2}+8 y^{2}+8 z^{2}$ (B) $2 x^{2}+8 y^{2}+8 z^{2}+8 x y z$ (C) $2 x^{2}+8 y^{2}+8 z^{2}-8 y z$ (D) $2 x^{2}+8 y^{2}+8 z^{2}+16 y z$

~~ 5. The value of $25 x^{2}+16 y^{2}+40 x y$ at $x=1$ and $y=-1$ is :

(A) 81 (B) -49 (C) 1 (D) None of these

~~ 6. On simplifying $(a+b)^{3}+(a-b)^{3}+6 a(a^{2}-b^{2})$ we get :

(A) $8 a^{2}$ (B) $8 a^{2} b$ (C) $8 a^{3} b$ (D) $8 a^{3}$

~~ 7. Find the value of $\frac{a^{3}+b^{3}+c^{3}-3 a b c}{a b+b c+c a-a^{2}-b^{2}-c^{2}}$, when $a=-5,5=-6, c=10$.

(A) 1 (B) -1 (C) 2

(D) -2

~~ 8. If $(x+y+z)=1, x y+y z+z x=-1 x y z=-1$ then value of $x^{3}+y^{3}+z^{3}$ is :

(A) -1 (B) 1 (C) 2 (D) -2

~~ 9. In method of factorisation of an algebraic expression. Which of the following statement is false ?

(A) Taking out a common factor from two or more terms.

(B) Taking out a common factor from a group of terms.

(D) By using standard identities.

~~ 10. Factors of $(a+b)^{3}-(a-b)^{3}$ is :

(A) $2 a b(3 a^{2}+b^{2})$ (B) $a b(3 a^{2}+b^{2})$ (C) $2 b(3 a^{2}+b^{2})$ (D) $3 a^{2}+b^{2}$

~~ 11. Degree of zero polynomial is :

(A) 0 (B) 1 (C) Both $0 \And 1$

(D) Not defined

~~ 12. If $a^{4}+\frac{1}{a^{4}}=119$, then find the value of $-\frac{1}{a^{3}}$.

~~ 13. If $x=152, y=-91$ find the value of $9 x^{2}+30 x y+25 y^{2}$.

~~ 14. Evaluate :

(i) $(5 x+4 y)^{2}$ (ii) $(4 x-5 y)^{2}$ (iii) $(2 x-\frac{1}{x})^{2}$

~~ 15. If $x+y=3$ and $x y=-18$, find the value of $x^{3}+y^{3}$.

~~ 16. If $x^{2}+\frac{1}{x^{2}}=51$ find the value of $x^{3}-\frac{1}{x^{3}}$..

~~ 17. Evaluate :

(i) $25^{3}-75^{3}+50^{3}$ (ii) $(\frac{1}{2})^{3}+(\frac{1}{3})^{3}-(\frac{5}{6})^{3}$ (iii) $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$

~~ 18. Find the product of :

(i) $(x+4)(x+7)$ (ii) $(x+\frac{1}{5})(x+5)$ (iii) $(P^{2}+16)(P^{2}-\frac{1}{4})$

~~ 19. Evaluate :

(i) $102 \times 106$ (ii) $994 \times 1006$ (iii) $34 \times 36$

~~ 20. Factorise : $4 x^{4}+(7 a)^{4}$.

~~ 21. Factorise : $x^{12}=1$.

~~ 22. Evaluate $\frac{(a-b)^{2}}{(b-c)(c-a)}+\frac{(b-c)^{2}}{(a-b)(c-a)}+\frac{(c-a)^{2}}{(a-b)(b-c)}$.

~~ 23. Write the following polynomials in standard forms :

(i) $x^{6}-3 a^{4}+\sqrt{2} x+5 x^{2}+7 x^{5}+4$

(ii) $m^{7}+8 m^{5}+4 m^{6}+6 m-3 m^{2}-11$

~~ 13. Factorise: $(x+1)(x+2)(x+3)(x+4)-3$.

~~ 14. Factorise : $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$.

~~ 15. Factorise : $x^{4}+2 x^{3} y-2 x y^{3}-y^{4}$.

$\ggg$
POLYNOMIALS

$\lll$

ML - 5

ZEROS OR ROOTS OF A POLYNOMIAL

A real number $\alpha$ is a root or zero of polynomial $f(x)=a_n x^{n}+a_{n-1} x^{n-1}+a_{n-2} x^{n-2}+\ldots \ldots+a_1 x+a_0$ ，

if $f(\alpha)=0$. i.e. $a_n \alpha^{n}+a_{n-1}+a_{n-2} \alpha^{n-2}+\ldots \ldots+a_1 \alpha+a_0=0$.

For example $x=3$ is root of the polynomial $f(x)=x^{3}-6 x^{2}+11 x-6$, because

$f(3)=(3)^{3}-6(3)^{2}+11(3)-=27-54+33-6=0$.

but $x=-$ is not a root of the above polynomial,

$ \begin{aligned} \because \quad f(-2) & =(-2) \cdot 3-6(-2)^{2}+11(-2)-6 \\ f(-2) & =-8-24-22-6 \\ f(-2) & =-60 \neq 0 \end{aligned} $

(a) Value of a Polynomial :

The value of a polynomial $f(x)$ at $x=\alpha$ is obtained by substituting $x=\alpha$ in the given polynomial and is denoted by $f(\alpha)$. e.g. If $f(x)=2 x^{3}-13 x^{2}+17 x+12$ then its value at $x=1$ is.

$ \begin{aligned} f(1) & =2(1)^{3}-13(1)^{2}+17(1)+12 \\ & =2-13+17+12=18 \end{aligned} $

Ex. 1 Show that $x=2$ is a root of $2 x^{3}+x^{2}-7 x-6$.

Sol. $p(x)=2 x^{3}+x^{2}-7 x-6$ then,

$p(2)=2(2)^{3}+(2)^{2} 7(2)-6=16+4-14-6=0$

Hence $x=2$ is a root of $p(x)$.

Ans.

Ex. 2 If $x=\frac{4}{3}$ is a root of the polynomial $f(x)=6 x^{3}-11 x^{2}+k x-20$ then find the value of $k$.

Sol. $\quad f(x)=6 x^{3}-11 x^{2}+k x-20$

$ \Rightarrow \quad f\Big(\frac{4}{3}\Big)=6\Big(\frac{4}{3}\Big)^{3}-11\Big(\frac{4}{3}\Big)^{2}+k\Big(\frac{4}{3}\Big)-20=0 $

$ \Rightarrow \quad 6 \cdot \frac{64}{9.3}-11 \cdot \frac{16}{9}+\frac{4 k}{3}-20=0 $

$ \Rightarrow \quad 128-176+12 k-180=0 $

$ \Rightarrow \quad 12 k+128-356=0 $

$ \Rightarrow \quad 12 k=228 \quad \text{Ans.}$

Ex. 3 If $x=2 \And x=0$ are two roots of the polynomial $f(x)=2 x^{3}-5 x^{2}+a x+b$. Find the values of and $b$.

Sol. $\quad f(x)=2(2)^{3}-5(2)^{2}+a(2)+b=0$

$\Rightarrow 16-20+2 a+b=0$

$\Rightarrow 2 a+b=4$

$\Rightarrow f(0)=2(0)^{3}-5(0)^{2}+a(0)+b=0$

$\Rightarrow b=0$

So, $2 a=4$

Hence, $a=2, b=0 \quad$Ans.

REMAINDER THEOREM

Let ’ $p(x)^{\prime}$ be any polynomial of degree greater than or equal to one and a be any real number and If $p(x)$ is divided by $(x-a)$, then the remainder is equal to $p(a)$.

Let $q(x)$ be the quotient and $r(x)$ be the remainder when $p(x)$ is divided by $(x-a)$ then

Dividend $\boldsymbol{=}$ Divisor $\times$ Quotient + Remainder

$p(x)=(x-a) \times q(x)+[r(x)$ or $r]$, where $r(x)=0$ or degree of $r(x)<$ degree of $(x-)$. But $(x-a)$ is a polynomial of degree 1 and a polynomial of degree less than 1 is a constant. Therefore, either $r(x)=0$ or $r(x)=$ Constant.

Let $r(x)=r$, then $p(x)=(x-a) q(x)+r$,

putting $x=a$ in above equation $p(a)=(a-a) q(a)+r=0 . q(a)+r$

$p(a)=0+r$

$\Rightarrow \quad p(a)=r$

This shows that the remainder is $p(a)$ when $p(x)$ is divided by $(x-a)$.

REMARK : If a polynomial $p(x)$ is divided by $(x+a),(a x-b),(x+b),(b-a x)$ then the remainder in the value of $p(x)$ at $x=-a, \frac{b}{a},-\frac{b}{a}, \frac{b}{a}$ i.e. $p(-a), p\Big(\frac{b}{a}\Big), p\Big(-\frac{b}{a}\Big), p\Big(\frac{b}{a}\Big)$ respectively.

Ex. 4 Find the remainder when $f(x)=x^{3}-6 x^{2}+2 x-4$ is divided by $g(x)=1-2 x$.

Sol. $1-2 x=0 \Rightarrow 2 x=1 \Rightarrow x=\frac{1}{2}$

$ \begin{aligned} f\Big(\frac{1}{2}\Big) & =\Big(\frac{1}{2}\Big)^{3}-6\Big(\frac{1}{2}\Big)^{2}+2\Big(\frac{1}{2}\Big)-4 \\ & =\frac{1}{8}-\frac{3}{2}+1-4 \\ & =\frac{1-12+8-32}{8}=-\frac{35}{8} \end{aligned} $

Ans.

Ex. 5 The polynomials $a x^{3}+3 x^{2}-13$ and $2 x^{3}-5 x+a$ are divided by $x+2$ if the remainder in each case is the same, find the value of a.

Sol. $\quad p(x)=ax+3 x^{2}-13$ and $q(x)=2 x^{3}-5 x+a$

when $p(x) \And q(x)$ are divided by $x+2=0 \Rightarrow x=-2$

$p(-2)=q(-2)$

$\Rightarrow a(-2)^{3}+3(-2)^{2}-13=2(-2)^{3}-5(-2)+a$

$\Rightarrow-8 a+12-13=-16+10+a$

$\Rightarrow-9 a=-5$

$\Rightarrow a=\frac{5}{9} \quad$Ans.

(a) Factor Theorem :

Let $p(x)$ be a polynomial of degree greater than or equal to 1 and ’ $a$ ’ be a real number such that $p(a)=0$, than $(x-a)$ is a factor of $p(x)$. Conversely, if $(x-a)$ is a factor of $p(x)$, then $p(a)=0$.

Ex. 6 Show that $x+1$ an $d 2 x-3$ are factors of $2 x^{3}-9 x^{2}+x+12$.

Sol. To prove that $(x+1)$ and $(2 x-3)$ are factors of $2 x^{3}-9 x^{2}+x+12$ it is sufficient to show that $p(-1)$ and $p\Big(\frac{3}{2}\Big)$ both are equal to zero.

$p(-1)=2(-1)^{3}-9(-1)^{2}+(-1)+12=-2-9-1+12=-12+12=0$

And, $p\Big(\frac{3}{2}\Big)=2\Big(\frac{3}{2}\Big)^{3}-9\Big(\frac{3}{2}\Big)+\Big(\frac{3}{2}\Big)+12$

$ =\frac{27}{4}-\frac{81}{4}+\frac{3}{2}+12=\frac{27-81+6+48}{4}=\frac{-81+81}{4}=0 $

Hence, $(x+1)$ and $(2 x-3)$ are the factors $2 x^{3}-9 x^{2}+x+12$.

Ans.

Ex. $7 \quad$ Find $\underline{a}$ and $\beta$ if $x+1$ and $x+2$ are factors of $p(x)=x^{3}+3 x^{2}-2 \alpha x+\beta$.

Sol. When we put $x+1=0$ or $x=-1$ and $x+2=0$ or $x=-2$ in $p(x)$

Then, $p(-1)=0$ \And $p(-2)=0$

Therefore, $p(-1)=(-1)^{3}+3(-1)^{2}-2 \alpha(-1)+\beta=0$

$\Rightarrow-1+3+2 \alpha+\beta=0 \Rightarrow \beta=-2 \alpha-2 \ldots .(i)$

And, $p(-2)=(-2)^{3}+3(-2)^{2}-2 \alpha(-2)+\beta=0$

$\Rightarrow-8+12+4 \alpha+\beta=0 \quad \Rightarrow \beta=-4 \alpha-4$

From equation (i) and (ii)

$ -2 \alpha-2=-4 \alpha-4 $

$\Rightarrow 2 \alpha=-2 \Rightarrow \alpha=-1$

Put $\alpha=-1$ in equation (i) $\Rightarrow \beta=-2(-1)-2=2-2=0$.

Hence, $\alpha=-1 \beta=0$.

Ans.

Ex. 8 What must be added to $3 x^{3}+x^{2}-22 x+9$ so that the result is exactly divisible by $3 x^{2}+7 x-6$.

Sol. Let $p(x)=3 x^{3}+x^{2}-22 x+9$ and $q(x)=3 x^{2}+7 x-6$.

We know if $p(x)$ is divided by $q(x)$ which is quadratic polynomial therefore if $p(x)$ is not exactly divisible by $q(x)$ then the remainder be $r(x)$ and degree of $r(x)$ is less than $q(x)$ (or Divisor)

$\therefore$ By long division method

Let we added $ax+b$ (linear polynomial) is $p(x)$, so that $p(x)+ax+b$ is exactly divisible by $3 x^{2}+7 x-6$.

Hence $p(X)+a x+b=s(x)=3 x^{3}+x^{2}-22 x+9 a x+b$

$ =3 x^{3}+x^{2}-x(22-a)+(9+b) $

alt text

Hence, $x(a-2+b-3=0 \cdot x+0)$

$\Rightarrow a-2=0 \And b-3=0$ $\Rightarrow \quad a=2$ or $b=3$

Ans.

Hence, if we add $ax+b$ or $2 x+3$ in $p(x)$ then it is exactly divisible by $3 x^{2}+7 x-6$.

Ex. 9 Using factor theorem, Factorise :

$ p(x)=2 x^{4}-7 x^{3}-13 x^{2}+63 x-45 $

Sol. $\quad 45 \Rightarrow \pm 1, \pm 3, \pm 5, \pm 9, \pm 15, \pm 45$

if we put $x=1$ in $p(x)$

$ \begin{aligned} & p(1)=2(1)^{4}-7(1)^{3}-13(1)^{2}+63(1)-45 \\ & 2-7-13+63-45=65-65=0 \end{aligned} $

$\therefore \quad x=1$ or $x-1$ is a factor of $p(x)$.

Similarly, if we put $x=3$ in $p(x)$

$ \begin{aligned} & p(3)=2(3)^{4}-7(3)^{3}-13(3)^{2}+63(3)-45 \\ & 162-189-117+189-45=162-162=0 \end{aligned} $

Hence, $x=3$ or $x-3=0$ is the factor of $p(x)$.

$ \begin{aligned} & p(x)=2 x^{4}-7 x^{3}-13 x^{2}+63 x-45 \\ & \therefore \quad p(x)=2 x^{3}(x-1)-5 x^{2}(x-1)-18(x-1)+45(x-1) \\ & 2 x^{4}-2 x^{3}(x-1)-5 x^{2}-18 x^{2}+18 x+45 x-54 \\ & \Rightarrow p(x)=(x-1)(2 x^{3}-5 x^{2}-18 x+45) \\ & \Rightarrow \quad p(x)=(x-1)(2 x^{3}-5 x^{2}-18 x+45) \\ & \Rightarrow \quad p(x)=(x-1)[2 x^{2}(x-3)+x(x-3)-15(x-3)] \\ & \Rightarrow \quad p(x)=(x-1)[2 x^{3}-6 x^{2}+x^{2}-3 x-15 x+45] \\ & \Rightarrow \quad p(x)=(x-1)(x-3)(2 x^{2}+x-15) \\ & \Rightarrow \quad p(x)=(x-1)(x-3)(2 x^{2}+6 x-5 x-15) \\ & \Rightarrow \quad p(x)=(x-1)(x-3)[2 x(x+3)-5(x+3)] \\ & \Rightarrow \quad p(x)=(x-1)(x-3)(x+3)(2 x-5) \end{aligned} $

FACTORISATION OFA QUADRATIC POLYNOMIAL

For factorisation of a quadratic expression $ax^{2}+bx+a$ where $a \neq 0$, there are two method.

(a) By Method of Completion of Square :

In the form $ax^{2}+bx+c$ where $a \neq 0$, firstly we take ’ $a$ ’ common in the whole expression then factorise by converting the expression $a\Big[x^{2}+\frac{b}{a} x+\frac{c}{a}\Big]$ as the difference of two squares.

Ex. 10 Factorise $x^{2}-31 x+220$.

Sol. $\quad x^{2}-31 a+220$

$ \begin{aligned} & =x^{2}-2 \cdot \frac{31}{2} \cdot x+\Big(\frac{31}{2}\Big)^{2}-\Big(\frac{31}{2}\Big)^{2}+220 \\ & =\Big(x-\frac{31}{2}\Big)^{2}-\frac{961}{4}+220=\Big(x-\frac{31}{2}\Big)^{2}-\frac{81}{4} \\ & =\Big(x-\frac{31}{2}\Big)-\Big(\frac{9}{2}\Big)^{2}=\Big(x-\frac{31}{2}+\frac{9}{2}\Big)\Big(x-\frac{31}{2}-\frac{9}{2}\Big) \end{aligned} $

$=(x-11)(x-20) \quad$Ans.

Ex. 11 Factorise : $-10 x^{2}+31 x-24$

Sol. $-10 x^{2}+31 x-24$

$ \begin{aligned} & =-[10 x^{2}-31 x+24]=-10\Big[x^{2}-\frac{31}{10} x+\frac{24}{10}\Big] \\ \\ & =-10\Big[x^{2}-2 \cdot \frac{31}{20} \cdot x+\Big(\frac{31}{20}\Big)^{2}-\Big(\frac{31}{20}\Big)^{2}+\frac{24}{10}\Big] \\ \\ & =-10\Big[\Big(x-\frac{31}{20}\Big)^{2}-\frac{961}{400}+\frac{24}{10}\Big]=-10\Big[\Big(x-\frac{31}{20}\Big)^{2}-\frac{1}{400}\Big] \\ \\ & =-10\Big[\Big(x-\frac{31}{20}\Big)^{2}-\Big(\frac{1}{20}\Big)^{2}\Big]=-10\Big[x-\frac{31}{20}+\frac{1}{20}\Big][x-\frac{31}{20}-\frac{1}{20}\Big] \end{aligned} $

$=-10\Big(\frac{2 x-3}{2}\Big)\Big(\frac{5 x-8}{5}\Big)=-(2 x-3)(5 x-8)=(3-2 x)(5 x-8) \quad$Ans.

(b) By Splitting the Middle Term :

In the quadratic expression $a x^{2}+b x+c$, where $a$ is the coefficient of $x^{2}, b$ is the coefficient of $x$ and $c$ is the constant term. In the quadratic expression of the form $x^{2}+b x+c, a=1$ is the multiple of $x^{2}$ and another terms are the same as above.

There are four types of quadratic expression :

(i) $a x^{2}+b x+c$ (ii) $a x^{2}-b x+c$

(iii) $a x^{2}-b x-c$ (iv) $a x^{2}+b x-c$

Ex. 12 Factorise : $2 x^{2}+12 \sqrt{2 x}+35$.

Sol. $\quad 2 x^{2}+12 \sqrt{2 x}+35$

Product $ac=70 \And b=12 \sqrt{2}$

$\therefore \quad$ Split the middle term as $7 \sqrt{2} \And 5 \sqrt{2}$

$ \begin{aligned} \Rightarrow 2 x^{2}+12 \sqrt{2} x & +35=2 x^{2}+7 \sqrt{2} x+5 \sqrt{2} x+35 \\ & =\sqrt{2} x\lfloor\sqrt{2} x+7\rfloor+5\lfloor\sqrt{2} x+7\rfloor \\ & =\lfloor\sqrt{2} x+5\rfloor(\sqrt{2} x+7\rfloor \quad \text{Ans.} \end{aligned} $

Ex. 13 Factorise : $x^{2}-14 x+24$.

Sol. Product $ac=24 \And b=-14$

$\therefore \quad$ Split the middle term as $-12 \And -2$

$\Rightarrow x^{2}-14 x+24=x^{2}-12-2 x+24$

$\Rightarrow x(x-12)-2(x-12)=(x-12)(x-2)$

Ans.

Ex. 14 Factorise : $\quad x^{2}-\frac{13}{24} x-\frac{1}{12}$.

Sol. $\quad x^{2}-\frac{13}{24} x-\frac{1}{12}=\frac{1}{24}[24 x^{2}-13 x-2]$

Product $ac=-48 \And b=-13 \therefore$ We split the middle term as $-16 x+3 x$.

$=\frac{1}{24}[24 x^{2}-16 x+3 x-2]$

$=\frac{1}{24}[8 x(3 x-2)+1(3 x-2)]$

$=\frac{1}{24}(3 x-2)(8 x+1)$

Ans.

Ex. 15 Factorise : $\quad \frac{3}{2} x^{2}-8 x-\frac{35}{2}$

Sol. $\quad \frac{3}{2} x^{2}-8 x-\frac{35}{2}=\frac{1}{2}(3 x^{2}-16 x-35)=\frac{1}{2}(3 x^{2}-21 x+5 x-35)$

$=\frac{1}{2}[3 x(x-7)+5 x(x-7)]=\frac{1}{2}(x-7)(3 x+5) \quad$Ans.

(c) Integral Root Theorem :

If $f(x)$ is a polynomial with integral coefficient and the leading coefficient is 1 , then any integer root of $f(x)$ is a factor of the constant term. Thus if $f(x)=x^{3}-6 x^{2}+11 x-6$ has an Integral root, then it is one of the factors of 6 which are $\pm 1, \pm 2, \pm 3, \pm 6$. .

Now Infect

$ \begin{aligned} & f(1)=(1)^{3}-6(1)^{2}+11(1)-6=1-6+11-6=0 \\ & f(2)=(2)^{3}-6(2)^{2}+11(2)-6=8-24+22-6=0 \\ & f(3)=(3)^{3}-6(3)^{2}+11(3)-6=27-54+33-6=0 \end{aligned} $

Therefore Integral roots of $f(X)$ are $1,2,3$.

(d) Rational Root Theorem :

Let $\frac{b}{c}$ be a rational fraction in lowest terms. If $\frac{b}{c}$ is a root of the polynomial $f(x)=$ $a_n x^{n}+a_{n-1} x^{n-1}+\ldots .+a_1 x+a_0, a_n \neq 0$ with integral coefficients. Then $b$ is a factor of constant term $a_0$ and $c$ is a factor of the leading coefficient $a_n$.

For example : If $\frac{b}{c}$ is a rational root of the polynomial $f(x)=6 x^{3}+5 x^{2}-3 x-2$, then the values of $b$ are limited to the factors of -2 which are $\pm 1, \pm 2$ and the value of $c$ are limited to the factors of 6 , which are $\pm 1, \pm 2, \pm 3, \pm 6$ Hence, the possible rational roots of $f(x)$ are $\pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm \frac{2}{3} . \pm 1, \pm 2, \pm 3, \pm 6$. Infect -

1 is a Integral root and $\frac{2}{3},-\frac{1}{2}$ are the rational roots of $f(x)=6 x^{3}+5 x^{2}-3 x-2$.

NOTE:

(i) An $n^{\text {th }}$ degree polynomial can have at most $n$ real roots.

(ii) Finding a zero or root of polynomial $f(x)$ means solving the polynomial equation $f(x)=0$. It follows from the above discussion that if $f(x)=a x+b, a \neq 0$ is a linear polynomial, then it has only one root given by $f(x)$

$=0$ i.e. $f(x)=a x+b=0$

$\Rightarrow \quad a x=-b$

$\Rightarrow \quad x=-\frac{b}{a}$

Thus $\quad a=-\frac{b}{a}$ is the only $root$ of $f(x)=a x+b$.

Ex. 16 If $f(x)=2 x^{3}-13 x^{2}+17 x+12$ then find out the value of $f(-2) \And f(3)$.

Sol. $f(x)=2 x^{3}-13 x^{2}+17 x+12$

$ \begin{matrix} f(-2) & =2(-2)^{3}-13(-2)^{2}+17(-2)+12 & \\ & =-16-52-34+12=-90 & \text{Ans.} \\ f(3) & =2(3)^{3}-13(3)^{2}+17(3)+12 & \\ & =54-117+51+12=0 & \text{Ans.} \end{matrix} $

(e) Factorisation of an Expression Reducible to A Quadratic Expression :

Ex. 17 Factorise :- $8+9(a-b)^{6}-(a-b)^{12}$

Sol. $\quad-8+9(a-b)^{6}-(a-b)^{12}$

Let $(a-b)^{6}=x$

Then $-8+9 x-x^{2}=-(x^{2}-9 x+8)=-(x^{2}-8 x-x+8)$

$$ \begin{aligned} & =-(x-8)(x-1) \\ & =-[(a-b)^{6}-8][(a-b)^{6}-1] \\ & =[1-(a-b)^{6}][(a-b)^{6}-8] \\ & =[(1)^{3}-{(a-b)^{2}}^{3}][{(a-b)^{2}}^{3}-(2)^{3}] \\ & =[1-(a-b)^{2}][1+(a-b)^{4}+(a-b)^{2}][(a-b)^{2}-2][(a-b)^{4}+4+2(a-b)^{2}] \quad \text{Ans.} \end{aligned} $$

Ex. 18 Factorise : 6x $x^{2}-5 x y-4 y^{2}+x+17 y-15$

Sol. $\quad 6 x^{2}+x[1-5 y]-[4 y^{2}-17 y+15]$

$=6 x^{2}+x[1-5 y]-[4 y^{2}-17 y+15]$

$=6 x^{2}+x[1-5 y]-[4 y(y-3)-5(y-3)]$

$=6 x^{2}+x[1-5 y]-(4 y-5)(y-3)$

$=6 x^{2}+3(y-3) x-2(4 y-5) x-(4 y-)(y-3)$

$=3 x[2 x+y 3]-(4 y-5)(2 x+y-3)$

$=(2 x+y-3)(3 x-4 y+5) \quad$Ans.

$$EXERCISE$$

OBJECTIVE DPP - 5.1

~~ 1. Factors of $(42-x-x^{2})$ are:

(A) $(x-7)(x-6)$

(B) $(x+7)(x-6)$

(D) $(x+7)(x+6)$

~~ 2. Factors of $\Big(x^{2}+\frac{x}{6}-\frac{1}{6}\Big)$ are :

(A) $\frac{1}{6}(2 x+1)(3 x+1)$

(B) $\frac{1}{6}(2 x+1)(3 x-1)$

(D) $\frac{1}{6}(2 x-1)(3 x+1)$

~~ 3. Factors of polynomial $x^{3}-3 x^{2}-10 x+2 x$ are :

(A) $(x-2)(x+3)(x-4)$

(B) $(x+2)(x+3)(x+4)$

(D) $(x-2)(x-3)(x-4)$

~~ 4. If $(x+a)$ is a factor of $x^{2}+p x+q$ and $x^{2}+m x+n$ then the value of $a$ is :

(A) $\frac{m-p}{n-q}$

(B) $\frac{n-q}{m-p}$

(D) $\frac{m+p}{n+q}$

SUBJECTIVE DPP - 5.2

~~ 1. Factorise : $8 x^{3}+16-9$.

~~ 2. Factorise : $x^{4}+x^{3}-7 x^{2}-x+6$.

~~ 3. Factorise : $9 z^{3}-27 z^{2}-100 z+300$.

~~ 4. Determine whether $x-3$ is a factor of polynomial $p(x)=x^{3}-3 x^{2}+4 x-12$.

~~ 5. Using factor theorem, prove that $p(x)$ is divisible by $g(x)$ if $P(x)=4 a^{4}+5 x^{3}-12 x^{2}-11 x+5, g(x)=4 x+5$.

~~ 6. Determine if $(x+1)$ is a factor of $x^{3}-x^{2}-(2-\sqrt{2}) x+\sqrt{2}$.

~~ 7. $x^{3}-23 x^{2}+142 x-120$.

~~ 8. $x^{3}+13 x^{2}+32 x+20$.

~~ 9. $2 x^{3}+y^{2}-2 y-1$.

~~ 10. $4 z^{3}+20 z^{2}+33 z+18$.

~~ 11. $x^{4}+5 x^{2}+4$.

~~ 12. $x^{3}-10 x^{2}-53 x-42$.

$$ANSWER KEY$$

(Objective DPP # 4.1)

Qus.	1	2	3	4	5	6	7	8	9	10	11
Ans.	$A$	$C$	$C$	$D$	$C$	$D$	$A$	$B$	$C$	$C$	$D$

(Subjective DPP # 4.2)

~~ 1. 36

~~ 2. 1

~~ 3. (i) $25x^2 + 40xy + 16y^2$

(ii) $16x^2-40xy+25y^2$

(iii) $4x^2-4+\frac{1}{x^2}$

~~ 4. 189

~~ 5. 364

~~ 6. (i) -281250

(ii) $-\frac{5}{12}$

(iii) -0.018

~~ 7. (i) $x^2+11x+28$

(ii) $x^2+\frac{26}{5}x+1$

(iii) $p^4+\frac{63}{4}p^2-4$

~~ 8. (i) 10812

(ii) 999964

(iii) 1224

~~ 9. $(2x^2+49a^2+14ax)(2x^2+49a^2-14ax)$

~~ 10. $(x-1)(x+1)(x^2+1)(x^2+x+1)(x^2-x+1)(x^4-x^2+1)$

~~ 11. 3

~~ 12. (i) $x^6+7x^5-3^4+5x^2+ \sqrt{2}x +4$

(ii) $m^7+4m^6+8m^5-3m^2+6m-11$

~~ 13. (i) $(x^2+5x+3)(x^2+5x+7)$

~~ 14. $(4a-3b)^3$

~~ 15. $(x-y)(x+y)^3$

(Objective DPP # 5.1)

Qus.	1	2	3	4
Ans.	C	B	A	B

(Objective DPP # 5.2)

~~ 1. $(2 x-1)(4 x^{2}+2 x+9)$

~~ 2. $(x+1)(x-1)(x+3)(x-2)$

~~ 3. $(3 z+10)(z-3)(3 z-10)$

~~ 4. Yes

~~ 5. $(x+1)$ is a factor.

~~ 6. $(x-1)(x-10)(x-12)$

~~ 7. $(x+1)(x+2)(x+10)$

~~ 8. $(y-1)(y+1)(2 y+1)$

~~ 9. $(z+2)(2 z+3)(2 z+3)$

~~ 10. $(x-1)(x+1)(x-2)(x+2)$

~~ 11. $(x+1)(x-14)(x+3)$

$ \ggg \text{COORDINATE GEOMETRY} \lll $

ML - 6

CO-ORDINATE SYSTEM

In two dimensional coordinate geometry, we u se generally two types of co-ordinate system.

(i) Cartesian or Rectangular co-ordinate system.

(ii) Polar co-ordinate system.

In cartesian co-ordinate system we represent any point by ordered pair $(x, y)$ where $x$ and $y$ are called $X$ and Y co-ordinate of that point respectively.

In polar co-ordinate system we represent any point by ordered pair $(r, \theta)$ where ’ $r$ ’ is called radius vector and ’ $\theta$ ’ is called vectorial angle of that point.

CARTESIAN CO-ORDINATE SYSTEM

(a) Rectangular Co-ordinate Axes :

Let $X^{\prime} OX$ and $Y^{\prime} OY$ are two lines such that $X^{\prime} OX$ is horizontal and $Y^{\prime} OY$ is vertical lines in the same plane and they intersect each other at $O$. This intersecting point is called origin. Now choose a convenient unit of length and starting from origin as zero, mark off a number scale on the horizontal line $X^{\prime} OX$, positive to the right of origin $O$ and negative to the left of origin $O$. Also mark off the same scale on the vertical line $Y^{\prime} OY$, positive upwards and negative downwards of the origin. The line $X^{\prime} OX$ is called $X$-axis and the line $Y^{\prime} OY$ is known as $Y$-axis and the two lines taken together are called the co-ordinate axis.

(b) Quadrants :

The co-ordinates axes $X^{\prime} OX$ and $Y^{\prime} OY$ divide the place of graph paper into four parts $XOY, X^{\prime} OY, X^{\prime} OY$ and $XOY^{\prime}$. These four parts are called the quadrants. The part $XOY, X^{\prime} O, X^{\prime} O Y^{\prime}$ and $XOY^{\prime}$ are known as the first, second, third and fourth quadrant respectively.

(c) Cartesian Co-ordinates of a Point :

Let $X^{\prime} OX$ and $Y^{\prime} OY$ be the co-ordinate axis and $P$ be any point in the plane. To find the position of $P$ with respect of $X^{\prime} OX$ and $Y ; OY$, we draw two perpendiculars from $P$ on both co-ordinate axes. Let $PM$ and $PN$ be the perpendiculars on $X$-axis and $Y$-axis reservedly. The length of the line segment $OM$ is called the $x$ coordinate be the or abscissa of point $P$. Similarly the length of line segment $ON$ is called they-coordinate or ordinate of point $P$.

Let $OM=x$ and $ON=y$. The position of the point $P$ in the plane with respect to the coordinate axis is represented by the ordered pair $(x, y)$. The ordered pair $(x, y)$ is called the coordinates of point $P$. “Thus, for a given point, the abscissa and ordinate are the distance of the given point from $Y$-axis and $X$-axis respectively”.

The above system of coordinating on ordered pair $(x, y)$ with every point in plane is called the Rectangular Cartesian coordinates system.

(d) Convention of Signs :

As discussed earlier that regions $XOY, X^{\prime} OY, X^{\prime} OY^{\prime}$ and $XOY^{\prime}$ are known as the first, second, third and fourth quadrants respectively. The ray $O X$ is taken as positive $X$-axis, $OX^{\prime}$ as negative $X$-axis, $OY$ as positive $Y$-axis and $OY^{\prime}$ as negative $Y$-axis. Thus we have,

In first quadrant: $\quad X>0, \quad y>0 \quad$ (Positive quadrant)

In second quadrant: $\quad X<0, \quad Y>0$

In third quadrant: $\quad X<0, Y<0 \quad$ (Negative quadrant)

In fourth quadrant: $\quad X>0, \quad Y<0$

(e) Points on Axis :

In point $P$ lies on $X$-axis then clearly its distance from $X$-axis will be zero, therefore we can say that its coordinate will be zero. In general, if any point lies on $X$-axis then its $y$-coordinate will be zero. Similarly if any point $Q$ lies on $Y$-axis, then its distance from $Y$-axis will be zero therefore we can say its $x$-coordinate will be zero. In general, if any point lies on Y-axis then its x-coordinate will be zero.

(f) Plotting of Points :

In order to plot the points in a plane, we may use the following algorithm $m$.

Step I: Draw two mutually perpendicular lines on the graph paper, one horizontal and other vertical.

Step II: Mark their intersection point as O (origin).

Step III: Choose a suitable scale on X-axis and Y-axis and mark the points on both the axis.

Step IV: Obtain the coordinates of the point which is to be plotted. Let the point be $P(a, b)$. To plot this point start from the origin and $|a|$ units move along $OX, OX^{\prime}$ according as ’ $a$ ’ is positive or negative respectively. Suppose we arrive at point M. From point M move vertically upward or downward $|b|$ through units according as ’ $\mathbf{b}$ ’ is positive or negative. The point where we arrive finally is the required point $P(a, b)$.

ILLUSTRATIONS :

Ex. 1 Plot the point $(3,4)$ on a graph paper.

Sol. let $X^{\prime} I X$ and $Y^{\prime} O Y$ be the coordinate axis. Here given point is $P(3,4)$, first we move 3 units along $O X$ as 3 is positive then we arrive a point $M$. Now from $M$ we move vertically upward as 4 is positive. Then we arrive at $P(3,4)$.

Ex. 2 Write the quadrants for the following points. (i) $A(3,4)$ (ii) $B(-2,3)$ (iii) $C(-5,-2)$ (iv) $D(4,-3)$ (v) $E(-5,-5)$

Sol. (i) Here both coordinates are positive therefore point A lies in $It^{\text {st }}$ quadrant.

(ii) Here $x$ is negative and $y$ is positive therefore point $B$ lies in IInd quadrant.

(iii) Here both coordinates are negative therefore point $C$ lines in IIIrd quadrant.

(iv) Here $x$ is positive and $y$ is negative therefore point $D$ lies in $I V^{\text {th }}$ quadrant.

(v) Point E lies in III quadrant.

Ex. 3 Plot the following points on the graph paper. (i) $A(2,5)$ (ii) $B(-5,-7)$ (iii) $C(3,-2)$ (iv) D $(0,5)$ (v) $E(5,0)$

Sol. Let $X O X^{\prime}$ and $Y O Y^{\prime}$ be the coordinate axis. Then the given points may be plotted as given below :

DISTANCE BETWEEN TWO POINTS

If there are two points $A(x_1, y_1)$ and $B(x_2, y_2)$ on the $XY$ plane, the distance between them is given by $AB=$

$d=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Ex. 4 Find the distance between (i) $(5,3)$ and $(3,2)$ (ii) $(-1,4)$ and $(2,-3)$ (iii) $(a, b)$ and $(-b, a)$

Sol. Let $d_1, d_2, d_3$ be the required distances. By using the formula, we have

(i) $\quad d_1=\sqrt{(5-3)^{2}+(3-2)^{2}}=\sqrt{2^{2}+1^{2}=\sqrt{5}}$

(ii) $\quad d_2=\sqrt{(-1-2)^{2}+{4-(-3)}^{2}}=\sqrt{(-3)^{2}+7^{2}}=\sqrt{58}$

(iii) $\quad d_3=\sqrt{{a+(-b)}^{2}+(b-a)^{2}}=\sqrt{(a+b)^{2}+(a-b)^{2}}=\sqrt{2 a^{2}+2 b^{2}}$

$$EXERCISE$$

OBJECTIVE DPP - 6.1

~~ 1. The abscissa of a point is distance of the point from : (A) $X$-axis (B) Y-axis (C) Origin (D) None of these

~~ 2. The $y$ co-ordinate of a point is distance of that point from :

(A) $X$-axis (B) Y-axis (C) Origin (D) None of these

~~ 3. If both co-ordinates of any point are negative then that point will lie in :

(A) First quadrant (B) Second quadrant (C) Thirst quadrant (D) Fourth quadrant

~~ 4. If the abscissa of any point is zero then that point will lie :

(A) on $X$-axis (B) on $Y$-axis (C) at origin (D) None of these

~~ 5. The co-ordinates of one end point of a diameter of a circle are $(4,-1)$ and coordinates of the centre of the circle are $(1,-3)$ then coordinates of the other end of the diameter are :

(A) $(2,5)$ (B) $(-2,-5)$ (C) $(3,2)$ (D) $(-3,-2)$

~~ 6. The point $(-2,-1),(1,0),(4,3)$ and $(1,2)$ are the vertices of a :

(A) Rectangle (B) Parallelogram (C) Square (D) Rhombus

~~ 7. The distance of the point $(3,5)$ from $X$-axis is :

(A) $\sqrt{34}$ (B) 3 (C) 5 (D) None of these

SUBJECTIVE DPP - 6.2

~~ 1. Plot the points in the plane if its co-ordinates are given as $A(5,0), B(0,3) C(7,2), D(-4,3), E(-3,-2)$ and $F(3,-2)$.

~~ 2. In which quadrant do the following points lie $A(2,3), B(-2,3), C(-3,-5), D(3,-1)$. Explain with reasons.

~~ 3. Plot the following pairs of numbers as points in the Cartesian plane.

$x$	-3	-2	8	4	0
$y$	5	0	3	8	-2

~~ 4. With rectangular exes, plot the points $O(0,0), A(4,0)$ and $C(0,6)$. Find the coordinates of the fourth points $B$ such the $OABC$ forms a rectangle.

~~ 5. Plot the points $P(-3,1)$ and $Q(2,1)$ in rectangular coordinate system and find all possible coordinates of other two vertices of a square having $P$ and $Q$ as two adjacent vertices".

~~ 6. Find the value of $x$, if the distance between the points $(x,-1)$ and $(3,2)$ is 5 .

~~ 7. The base $A B$ two equilateral triangles $A B C$ and $A B C^{\prime}$ with side $2 a$, lies along the $x$-axis such that the mid point of $A B$ is at origin. Find the coordinates of the vertices $C$ and $C^{\prime}$ of the triangles.

$$\text{ANSWER KEY}$$

(Objective DPP # 6.1)

Qus.	1	2	3	4	5	6	7
Ans.	$B$	$A$	$C$	$B$	$B$	$B$	$C$

(Subjective DPP # 6.2)

~~ 2. A- Ist quadrant B - II ${ }^{\text {nd }}$ quadrant C IIIrd quadrant D - IV ${ }^{\text {th }}$ quadrant

~~ 4. $(4,6)$

~~ 5. $(-3,6),(2,6) \And (-3,-4),(2,-4)$

~~ 6. 7 or -1

~~ 7. $C(0, \sqrt{3} a), C^{\prime}(0,-\sqrt{3} a)$

LINEAR EQUATION IN $> > >$ TWO VARIABLES $< < <$

ML - 7

LINEAR EQUATIONS IN ONE VARIABLE

An equation of the form $a x+b=0$ where $a$ and $b$ are real numbers and ’ $x$ ’ is a variable, is called a linear equation in one variable.

Here ’ $a$ ’ is called coefficient of $x$ and ’ $b$ ’ is called as a constant term. i.e. $3 x+5=0,7 x-2=0$ etc.

LINEAR EQUATION IN TWO VARIABLES

An equation of the form $a x+b y+c=0$ where $a, b, c$ are real numbers and $a, b \neq 0$, and $x$, $y$ are variable, is called a linear equation in two variables, here ’ $a$ ’ is called coefficient of $x$, ’ $b$ ’ is called coefficient of $y$ and ’ $c$ ’ is called constant term.

Any pair of values of $x$ and $y$ which satisfies the equation $a x+b y+c=0$, is called a solution of it.

Ex. 1 Prove that $x=3, y=2$ is a solution of $3 x-2 y=5$.

Sol. $x=3, y=2$ is a solution of $3 x-2 y=5$, because L.H.S. $=3 x-2 y=3 \times 3-2 \times 2=9-4=5=$ R.H.S.

i.e. $x=3, y=2$ satisfied the equation $3 x-2 y=5$.

$\therefore \quad$ it is solution of the given equation.

Ex. 2 Prove that $x=1, y=1$ as well as $x=2, y=5$ is a solution of $4 x-y-3=0$.

Sol. Given eq. is $4 x-y-3=0 \quad \ldots \text{(i)}$

First we put $x=1, y=1$ in L.H.S. of eq…(i)

Here L.H.S. $=4 x-y-3=4 \times 1-1-3=4-4=0=$ R.H.S.

Now we put $x=2, y=5$ in eq. (i)

L.H.S. $=4 x-y-3=4 \times 2-5-3=8-8=0=$ R.H.S.

Since, $x=1, y=1$ and $x=2, y=5$ both pair satisfied in given equation therefore they are the solution of given equation.

Ex. 3 Determine whether the $x=2, y=-1$ is a solution of equation $3 x+5 y-2=0$.

Sol. Given eq, is $3 x+5 y-2=0$

Taking L.H.S. $=3 x+5 y-2=3 \times 2+5 \times(-1)-2=6-5-2=1 \neq 0$

Here L.H.S. $\neq$ R.H.S. therefore $x=2, y=-$ is not a solution of given equations.

GRAPH OF A LINEAR EQUATION

(A) in order to draw the graph of a linear equation in one variable we may follow the following algorithm.

Step I: Obtain the linear equation.

Step II: If the equation is of the form $a x=b, a \neq 0$, then plot the point $\Big(\frac{b}{a}, 0\Big)$ and one more point $\Big(\frac{b}{a}, \alpha\Big)$ when $\alpha$ is any real number. If the equation is of the form $a y=b, a \neq 0$, then plot the point $\Big(0, \frac{b}{a}\Big)$ and $\Big(\beta, \frac{b}{a}\Big)$ where $\beta$ is any real number.

Step III : Joint the points plotted in step II to obtain the required line.

NOTE :

If eq. is in form $a x=b$ then we get a line parallel to $Y$-axis and if eg. is in form $a y=b$ then we get a line parallel to X-axis.

Ex. 4 Draw the graph of (i) $2 x+5=0$ (ii) $3 y-15=0$

Sol. (i) Graph of $2 x+5=0$

On simplifying it we get $2 x=-5 \Rightarrow x=-\frac{5}{2}$

First we plot point $A_1\Big(-\frac{5}{2}, 0\Big)$ \And then we plot any other point $A_2\Big(-\frac{5}{2}, 2\Big)$ on the graph paper, then we join these two points we get required line $\ell$ as shown in figure below.

(ii) Graph of $3 y-15=0$

On simplifying it we get $3 u=15 \Rightarrow y=\frac{15}{3} f=5$.

First we plot the point $B_1(0,5)$ \And then we plot any other point $B_2(3,5)$ on the graph paper, then we join these two points we get required line $m$ as shown in figure.

NOTE :

A point which lies on the line is a solution of that equation. A point not lying on the line is not a solution of the equation.

(B) In order to draw the graph of a linear equation $a x+b y+c=0$ may follow the following algorithm.

Step I : Obtain the linear equation $a x+b y+c=0$.

Step II : Express $y$ in terms of $x$ i.e. $y=-\Big(\frac{a b+c}{b}\Big)$ or $x$ in terms of $y$ i.e. $x=-\Big(\frac{b y+c}{a}\Big)$.

Step III : Put any two or three values for $x$ or $y$ and calculate the corresponding values of $y$ or $x$ respectively from the expression obtained in Step II. Let we get points as $(\alpha_1, \beta_1),(\alpha_2, \beta_2),(\alpha_3, \beta_3)$.

Step IV : Plot the points $(\alpha_1, \beta_1),(\alpha_2, \beta_2),(\alpha_3, \beta_3)$ on graph paper.

Step V : Joint the pints marked in step IV to obtain. The line obtained is the graph of the equation $a x+b y+c=0$.

Ex. 5 Draw the graph of the line $x-2 y=3$, from the graph find the coordinate of the point when (i) $x=-5$ (ii) $y=0$

Sol. Here given equation is $x-2 y=3$.

Solving it for $y$ we get $2 y=x-3 \Rightarrow y=\frac{x-3}{2}$

Let $x=0$, then $y=\frac{0-3}{2}=\frac{-3}{2}$

$x=3$, then $y=\frac{3-3}{2}=0$

$x=-2$, then $y=\frac{-2-3}{2}=\frac{-5}{2}$ Hence we get

$x$	0	3	-2
$y$	$-\frac{3}{2}$	0	$-\frac{5}{2}$

Clearly when $x=-5$ then $y=-4$ and when $y=0$ then $x=3$.

Ex. 6 Draw the graphs of the lines represented by the equations $x+y=4$ and $2 x-y=2$ in the same graph. Also find the coordinate of the point where the two lines intersect.

Sol. Given equations are

$x+y=4 \ldots \ldots$ (i) \And $2 x-y=2 \ldots \ldots$ (ii)

(i) We have $y=4-x$

$x$	0	2	4
$y$	4	2	0

(ii) We have $y=2 x-2$

$x$	1	0	3
$y$	0	-2	4

By drawing the lines on a graph paper, clearly we can say that $P$ is the point of intersection where coordinates are $x=2, y=2$

DIFFERENCE FORMS OF A LINE

(a) Slope of a Line :

If a line makes an angle $\theta$ with positive direction of $x$-axis then tangent of this angle is called the slope of a line, it is denoted by $m$ i.e. $m=\tan \theta$.

(i) Slope - intercept from is $\mathbf{y}=\mathbf{m} \mathbf{x}+\mathbf{c}$ where $\mathbf{m}$ is the slope of line and $c$ is intercept made by line with Y-axis.

(ii) The equation of a line passing through origin is $y=m x$. Here $c=0$ then the line passes always from origin.

(iii) Intercept from of line is $\frac{x}{a}+\frac{y}{b}=1$ where $a \And b$ are intercepts on positive direction of $x$-axis and $y$-axis respectively made by line.

SOLUTION OF LINEAR EQUATION IN ONE VARIABLE

Let $a x+b=0$ is one equation then $a x+b=0 \Rightarrow a x+-b \Rightarrow x=-\frac{b}{a}$ is a solution.

Ex. 7 Solve : $\frac{x}{2}=3+\frac{x}{3}$

Sol. Given $\frac{x}{2}=3+\frac{x}{3} \Rightarrow \frac{x}{2}-\frac{x}{3}=3$

$\Rightarrow \frac{3 x-2 x}{6}=3$

$\Rightarrow \frac{x}{6}=3$

$\Rightarrow x=18 \quad$Ans.

SOLUTION OF LINEAR EQUATIONS IN TWO VARIABLE

(a) By Elimination of Making Equal Coefficient :

Ex. 8 Solve the following equations

$2 x-3 y=5$

$3 x+2 y=1$

Sol. Given eq. are $2 x-3 y=5$…..(i)

$ 3 x+2 y=1 $

Multiplying 1 eg.(i) by 3 and eg. (ii) by 2 we get

On subtraction $\begin{aligned} & 6 x-9 y=15 \\ & -6 x+_4 y=_2\end{aligned}$

$\Rightarrow \quad-13 y=13$

$\Rightarrow y=\frac{13}{-13}$

$\Rightarrow y=-1$

Put the value of $y$ in eg. (i) we get

$ \begin{aligned} & 2 x-(3) \times(-1)=5 \\ & 2 x+3=5 \\ \Rightarrow \quad & 2 x=5-3 \\ \Rightarrow \quad & 2 x=2 \\ \Rightarrow \quad & x=1 \\ \therefore \quad & x=1, y=1 \quad \text{Ans.} \end{aligned} $

(b) Substitution Method :

Ex. 9 Solve $x+4 y=14$

$7 x-3 y=$

Sol. From equation (i) $x=14-4 y$

Substitute the value of $x$ in equation (ii)

$ \begin{aligned} & \Rightarrow \quad 7(14-4 y)-3 y=5 \\ & \Rightarrow \quad 98-28 y-3 y=5 \\ & \Rightarrow \quad 98-31 y=5 \\ & \Rightarrow \quad 93=31 y \\ & \Rightarrow \quad y=\frac{93}{31} \\ & \Rightarrow y=3 \end{aligned} $

Now substitute value of $y$ in equation (ii)

$ \begin{aligned} & \Rightarrow \quad 7 x-3(3)=5 \\ & \Rightarrow \quad 7 x-3(3)=5 \\ & \Rightarrow \quad 7 x=14 \\ & \Rightarrow \quad x=\frac{14}{7}=2 \end{aligned} $

So, solution is $x=2$ and $y=3$.

Ans.

$EXERCISE$

OBJECTIVE DPP # 7.1

~~ 1. Which of the following equation is not linear equation ?

(A) $2 x+3=7 x-2$ (B) $\frac{2}{3} x+5=3 x-4$ (C) $x^{2}+3=5 x-3$ (D) $(x-2)^{2}=x^{2}+8$

~~ 2. Solution of equation $\sqrt{3} x-2=2 \sqrt{3}+4$ is

(A) $2(\sqrt{3}-1)$ (B) $2(1-\sqrt{3})$ (C) $1+\sqrt{3}$ (D) $2(1+\sqrt{3})$

~~ 3. The value of $x$ which satisfy $\frac{6 x+5}{4 x+7}=\frac{3 x+5}{2 x+6}$ is

(A) -1 (B) 1 (C) 2

(D) -2

~~ 4. Solution of $\frac{x-a}{b+c}+\frac{x-b}{c+a}+\frac{x-c}{a+b}=3$ is

(A) $a+b-c$ (B) $a-b+c$ (C) $-a+b+c$ (D) $a+b+c$

~~ 5. A man is thrice as old as his son. After 14 years, the man will be twice as old as his son, then present age of this son.

(A) 42 years (B) 14 years (C) 12 years (D) 36 years

~~ 6. One forth of one third of one half of a number is 12 , then number is

(A) 284 (B) 286 (C) 288 (D) 290

~~ 7. A linear equation in two variables has maximum

(A) only one solution (B) two solution (C) infinite solution (D) None of these

~~ 8. Solution of the equation $x-2 y=2$ is/are

(A) $x=4, y=1$ (B) $x=2, y=0$ (C) $x=6, y=2$ (D) All of these

~~ 9. The graph of line $5 x+3 y=4$ cuts $Y$-axis at the point

(A) $\Big(0, \frac{4}{3}\Big)$ (B) $\Big(0, \frac{3}{4}\Big)$ (C) $\Big(\frac{4}{5}, 0\Big)$ (D) $\Big(\frac{5}{4}, 0\Big)$

~~ 10. If $x=1, y=1$ is a solution of equation $9 a x+12 a y=63$ then, the value of $a$ is

(A) -3 (B) 3 (C) 7 (D) 5

SUBJECTIVE DPP - 7.2

Solve the following linear equations in one variable

~~ 1. If $\frac{2 x+7}{x+2}=\frac{4 x+3}{2 x-7}$, find the value of $x^{3}+x^{2}+x+1$.

~~ 2. Determine whether $x=5, y=4$ is a solution of the equation $x-2 y=-3$

Solve the following linear equations in two variable.

~~ 3. $8 x-5 y=34,3 x-2 y=13$

~~ 4. $20 x+3 y=7,8 y-15 x=5$

~~ 5. $\quad 2 x-3 y-3=0, \frac{2 x}{3}+4 y+\frac{1}{2}=0$

~~ 6. Draw the graph of $2 x+3 y=6$ and use it to find the area of triangle formed by the line and co-ordinate axis.

~~ 7. Draw the graph of the lines $4 x-y=5$ and $5 y-4 x=7$ on the same graph paper and find the coordinates of their point of intersection.

~~ 8. Find two numbers such that five times the greater exceeds four times the lesser by 22 and three times the greater together with seven times the lesser is 32 .

~~ 9. Draw the graph of $x-y+1=0$ and $3 x+2 y-12=0$ on the same graph. Calculate the area bounded by these lines \And X-axis.

alphabet~~ 10. If $p=3 x+1, q=\frac{1}{3}(9 x+13)$ and $p: q=6: 5$ then find $x$.

$$\text{ANSWER KEY}$$

(Objective DPP # 7.1)

Qus.	1	2	3	4	5	6	7	8	9	10
Ans.	$C$	$D$	$B$	$D$	$B$	$C$	$C$	$D$	$A$	$B$

(Subjective DPP # 7.2)

~~ 1. -104

~~ 2. Yes

~~ 3. $x=3, y=-2$

~~ 4. $x=\frac{1}{5}, y=1$

~~ 5. $x=\frac{21}{20}, y=-\frac{3}{10}$

~~ 6. Area $=3$ sq. units

~~ 7. $x=2, y=3$

~~ 8. 6,2

~~ 9. $\quad 7.5$ sq. units

~~ 10. -7

$> > >$ INTRODUCTION $< < <$

TO EUCLID’S GEOMETRY

ML - 8

INTRODUCTION

The credit for introducing geometrical concepts goes to the distinguished Greek mathematician ‘Euclid’ who is known as the “Father of Geometry” and the word ‘geometry’ comes from the Geek words ‘geo’ which means ‘Earth’ and ‘metreon’ which means ‘measure’.

BASIC CONCEPTS IN GEOMETRY

A ‘point’, a ’line’ and a ‘plane’ are the basic concepts to be used in geometry.

(a) Axioms :

The basic facts which are granted without proof are called axioms.

(b) Euclid’s Definitions :

(i) A point is that which has not part.

(ii) A line is breathless length.

(iii) The ends of a line segment are points.

(iv) A straight line is that which has length only.

(v) A surface is that which has length and breadth only.

(vi) The edges of surface are lines.

(vii) A plane surface is that which lies evenly with the straight lines on itself.

(c) Euclid’s Five Postulates :

(i) A straight line may be drawn from any one point to any other point.

(ii) A terminated line or a line segment can be produced infinitely.

(iii) A circle can be drawn with any centre and of any radius.

(iv) All right angles are equal to one another.

(v) If a straight line falling on two straight lines makes the exterior angles on the same side of it taken together less than two right angles, then the two straight lines if produced infinitely meet on that side on which the sum of angles are less than two right angles.

(d) Important Axioms :

(i) A line is the collection of infinite number of points.

(ii) Through a given point, an infinite lines can be drawn.

(iii) Given two distinct points, there is one and only one line that contains both the points.

(iv) If $P$ is a point outside a line $\ell$, then one and only one line can be drawn through $P$ which is parallel to

(v) Two distinct lines can not have more than one point in common.

(vi) Two lines which are both parallel to the same line, are parallel to each other.

i.e. $\ell||n, m|| n \Rightarrow \ell || m$

SOME IMPORTANT DEFINITOINS

(i) Collinear points : Three or more points are said to be collinear if there is a line which contains all of them.

(ii) Concurrent Lines : Three or more lines are said to be concurrent if there is a point which lies on all of them.

(iii) Intersecting lines : Two lines are intersecting if they have a common point. The common point is called the “point of intersection”.

(iv) Parallel lines : Two lines $I$ and $m$ in a plane are said to be parallel lines if they do not have a common point.

(v) Line Segment : Given two points A and B on a line I, the connected part (segment) of the line with end points at $A$ and $B$, is called the line segment $A B$.

(vi) Interior point of a line segment : A point $R$ is called an interior point of a line segment $P Q$ if $R$ lies between $P$ and $Q$ but $R$ is neither $P$ nor $Q$.

$ P \longrightarrow R $

(vii) Congruence of line segment : Two line segments $A B$ and $C D$ are congruent if trace copy of one can be superposed on the other so as to cover it completely and exactly in this case we write $A B \cong C D$. In other words we can say two lines are congruent if their lengths is same.

(viii) Distance between two points : The distance between two points $P$ and $Q$ is the length of line segment PQ

(ix) Ray : Directed line segment is called a ray. If $AB$ is a ray then it is denoted by $\overline{A B}$. Point $A$ is called initial point of ray.

(x) Opposite rays : Two rays $AB$ and $AC$ are said to be opposite rays if they are collinear and point $A$ is the only common point of the two rays.

Ex. 1 If a point $C$ lies between two points $A$ and $B$ such that $A C=B C$, then prove that $A C=\frac{1}{2} A B$. Explain by drawing the figure.

Sol. According to the given statement, the figure will be as shown alongside in which the point $C$ lies between two points $A$ and $B$ such that $AC=BC$.

Clearly, $\quad AC+BC=AB$

$\Rightarrow AC+AC=AB$

$[\because AC=BC]$

$\Rightarrow \quad 2 AC=AB$

And, $\quad AC=\frac{1}{2} AB$

Ex. 2 Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they, and how might you define them? (i) parallel lines (i) perpendicular lines (iii) line segment (iv) radius

Sol. (i) Parallel lines : Lines which don’t intersect any where are called parallel lines.

(ii) Perpendicular lines : Two lines which are at a right angle to each other are called perpendicular lines.

(iii) Line segment : it is a terminated line.

(iv) Radius : The length of the line-segment joining the centre of a circle to any point on its circumference is called its radius.

Ex. 3 How would you rewrite Euclid’ fifth postulate so that it would be easier to understand ?

Sol. Two distinct intersecting lines cannot be parallel to the same line.

Ex. 4 Does Euclid’s fifth postulate imply the existence of parallel lines ? Explain.

Sol. if a straight line $\ell$ falls on two straight lines $m$ and $n$ such that sum of the interior angles on one side of $\ell$ is two right angles, then by Euclid’s fifth postulate the line will not meet on this side of $\ell$. Next, we know that the sum of the interior angles on the other side of line $\ell$ also be two right angles. Therefore they will not meet on the other side. So, the lines $m$ and $n$ never meet and are, therefore parallel.

Theorem 1 : If $\ell, m, n$ are lines in the same plane such that $\ell$ intersects $m$ and $n || m$, then $\ell$ intersects $n$ also.

Given : Three lines $\ell, m, n$ in the same plane s.t. $\ell$ intersects $m$ and $n || m$.

To prove : Lines $\ell$ and $n$ are intersecting lines.

Proof : Let $\ell$ and $n$ be non intersecting lines. Then. $\ell || n$.

But, $n || m \quad$ [Given]

$\therefore \quad \ell || n$ and $n||m \Rightarrow \ell|| m$

$\Rightarrow \quad \ell$ and $m$ are non-intersecting lines.

This is a contradiction to the hypothesis that $\ell$ and $m$ are intersecting lines.

So our supposition is wrong.

Hence, $\ell$ intersects line $n$.

Theorem 2 : If lines $AB, AC, AD$ and $AE$ are parallel to a line $\ell$, then points $A, B, C, D$ and $E$ are collinear.

Given : Lines $AB, AC, AD$ and $AE$ are parallel to a line $\ell$.

To prove : A, B, C, D, E are collinear.

Proof : Since AB, AC, AD and AE are all parallel to a line $\ell$ Therefore point $A$ is outside $\ell$ and lines $A B$,

$AC, AD, AE$ are drawn through $A$ and each line is parallel to $\ell$.

But by parallel lines axiom, one and only one line can be drawn through the point A outside it and parallel to $\ell$.

This is possible only when A, B, C, D, and E all lie on the same line. Hence, A, B, C, D and E are collinear.

$$\text{EXERCISE}$$

SUBJECTIVE DPP # 8.1

~~ 1. How many lines can pass through :

(i) one point (ii) two distinct points

~~ 2. Write he largest number of points in which two distinct straight lines may intersect.

~~ 3. A, B and C are three collinear points such that point A lines between B and C. Name all the line segments determined by these points and write the relation between them.

~~ 4. State, true of false :

(i) A point is a undefined term

(ii) A line is a defined term.

(iii) Two distinct lines always intersect at one point.

(iv) Two distinct point always determine a line.

(v) A ray can be extended infinitely on both the sides of its.

(vi) A line segments has both of its end-points fixed and so it has a definite length.

~~ 5. Name three undefined terms.

~~ 6. If $AB$ is a line and $P$ is a fixed point, outside $AB$, how many lines can be drawn through $P$ which are :

(i) parallel to $AB$

(ii) Not parallel to $AB$

~~ 7. Out of the three lines $AB, CD$ and $EF$, if $AB$ is parallel to $EF$ and $CD$ is also parallel to $EF$, then what is the relation between $AB$ and $CD$.

~~ 8. If $A, B$ and $C$ three points on a line, and $B$ lines between $A$ and $C$, then prove that :

$AB+BC=AC$.

~~ 9. In the given figure, if $AB=CD$; prove that $AC=BD$.

~~ 10. (i) How many lines can be drawn to pass through three given point if they are not collinear?

(ii) How many line segments can be drawn to pass through there two given points if they are collinear

$$\text{ANSWER KEY}$$

(Subjective DPP # 8.1)

~~ 1. (i) Infinite $\quad \quad $ (ii) Only one

~~ 2. One

~~ 3. $BA, AC \And BC ; BA+AC=BC$

~~ 4. (i) True (ii) False (iii) False (iv) True (v) False (vi) True (ii) Only one

~~ 5. Point, line and plane

~~ 6. (i) Only one

(ii) Infinite

~~ 7. $AB || CD$

~~ 10.

(i) Three lines (ii) one

$> > > \text{LINES AND ANGLES} < < <$

ML - 9

LINE

A line has length but no width and no thickness.

ANGLE

An angle is the union of two non-collinear rays with a common initial point. The common initial point is called the ‘vertex’ of the angle and two rays are called the ‘arms’ of the angles.

REMAK :

Every angle has a measure and unit of measurement is degree.

One right angle $=90^{\circ}$

$1^{0}=60^{\prime}$ (minutes)

$1^{\prime}=60^{\prime \prime}$ (Seconds)

Angle addition axiom : If $X$ is a point in the interior of $\angle BAC$, then $m \angle BAC=m \angle BAX+m \angle XAC$

(a) Types of Angles :

(i) Right angles : An angle whose measure is $90^{\circ}$ is called a right angle.

(ii) Acute angle : An angle whose measure is less than $90^{\circ}$ is called an acute angle.

(iii) Obtuse angle : An angle whose measure is more than $90^{\circ}$ but less than $180^{\circ}$ is called an obtuse angle.

$ 90^{\circ}<\angle AOB<180^{\circ} $

(iv) Straight angle An angle whose measure is $180^{\circ}$ is called a straight angle.

(v) Reflex angle : An angle whose measure is more than $180^{\circ}$ is called a reflex angle.

(vi) Complementary angles : Two angles, the sum of whose measures is $90^{\circ}$ are called complementary angles.

$\angle AOC \And \angle BOC$ are complementary as $\angle AOC+\angle BOC=90^{\circ}$

(vii) Supplementary angles : Two angles, the sum of whose measures is $180^{\circ}$, are called the supplementary angles.

$\angle AOC \And \angle BOC$ are supplementary as their sum is $180^{\circ}$.

(viii) Angle Bisectors : A ray OX is said to be the bisector of $\angle A O B$, if $X$ is a point in the interior of $\angle A O B$, and $\angle AOX=\angle BOX$.

(ix) Adjacent angles : Two angles are called adjacent angles, it

(A) they have the same vertex,

(B) they have a common arm,

$\angle AOX$ and $\angle BOX$ are adjacent angles, $OX$ is common arm, $OA$ and $OB$ are non common arms and lies on either side of $O X$.

(x) Linear pair of angles : Two adjacent angles are said to form a linear pair of angles, if their non common arms are two opposite rays.

(xi) Vertically opposite angles : Two angles are called a pair of vertically opposite angles, if their arms form two pairs of opposite rays.

$\angle AOC \And \angle BOD$ form a pair of vertically opposite angles. Also $\angle OD \And \angle BOC$ form a pair of vertically opposite angles.

(b) Angles Made by a Transversal with two Parallel Lines :

(i) Transversal : A line which intersects two or more give parallel lines at distinct points is called a transversal of the given lines.

(ii) Corresponding angles : Two angles on the same side of transversal are known as the corresponding angles if both lie either above the two lines or below the two lines, in figure $\angle 1 \And \angle 5, \angle 4 \And \angle 8, \angle 2 \And \angle 6, \angle 3 \And \angle 7$ are the pairs of corresponding angles.

(iii) Alternate interior angles : $\angle 3 \And \angle 5, \angle 2 \And \angle 8$, are the pairs of alternate interior angles.

(iv) Consecutive interior angles : The pair of interior angles on the same side of the transversal are called pairs of consecutive interior angles. In figure $\angle 2 \And \angle 5, \angle 3 \And \angle 8$, are the pair of consecutive interior angles.

(v) Corresponding angles axiom :

It a transversal intersects two parallel lines, then each pair of corresponding angles are equal. Conversely, if a transversal intersects two lines, making a pair of equal corresponding angles, then the lines are parallel.

(c) Important Facts to Remember :

(i) If a ray stands on line, then the sum of the adjacent angles so formed is $180^{\circ}$.

(ii) If the sum of two adjacent angles is $180^{\circ}$, then their non common arms are two apposite rays.

(iii) The sum of all the angles round a point is equal to $360^{\circ}$

(iv) If two lines intersect, then the vertically opposite angles are equal.

(v) If a transversal interests two parallel lines then the corresponding angles are equal, each pair of alternate interior angles are equal and each pair of consecutive interior angles are supplementary.

(vi) if a transversal intersects two lines in such a way that a pair of alternet interior angles are equal, then the two lines are parallel.

(vii) If a transversal intersects two lines in such a way that a pair of consecutive interior angles are supplementary, then the two lines are parallel.

(viii) If two parallel lines are intersected by a transversal, the bisectors of any pair of alternate interior angles are parallel and the bisectors of an two corresponding angles are also parallel.

(ix) If a line i s perpendicular to one or two given parallel, lines, then it is also perpendicular to the other line.

(x) Two angles which have their arms parallel are either equal or supplementary.

(xi) Two angles whose arms are perpendicular are either equal or supplementary.

IMPORTANT THEOREMS

Theorem 1 : If two lines intersect each other, then the vertically opposite angles are equal.

Given : Two lines $A B$ and $C D$ intersecting at a point $O$.

To prove : (i) $\angle AOC=\angle BOD$

(ii) $\angle BOC=\angle AOD$

Proof : Since ray $O D$ stands on $A B$ $\therefore \angle AOD+\angle DOB=180^{\circ}$ [linear pair] again, ray $OA$ stands on $CD$ $\therefore \quad \angle AOC+\angle AOD=180^{\circ}$ [linear pair]

by (i) $\And $ (ii) we get

$\angle AOD+\angle DOB=\angle AOC+\angle AOD$

$\Rightarrow \angle DOB=\angle AOC$

$\Rightarrow \angle AOC=\angle DOB$

Similarly we can prove that $\angle BOC=\angle DOA$

Hence Proved.

Theorem 2 : If a transversal intersects two parallel lines, then each pair of alternate interior angles is equal.

Given : $AB$ and $CD$ are two parallel lines, Transversal $I$ intersects $AB$ and $CD$ at $P$ and $Q$ respectively making two pairs of alternate interior angles, $\angle 1, \angle 2 \And \angle 3, \angle 4$.

To prove : $\quad \angle 1=\angle 2$ and $\angle 3=\angle 4$

Proof : Clearly, $\angle 2=\angle 5 \quad$ [Vertically opposite angles]

And, $\angle 1=\angle 5 \quad$ [Corresponding angles]

$\therefore \quad \angle 1=\angle 2$

Also, $\quad \angle 3=\angle 6 \quad$ [Vertically opposite angles]

And, $\quad \angle 4=\angle 6 \quad$ [Corresponding angles]

$\therefore \quad \angle 3=\angle 4 \quad$Hence, Proved.

ILLUSTRATIONS

Ex. 1 Two supplementary angles are in ratio $4: 5$, find the angles,

Sol. Let angles are $4 x \And 5 x$.

$\therefore \quad$ Angles are supplementary

$\therefore 4 x+5 x=180^{\circ} \Rightarrow 9 x=180^{\circ}$

$\Rightarrow \quad x=\frac{180^{\circ}}{9}=20^{0}$

$\therefore \quad$ Angles are $4 \times 20^{0}, 5 \times 20^{0} \Rightarrow 80^{\circ} \And 100^{0}$

Ans.

Ex. 2 If an angle differs from its complement by 10, find the angle.

Sol. let angles is $x^{0}$ then its complement is $90-x^{0}$.

Now given $\quad x^{0}-(90-x^{0})=10$

$\Rightarrow x^{0}-90^{0}+x^{0}=10$

$\Rightarrow 2 x^{0}=10+90=100$

$\Rightarrow \quad x^{0}=\frac{100^{0}}{2}=50^{0}$

$\therefore \quad$ Required angle is $50^{\circ}$. Ans.

Ex. 3 In figure, $OP$ and $OQ$ bisects $\angle BOC$ and $\angle AOC$ respectively. Prove that $\angle POQ=90^{\circ}$.

Sol. $\therefore OP$ bisects $\angle BOC$

$\therefore \quad \angle POC=\frac{1}{2} \angle BOC \quad \ldots \text{(i)}$

Also OQ bisects $\angle A O C$

$\therefore \quad \angle COQ=\frac{1}{2} \angle AOC \quad \ldots \text{(ii)}$

$\therefore \quad OC$ stands on $AB$

$\therefore \quad \angle AOC+\angle BOC=180^{\circ} \quad \text{[Linear pair]}$

$\Rightarrow \frac{1}{2} \angle AOC+\frac{1}{2} \angle BOC=\frac{1}{2} \times 180^{0}$

$\Rightarrow \angle COQ+\angle POC=90 \circ \text{[Using (i) $\And$ (ii)]}$

$\Rightarrow \angle POQ=90^{\circ} \quad \text{[By angle sum property]}$

Hence Proved.

Ex. 4 In figure, lines $AB, CD$ and $EF$ intersect at $O$. Find the measures of $\angle AOC, \angle DOE$ and $\angle BOF$

Sol. Given $\angle AOE=40^{\circ} \And \angle BOD=35^{\circ}$

Clearly $\angle AOC=\angle BOD \quad \text{[Vertically opposite angles]}$

$\Rightarrow \angle A O C=35^{\circ} \quad$Ans.

$ \angle BOF=\angle AOE \quad \text { [Vertically opposite angles] } $

$\Rightarrow \angle B O F=40^{\circ}$Ans.

Now, $\angle DOE=\angle COF \quad$ [Vertically opposite angles]

$\therefore \quad \angle DOE=105^{\circ} \quad$Ans.

Ex. 5 In figure if $I||m, n|| p$ and $\angle 1=85^{\circ}$ find $\angle 2$

Sol. $\quad \therefore n || p$ and $m$ is transversal

$\therefore \angle 1=\angle 3=85^{\circ} \quad \text{[Corresponding angles]}$

Also $m || I \And p$ is transversal

$\therefore \quad \angle 2+\angle 3=180^{\circ}$

$[\because$ Consecutive interior angles]

$\Rightarrow \angle 2+85^{\circ}=180^{\circ}$

$\Rightarrow \angle 2+180^{\circ}-85^{0}$

$\Rightarrow \angle 2=95^{\circ}$

Ans.

$$\text{EXERCISE}$$

OBJECTIVE DPP # 9.1

~~ 1. If two lines intersected by a transversal, then each pair of corresponding angles so formed is "

(A) Equal (B) Complementary (C) Supplementary (D) None of these

~~ 2. Two parallel lines have :

(A) a common point (B) two common point (C) no any common point (D) infinite common points

~~ 3. An angle is $14^{0}$ more than its complementary angle then angle is :

(A) $38^{0}$ (B) $52^{0}$ (C) $50^{\circ}$ (D) none of these

~~ 4. The angle between the bisectors of two adjacent supplementary angles is :

(A) acute angle (B) right angle (C) obtuse angle (D) none of these

~~ 5. If one angle of triangle is equal to the sum of the other two then triangle is :

(A) acute a triangle (B) obtuse triangle (C) right triangle (D) none

~~ 6. $X$ lines in the interior of $\angle BAC$. If $\angle BAC=70^{\circ}$ and $\angle BAX=42^{\circ}$ then $\angle XAC=$

(A) $28^{0}$ (B) $29^{0}$ (C) $27^{0}$ (D) $30^{\circ}$

~~ 7. If the supplement of an angle is three times its complement, then angle is :

(A) $40^{\circ}$ (B) $35^{0}$ (C) $50^{\circ}$ (D) $45^{0}$

~~ 8. Two angles whose measures are $a \And b$ are such that $2 a-3 b=60^{0}$ then $\frac{4 a}{5 b}=$ ? If they form a linear pair :

(A) 0 (B) $\frac{8}{5}$ (C) $\frac{1}{2}$ (D) $\frac{2}{3}$

~~ 9. Which one of the following statements is not false :

(A) if two angles forming a linear pair, then each of these angles is of measure $90^{\circ}$

(B) angles forming a linear pair can both be acute angles

(D) bisectors of the adjacent angles form a right angle

~~ 10. Which one of the following is correct :

(A) If two parallel lines are intersected by a transversal, then alternate angles are equal

(B) If two parallel lines are intersected by a transversal then sum of the interior angles on the same side of transversal is $180^{\circ}$

(D) All of these

SUBJECTIVE DPP # 9.2

~~ 1. The supplement of an angle is one third of itself. Determine the angle and its supplement.

~~ 2. Two complementary angles are such that two times the measure of one is equal to three times measure of the other. Find the measure of the large angle.

~~ 3. Find the complement of each of the following angles.

(A) $36^{\circ} 40^{\prime}$

(B) $42^{0} 25^{\prime} 36^{\prime \prime}$

~~ 4. Write the supplementary angles of the following anglels .

(A) $54^{\circ} 28^{\prime}$

(B) $98^{\circ} 35^{\prime} 20^{\prime \prime}$

~~ 5. In figure, if $\angle BOC=7 x+20^{\circ}$ and $\angle COA=3 x$, then find the value of $x$ for which $AOB$ becomes a straight line.

~~ 6. In figure, if $x+y=w+z$ then prove that $AOB$ is a straight line.

~~ 7. If the bisectors of two adjacent angles form a right angle prove that their non common angles are in the same straight line.

~~ 8. In figure, find $\angle COD$ when $\angle AOC+\angle BOD=100^{\circ}$.

~~ 9. In figure $x: y: z=5: 4: 6$. if $X O Y$ is a straight line find the values of $x, y$ and $z$.

~~ 10. In the given figure, $AB$ is a mirror, $PO$ is the incident ray and $OR$, the reflected ray. If $\angle POR=112^{\circ}$ find $\angle POA$

~~ 11. In figure, if $A B||C D|| E F$ and $y: x=3: 7$ find $x$

~~ 12. In figure if $AB || CD, EF \perp CD$ and $\angle GED=126^{\circ}$, find $\angle AGE, \angle GEF$ and $\angle FGE$.

~~ 13. $\triangle ABC$ is an isosceles triangle in which $\angle B=\angle V$ and $L \And M$ are points on $AB \And AC$ respectively such that $LM || BC$. If $\angle A=50^{\circ}$ find $\angle LMC$.

~~ 14. In figure if $AB||DF, AD|| FG, \angle BAC=65^{\circ}, \angle ACB=55^{\circ}$. Find $\angle FGH$

~~ 15. In figure, $AB || ED$ and $\angle ABC=30^{\circ}, \angle EDC=70^{\circ}$ then find $x^{0}$.

$$\text{ANSWER KEY}$$

(Objective DPP # 9.1)

Qus.	1	2	3	4	5	6	7	8	9	10
Ans.	$D$	$C$	$B$	$B$	$C$	$A$	$D$	$B$	$C$	$D$

(Subjective DPP # 9.2)

~~ 1. $135^{\circ}, 45^{0}$

~~ 2. (A) $53^{\circ} 20^{\prime}$

~~ 3. (A) $125^{\circ} 32^{\prime}$

~~ 4. $16^{0}$

~~ 5. $60^{0}, 48^{0}, 72^{0}$

~~ 6. $126^{0}$

~~ 7. $115^{0}$

~~ 8. $260^{\circ}$

~~ 9. $54^{0}$

(B) $\quad 47^{0} 34^{\prime} 24^{\prime \prime}$

(B) $81^{0} 24^{\prime} 40^{\prime \prime}$

~~ 8. $80^{\circ}$

~~ 9. 34

~~ 10. $126^{0}, 36^{0}, 54^{0}$

~~ 11. $125^{0}$

$\quad \ggg$
TRIANGLES
$\quad \lll$

ML - 10

TRIANGLE

A plane figure bounded by three lines in a plane is called a triangle. Every triangle have three sides and three angels. If $ABC$ is any triangle then $AB, BC \And CA$ are three sides and $\angle A, \angle B$ and $\angle C$ are three angles.

(a) Types of Triangles :

(i) On the basis of sides we have three types of triangles:

(A) Scalene triangle : A triangle whose no two sides are equal is called a scalene triangle.

(B) Isosceles triangle - A triangle having two sides equal is called an isosceles triangle.

(C) Equilateral triangle - A triangle in which all sides are equal is called an equilateral triangle.

(ii) On the basis of angles we have three types of triangles:

(A) Right triangle - A triangle in which any one angle is right angle $(=90^{\circ})$ is called right triangle.

(B) Acute triangle - A triangle in which all angles are acute $(>90^{\circ})$ is called an acute triangle.

(C) Obtuse triangle - A triangle in which any one angle is obtuse $(<90^{\circ})$ is called an obtuse triangle.

CONGRUENT FIGURES

The figures are called congruent if they have same shape and same size. In order words, two figures are called congruent if they are having equal length, width and height.

Fig. (i)

Fig. (ii)

In the above figures {fig. (i) and fig. (ii)} both are equal in length, width and height, so these are congruent figures.

(a) Congruent Triangles :

Two triangles are congruent if and only if one of them can be made to superimposed on the other, so an to cover it exactly.

If two triangles $\triangle ABC$ and $\triangle DEF$ are congruent then there exist a one to one correspondence between their vertices and sides. i.e. we get following six equalities.

$\angle A=\angle D, \angle B=\angle E, \angle C=\angle F$ and $AB=DE, BC=EF, AC=DF$.

If two $\triangle ABC \And \triangle DEF$ are congruent under $A \leftrightarrow D, B \leftrightarrow E, C \leftrightarrow F$ one to one correspondence then we write $\triangle ABC \cong \triangle DEF$ we can not write as $\triangle ABC \cong \triangle DFE$ of $\triangle ABC \cong \triangle EDF$ or in other forms because $\triangle ABC \cong \triangle DFE$ have following one-one correspondence $A \leftrightarrow D, B \leftrightarrow F, C \leftrightarrow E$.

Hence we can say that “two triangles are congruent if and only if there exists a one-one correspondence between their vertices such that the corresponding sides and the corresponding angles of the two triangles are equal.

(b) Sufficient Conditions for Congruence of two Triangles :

(i) SAS Congruence Criterion :

Two triangles are congruent if two sides and the included angle of one are equal to the corresponding sides and the included angle of the other triangle.

(ii) ASA Congruence Criterion :

Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.

(iii) AAS Congruence Criterion :

If any two angles and a non included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.

(iv) SSS Congruence Criterion :

Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.

(v) RHS Congruence Criterion :

Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and one side of the other triangle.

(c) Congruence Relation in the Set of all Triangles :

By the definition of congruence of two triangles, we have following results.

(I) Every triangle is congruent to itself i.e. $\triangle ABC \cong \triangle ABC$

(II) If $\triangle ABC \cong \triangle DEF$ then $\triangle DEF \cong \triangle ABC$

(III) If $\triangle ABC \cong \triangle DEF$ and $\triangle DEF \cong \triangle PQR$ then $\triangle ABC \cong \triangle PQR$

NOTE : If two triangles are congruent then their corresponding sides and angles are also congruent by cpctc (corresponding parts of congruent triangles are also congruent).

Theorem-1 : Angles opposite to equal sides of an isosceles triangle are equal.

Given : $\triangle ABC$ in which $AB=AC$

To prove : $\angle B=\angle C$

Construction : We draw the bisector $AD$ of $\angle A$ which meets $BC$ in $D$.

Proof : In $\triangle ABD$ and $\triangle ACD$ we have

$AB = AC \quad \text{[Given]}$

$\angle BAD = \angle CAD \quad [\because \text{AD is bissector of }]$ $\angle A$

$\text{And, } \quad AD = AD \quad \text{[Common side]}$

$\therefore \quad $ By SAS criterion of congruence, we have

$\cong ABD = \cong ACD$

$\Rightarrow \cong B = \cong C \text{by cpctc}$

Hence Proved.

Theorem-2: if two angles of a triangle are equal, then sides opposite to them are also equal.

Given : $\triangle ABC$ in which $\angle B=\angle C$

To prove : $A B=A C$

Construction: We draw the bisector of $\angle A$ which meets $BC$ in $D$.

Proof : In $\triangle ABD$ and $\triangle ACD$ we have $\angle B=\angle C$ [Given] $\angle BAD=\angle CAD$ $[\because AD$ is bisector of $\angle A]$ $AD=AD$ [Common side]

$\therefore \quad$ By AAS criterion of congruence, we get

$\triangle ABD \cong \triangle ACD$

$\Rightarrow \quad AB=AC$

[By cpctc]

Hence, Proved.

Theorem-3 :if the bisector of the vertical angle bisects the base of the triangle, then the triangle is isosceles.

Given : $A \triangle A B C$ in which $AD$ is the bisector of $\angle A$ meeting $BC$ in $D$ such that $BD=CD$

To prove : $\triangle ACD$ is an isosceles triangle.

Construction : We produce $AD$ to $E$ such that $AD=DE$ and join $EC$.

Proof: In $\triangle ADB$ and $\triangle EDC$ we have

$AD=DE \quad \text{[By construction]}$

$\angle ADB=\angle CDE \text{[Vertically opposite angles]}$

$BD=DC \quad \text{[Given]}$

$\therefore \quad$ By SAS criterion of congruence, we get

$\triangle ADB \cong \Delta EDC \Rightarrow AB=EC$

And, $\angle BAD=\angle CED \quad \text{[By cpctc]}$

But, $\angle BAD=\angle CAD$

$\therefore \quad \angle CAD=\angle CED$

$\Rightarrow AC=EC \quad \text{[Sides opposite to equal angles are equal]}$

$\Rightarrow AC=AB \quad \text{[By eg. (i)]}$

Hence Proved.

Ex. 1 Prove that measure of each angle of an equilateral triangle is $60^{\circ}$.

Sol. Let $\triangle ABC$ be an equilateral triangle, then we have

$ \begin{aligned} & AB=BC=CA \quad \ldots \text{(i)} \\ & \therefore \quad AB=BC \end{aligned} $

$\therefore \quad \angle C=\angle A \quad \ldots \text{(ii)}$

Also, $\quad BC=CA$

$\therefore \angle A=\angle B \quad \ldots \text{(iii)} \quad \text{[Angles opposite to equal sides are equal]}$

By (ii) $\And $ (iii) we get $\angle A=\angle B=\angle C$

Now in $\triangle ABC \angle A+\angle B+\angle C=180^{\circ}$

$ \begin{aligned} & \Rightarrow 3 \angle A=180^{\circ} \\ & {[\therefore \angle A=\angle B=\angle C]} \\ & \Rightarrow \quad \angle A=60^{\circ}=\angle B=\angle C \end{aligned} $

Hence Proved.

Ex. 2 If $D$ is the mid-point of the hypotenuse $AC$ of a right triangle $ABC$, prove that $BD=\frac{1}{2} AC$.

Sol. Let $\triangle ABC$ is a right triangle such that $\angle B=90^{\circ}$ and $D$ is mid point of $AC$ then we have to prove that $BD=$ $\frac{1}{2} AC$ we produce $BD$ to $E$ such that $BD=AC$ and $EC$.

Now is $\triangle ADB$ and $\triangle CDE$ we have

$ \begin{aligned} & AD=DC \quad \text{[Given]}\\ & BD=DE \quad \text{[By construction]} \end{aligned} $

And, $\angle ADB=\angle CDE \quad \text{[Vertically opposite angles]}$

$\therefore \quad$ By SAS criterion of congruence we have

$\Rightarrow \quad EC=AB$ and $\angle CED=\angle ABD \quad \ldots \text{(i) [By cpctc]}$

But $\angle CED \And \angle ABD$ are alternate interior angles

$\therefore \quad CE || AB \Rightarrow \angle ABC+\angle ECB=180^{\circ} \quad \text{[Consecutive interior angles]}$

$\Rightarrow \quad 90+\angle ECB=180^{\circ}$

$\Rightarrow \angle ECB=90^{\circ}$

Now, In $\triangle ABC \And \triangle ECB$ we have

$AB=EC \quad \text{[By (i)]}$

$BC=BC \quad \text{[Common]}$

And, $\angle ABC=\angle ECB=90^{\circ}$

$\therefore \quad$ BY SAS criterion of congruence

$\triangle ABC \cong \triangle ECB$

$\Rightarrow AC=EB \quad \text{[By cpctc]}$

$\Rightarrow \frac{1}{2} AC=\frac{1}{2} EB$

$\Rightarrow BD=\frac{1}{2} AC$

Hence Proved.

Ex. 3 In a right angled triangle, one acute angle is double the other. Prove that the hypotenuse is double the smallest side.

Sol. Let $\triangle ABC$ is a right triangle such that $\angle B=90^{\circ}$ and $\angle ACB=2 \angle CAB$, then we have to prove $AC=2 BC$. we produce $CB$ to $D$ such that $BD=CB$ and join $AD$.

Proof : In $\triangle ABD$ and $\triangle ABC$ we have

$BD=BC \quad \text{[By construction]}$

$AB=AB \quad \text{[Common]}$

$\angle ABD=\angle ABC=90^{\circ}$

$\therefore \quad$ By SAS criterion of congruence we get $\triangle ABD \cong \triangle ABC$

$ \begin{aligned} & \Rightarrow AD=AC \text { and } \angle DAB=\angle CAB \quad[\text { By cpctc }] \\ & \Rightarrow AD=AC \text { and } \angle DAB=x \quad[\therefore \angle CAB=x] \end{aligned} $

Now, $\angle DAC=\angle DAB+\angle CAB=x+x=2 x$

$\therefore \quad \angle DAC=\angle ACD$

$\Rightarrow DC=AD \quad \text{[Side Opposite to equal angles]}$

$\Rightarrow 2 BC=AD \quad [\because DC=2 BC]$

$\Rightarrow 2 BC=AC \quad [[AD=AC]]$

Hence Proved.

Ex. 4 In figure, two sides $A B$ and $B C$ and the median $A M$ of a $\triangle A B C$ are respectively equal to sides $D E$ and $E F$ and the median $DN$ of $\triangle DEF$. Prove that $\triangle ABC \cong \triangle DEF$.

Sol. $\quad \therefore \quad AM$ and $DN$ are medians of $\triangle ABC \And \triangle DEF$ respectively

$ \therefore BM=MC \And EN=NF $

$ \Rightarrow BM=\frac{1}{2} BC \And EN=\frac{1}{2} EF$

$But, BC=EF \quad \therefore BM=EN \quad \ldots \text{(i)}$

In $\triangle ABM \And \triangle DEN$ we have

$AB=DE \quad \text{[Given]}$

$AM=DN \quad \text{[Given]}$

$BM=EN \quad \ldots \text{[By (i)]}$

$\therefore \quad$ By SSS criterion of congruence we have

$ \Delta ABM \cong \Delta DEN \Rightarrow \angle B=\angle E \ldots \text { (ii) } \quad[\text { By cpctc }] $

Now, In $\triangle ABC \And \triangle DEF$

$AB=DE \quad \text{[Given]}$

$\angle B=\angle E \quad \text{[By (ii)]}$

$BC=EF \quad \text{[Given]}$

$\therefore \quad$ By SAS criterion of congruence we get

$\triangle ABC \cong \triangle DEF$

Hence Proved.

SOME INEQUALITY RELATIONS IN A TRIANGLE

(i) If two sides of triangle are unequal, then the longer side has greater angle opposite to it. i.e. if in any $\triangle ABCAB>AC$ then $\angle C>\angle B$.

(ii) In a triangle the greater angle has the longer side opposite to it.

i.e. if in any $\triangle ABC \angle A>\angle B$ then $BC>AC$.

(iii) The sum of any two sides of a triangle is greater than the third side.

i.e. if in any $\triangle A B C, A B+B C>A C, B C+C A>A B$ and $A C+A B>B C$.

(iv) Of all the line segments that can be drawn to a given line, from a point, not lying on it, the perpendicular line segment is the shortest.

$P$ is any point not lying online $\ell, PM \perp$ then $PM<PN$.

(v) The difference of any two sides of a triangle is less than the third side.

i.e. In any $\triangle A B C, A B-B C<A C, B C-C A<A B$ and $A C<A B<B C$.

Ex. 5 In figure, $PQ=PR$, show that $PS>PQ$

Sol. In $\triangle PQR$

$ \therefore \quad PQ=PR $

$ \Rightarrow \quad \angle PRQ=\angle PQR \quad \ldots \text{(i)} \quad \text{[Angles opposite to equal sides]} $

$ \text { In } \triangle PSQ, SQ \text { is produced to } R $

$\therefore \quad$ Ext. $\angle PQR>\angle PSQ \quad \ldots \text{(ii)}$

$\angle PRQ>\angle PSQ \quad \text{[By eq. (i) and (ii)]}$

$\Rightarrow \angle PRS>\angle PSR$

$\Rightarrow PS>PR \quad \text{[Sides opposite to greater angles is larger]}$

But, $\quad PR=PQ$

$\therefore \quad PS>PQ \quad$Hence Proved.

Ex. 6 In figure, $T$ is a point on side $QR$ of $\triangle PQR$ and $S$ is a point such that $RT=ST$. Prove that $PQ+PR>QS$

Sol. In $\triangle PQR$ we have

$ P Q+P R>Q R $

$\Rightarrow PQ+PR>QT+TR$

$\Rightarrow PQ+PR>QT+ST \therefore RT=ST$

In $\triangle QST QT+ST>SQ$

$\therefore \quad PQ+PR>SQ$

Hence Proved.

$$EXERCISE$$

OBJECTIVE DPP # 10.1

~~ 1. In the three altitudes of a $\Delta$ are equal then triangle is :

(A) isosceles (B) equilateral (C) right angled (D) none

~~ 2. $A B C D$ is a square and $P, Q, R$ are points on $A B, B C$ and $C D$ respectively such that $A P=B Q=C R$ and $\angle PQR=90^{\circ}$, then $\angle QPR$

(A) $45^{0}$ (B) $50^{\circ}$ (C) $60^{\circ}$ (D) LM

~~ 3. In a $\triangle X Y Z, L M || Y Z$ and bisectors $Y N$ and $Z N$ of $\angle Y \And \angle Z$ respectively meet at $N$ on $L M$ then $Y L+Z M=$

(A) YZ (B) $X Y$ (C) XZ (D) LM

~~ 4. In a $\triangle PQR, PS$ is bisector of $\angle P$ and $\angle Q=70^{\circ} \angle R=30^{\circ}$, then

(A) $Q S>P Q>P R$ (B) $QS<PQ<PR$ (C) $PQ>QS>SR$ (D) $PQ<QS<SR$

~~ 5. If $D$ is any point on the side $BC$ of a $\triangle ABC$, then :

(A) $AB+BC+CA>2 AD$ (B) $AB+BC+CA<2 AD$ (C) $AB+BC+CA>3 AD$ (D) None

~~ 6. For given figure, which one is correct :

(A) $\triangle ABC \cong \triangle DEF$ (B) $\triangle ABC \cong \Delta FED$ (C) $\triangle ABC \cong \triangle DFE$ (D) $\triangle ABC \cong \triangle EDF$

~~ 7. In a right angled triangle. One acute angle is double the other then the hypotenuse is :

(A) Equal to smallest side (B) Double the smallest side (C) Triple the smallest side (D) None of these

SUBJECTIVE DPP - 10.2

~~ 1. In the $\triangle ABC$ given below, $BD$ bisects $\angle B$ and is perpendicular to $AC$. If the lengths of the sides of the triangle are expressed in terms of $x$ and $y$ as shown, find the values of $x$ and $y$.

~~ 2. In the figure, $AB=AD$ prove that $\angle BCD$ is a right angle.

~~ 3. If the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles.

~~ 4. $AD$ is meadian of $\triangle ABC$. Prove that $AB+AC>2 AD$.

~~ 5. $O$ is any point in the interior of a triangle $ABC$. Prove that $OB+OC<AB+AC$.

~~ 6. In figure, $\triangle ABC$ is a right angled triangle at $B$. $ADEC$ and $BCFG$ are square Prove that $AF=BE$.

~~ 7. In figure $CD$ is the diameter perpendicular to the chord $AB$ of a circle with centre $O$. Prove that

(a) $\angle CAO=\angle CBO$ (b) $\angle AOB=2 \angle ACB$

~~ 8. $\quad ABCD$ is a square and $EF || BD$. $E$ and $F$ are the mid point of $BC$ and $DC$ respectively. Prove that

(a) $BE=DF$ (b) AR bisects $\angle BAD$

~~ 9. In figure, $\triangle A B C$ is an equilateral triangle $P Q || A C$ and $A C$ is produced to $R$ such that $C R=P Q$. Prove that QR bisects PC.

~~ 10. In figure, the congruent parts of triangles have been indicated by line markings. Find the values of $x \And y$.

$$\text{ANSWER KEY}$$

(Objective DPP # 10.1)

Qus.	1	2	3	4	5	6	7
Ans.	B	A	D	B	A	C	B

(Subjective DPP # 10.2)

~~ 1. 16,8 ~~ 2. 71,9

$\ggg$
QUADRILATERIAL
$\lll$

ML - 11

QUADRILATERL

Aquadrilateral is a closed figure obtained by joining four points (with no three points collinear) In an order.

(I) Since, quad means four and lateral is forsides thereforequadrilateral meansa figure bounded by four sides.

(II) Every quadrilateral has :

(A) Four vertices,

(B) Four sides

(C) Four angles and

(D) Two diagonals.

(III) A diagonals is a line segment obtained on joining the opposite vertices.

(a) Sum of the Angles of a Quadrilateral :

Consider a quadrilateral $A B C D$ as shown alongside. Join $A$ and $C$ to get the diagonal $A C$ which divides the quadrilateral $ABCD$ into two triangles $ABC$ and $ADC$.

We know the sum of the angles of each triangle is $180^{\circ}$ ( 2 right angles).

$\therefore \quad$ In $\triangle ABC ; \angle CAB+\angle B+\angle BCA=180^{\circ}$ and

In $\triangle ADC ; \angle DAC+\angle D+\angle DCA=180^{\circ}$

On adding, we get : $(\angle CAB+\angle DAC)+\angle B+\angle D+(\angle BCA+\angle DCA)=180^{\circ}+180^{\circ}$

$\Rightarrow \angle A+\angle B+\angle D+\angle C=360^{\circ}$

Thus, the sum of the angles of a quadrilateral is $360^{\circ}$ (4-right angles).

Ex. 1 The angles of a quadrilateral are in the ratio $3: 5: 9: 13$. Find all the angles of the quadrilateral.

Sol. Given the ratio between the angles of the quadrilateral $=3: 5: 9: 13$ and $3+5+9+13=30$

Since, the sum of the angles of the quadrilateral $=360^{\circ}$

$\therefore \quad$ First angle of it $=\frac{3}{30} \times 360^{\circ}=36^{0}$,

Second angle $=\frac{5}{30} \times 360^{\circ}=60^{\circ}$,

Third angle $=\frac{9}{30} \times 360^{\circ}=108^{0}$,

And, $\quad$ Fourth angle $=\frac{13}{30} \times 360^{\circ}=156^{0}$

$\therefore \quad$ The angles of quadrilateral are $360^{\circ}, 60^{\circ}, 108^{\circ}$ and $156^{\circ}$.

ALTERNATE SOLUTION :

Let the angles be $3 x, 5 x, 9 x$ and 13 .

$\therefore \quad 3 x+5 x+9 x+13 x=360^{0}$

$\Rightarrow 30 x=360^{\circ}$ and $x=\frac{360^{\circ}}{30}=12^{\circ}$

$\therefore \quad 1^{\text {st }}$ angle $=3 x=2 \times 12^{0}=360^{\circ}$

$2^{\text {nd }}$ angle $=5 x=\times 12^{0}=60^{0}$

3rd angle $=9 x=9 \times 12^{0}=108^{0}$

And, $4^{\text {th }}$ angle $=13 \times 12^{0}=156^{0}$.

Ex. 2 Use the informations given in adjoining figure to calculate the value of $x$.

Sol. Since, $EAB$ is a straight line.

$\therefore \quad \angle DAE+\angle DAB=180^{\circ}$

$\Rightarrow 73^{\circ}+\angle DAB=180^{\circ}$

i.e., $\angle DAB=180^{\circ}-73^{\circ}=107^{\circ}$

Since, the sum of the angles of quadrilateral $ABCD$ is $360^{\circ}$

$\therefore 107^{0}+105^{\circ}+x+80^{0}=360^{\circ}$

$\Rightarrow 292^{0}+x=360^{0}$

$\Rightarrow x=360^{\circ}-292^{\circ}$

$\Rightarrow x=68^{\circ}$

Ans.

(b) Types of Quadrilaterals :

(i) Trapezium : It is a quadrilateral in which one pair of opposite sides are parallel. In the quadrilateral $ABCD$, drawn alongside, sides $AB$ and $DC$ are parallel, therefore it is a trapezium.

(ii) Parallelogram : It is a quadrilateral in which both the pairs of opposite sides are parallel. The adjoining figure shows a quadrilateral $A B C D$ in which $A B$ is parallel to $D C$ and $A D$ is parallel to $B C$, therefore $A B C D$ is a parallelogram.

(iii) Rectangle : it is a quadrilateral whose each angle is $90^{\circ}$

(A) $\angle A+\angle B=90^{\circ}+90^{\circ}=180^{\circ} \Rightarrow AD || BC$

(B) $\angle B+\angle C=90^{\circ}+90^{\circ}=180^{\circ} \Rightarrow AB || DC$

Rectangle $A B C D$ is a parallelogram Also.

(iv) Rhombus : It is a quadrilateral whose all the sides are equal. The adjoining figure shows a quadrilateral $ABCD$ in which $AB=BC=CD=DA$; therefore it is a rhombus.

(v) Square : It is a quadrilateral whose all the sides are equal and each angle is $90^{\circ}$. The adjoining figure shows a quadrilateral $ABCD$ in which $AB=BC=CD=DA$ and $\angle A=\angle B=\angle C=\angle D=90^{\circ}$, therefore $ABCD$ is a square.

(vi) Kite : It is a quadrilateral in which two pairs of adjacent sides are equal. The adjoining figure shows a quadrilateral $A B C D$ in which adjacent sides $A B$ and $A D$ are equal i.e., $A B=A D$ and also the other pair of adjacent sides are equal i.e., $BC=CD$; therefore it is a kite or kite shaped figure.

REMARK :

(i) Square, rectangle and rhombus are all parallelograms.

(ii) Kite and trapezium are not parallelograms.

(iii) A square is a rectangle.

(iv) A square is a rhombus.

(v) A parallelogram is a trapezium.

PARALLELOGRAM

A parallelogram is a quadrilateral in which both the pairs of opposite sides are parallel.

Theorem 1 : A diagonal of a parallelogram divides the parallelogram into two congruent triangles.

Given : A parallelogram ABCD.

To prove : A diagonal divides the parallelogram into two congruent triangles

i.e., if diagonal $AC$ is drawn then $\triangle ABC \cong \triangle CDA$

and if diagonal $BD$ is drawn then $\triangle ABD \cong \triangle CDB$

Construction : Join A and C

Proof : Sine, $A B C D$ is a parallelogram

$AB || DC$ and $AD || BC$

In $\triangle ABC$ and $\triangle CDA$

And, $AC=AC \quad$ [Common side]

$\therefore \quad \triangle ABC \cong \triangle CDA \quad \text{[By ASA]}$

Similarly, we can prove that

$\triangle ABD \cong \triangle CDB$

Theorem 2 : In a parallelogram, opposite sides are equal.

Given : A parallelogram $A B C D$ in which $A B || D C$ and $A D || B C$.

To prove : Opposite sides are equal i.e., $AB=DC$ and $AD=BC$

Construction : Join A and C

Proof : In $\triangle ABC$ and $\triangle CDA$

$\angle BAC=\angle DCA \quad \text{[Alternate angles]}$

$\angle BCA=\angle DAC \quad \text{[Alternate angles]}$

$AC=AC \quad \text{[Common]}$

$\therefore \quad \triangle ABC \cong \triangle CDA \quad \text{[By ASA]}$

$\Rightarrow AB=DC$ and $AD=BC \quad \text{[By cpctc]}$

Hence Proved.

Theorem 3: If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Given : A quadrilateral ABCD in which

To Prove: $A B C D$ is a parallelogram i.e., $A B$ || DC and AD || BC

Construction : Join A and C

Proof : In $\triangle ABC$ and $\triangle CDA$

	$A B=D C$	[Given]
	$AD=BC$	[Given]
And	$A C=A C$	[Common]
$\therefore$	$\Delta ABC \cong \Delta CDA$	[By SSS]
$\Rightarrow$	$\angle 1=\angle 3$	[By cpctc]
And	$\angle 2=\angle 4$	[By cpctc]

But these are alternate angles and whenever alternate angles are equal, the lines are parallel.

$\therefore \quad AB || DC$ and $AD || BC$

$\Rightarrow \quad$ ABCD is a parallelogram. Hence Proved.

Theorem 4 : In a parallelogram, opposite angles are equal.

Given : A parallelogram $ABCD$ in which $AB || DC$ and $AD || BC$.

To prove : Opposite angles are equal

i.e. $\angle A=\angle C$ and $\angle B=\angle D$

Construction : Draw diagonal $A C$

Proof : In $\triangle ABC$ and $\triangle CDA$ :

$ \begin{aligned} & \angle BAC=\angle DCA \quad \text { [Alternate angles] } \\ \\ & \angle BCA=\angle DAC \quad \text { [Alternate angles] } \\ \\ & AC=AC \quad[\text { Common] } \\ & \therefore \quad \Delta ABC \cong \Delta CDA[By ASA]\\ \\ & \Rightarrow \quad \angle B=\angle D \quad[\text { By cpctc }] \\ \\ & \text { And, } \quad \angle BAD=\angle DCB \text { i.e., } \quad \angle A=\angle C \\ \\ & \text { Similarly, we can prove that } \quad \angle B=\angle D \end{aligned} $

Hence Proved.

Theorem 5: If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.

Given : A quadrilateral $ABCD$ in which opposite angles are equal.

i.e., $\angle A=\angle C$ ad $\angle B=\angle D$

To prove : $ABCD$ is a parallelogram i.e, $AB || DC$ and $AD || BC$.

Proof : Since, the sum of the angles of quadrilateral is $360^{\circ}$

$\Rightarrow \angle A+\angle B+\angle C+\angle D=360^{\circ}$

$\Rightarrow \quad \angle A+\angle D+\angle A+\angle D=360 \quad[\angle A=\angle C$ and $\angle B=\angle D]$

$\Rightarrow 2 \angle A=2 \angle D=360^{\circ}$

$\Rightarrow \angle A+\angle D=180^{\circ} \quad \text{[Co-interior angle]}$

$\Rightarrow AB || DC$

Similarly,

$ \angle A+\angle B+\angle C+\angle D=360^{\circ} $

$\Rightarrow \angle A+\angle B+\angle A+\angle B=360^{\circ} \quad[\angle A=\angle C$ and $\angle B=\angle D]$

$\Rightarrow 2 \angle A+2 \angle V=360^{\circ}$

$\Rightarrow \angle A+\angle B=180^{\circ} \quad[\because$ This is sum of interior angles on the same side of transversal $A B]$

$\therefore \quad AD || BC$

So, $\quad AB || DC$ and $AD || BC$

$\Rightarrow ABCD$ is a parallelogram.

Hence Proved.

Theorem 6 : The diagonal of a parallelogram bisect each other.

Given : A parallelogram $ABCD$. Its diagonals $AC$ and $BD$ intersect each other at point $O$.

To prove : Diagonals $AC$ and $BD$ bisect each other i.e., $OA=OC$ and $OB=OD$.

Proof : In $\triangle AOB$ and $\triangle COD$

$\because \quad AB || DC$ and $BD$ is a transversal line.

$\therefore \quad \angle ABO=\angle DCO \quad \text{[Alternate angles]}$

$\therefore \quad AB || DC$ and $AC$ is a transversal line.

$\therefore \quad \angle BAO=\angle DCO \quad \text{[Alternate angles]}$

And,$\quad AB=DC$

$\Rightarrow \triangle AOB \cong \triangle COD \quad \text{[By ASA]}$

$\Rightarrow OA=OC$ and $OB=OD \quad \text{[By cpctc]}$

Hence Proved.

Theorem 7 : If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Given : A quadrilateral $ABCD$ whose diagonals $AC$ and $BD$ bisect each other at point $O$.

i.e., $OA=OC$ and $OB=OD$

To prove : $ABCD$ is a parallelogram

i.e., $AB || DC$ and $AD || BC$.

Proof : In $\triangle AOB$ and $\triangle COD$

$OA = OC \quad \text{[Given]}$

$OB = OD \quad \text{[Given]}$

And, $\angle AOB=\angle COD \quad \text{[Vertically opposite angles]}$

$\Rightarrow \Delta AOB \cong \triangle COD \quad \text{[By SAS]}$

$\Rightarrow \angle 1=\angle 2 \quad \text{[By cpctc]}$

But these are alternate angles and whenever alternate angles are equal, the lines are parallel.

$\therefore \quad AB$ is parallel to $DC$ i.e., $AB || DC$

Similarly, $\Delta AOD \cong \triangle COB \quad \text{[By SAS]}$

$\Rightarrow \angle 3=\angle 4$

But these are also alternate angles $\quad \Rightarrow \quad AD || BC$

$AB || DC$ and $AD || BC \quad \Rightarrow \quad ABCD$ is parallelogram.Hence Proved.

Theorem 8 : A quadrilateral is a parallelogram, if a pair of opposite sides is equal and parallel.

Given : A quadrilateral $ABCD$ in which $AB || DC$ and $AB=DC$.

To prove : $A B C D$ is a parallelogram

i.e., $AB || DC$ and $AD || BC$.

Construction : Join A and C.

Proof : Since $A B$ is parallel to $D C$ and $A C$ is transversal

$ \angle BAC=\angle DCA \quad \text{[Alternate angles]}$

$ AB=DC \quad \text{[Given]}$

$\text { And } AC=AC {[\text { Common Side }]} $

$\Rightarrow \triangle BAC \cong \triangle DCA \quad \text{[By SAS]}$

$\Rightarrow \angle BCA=\angle DAC \quad \text{[By cpctc]}$

But these are alternate angles and whenever alternate angles are equal, the lines are parallel.

$\Rightarrow AD || BC$

Now, AB || DC (given) and AD || BC $\quad$ [Proved above]

$\Rightarrow \quad ABCD$ is a parallelogram

Hence Proved.

REMARKS :

In order to prove that given quadrilateral is parallelogram, we have to prove that :

(i) Opposite angles of the quadrilateral are equal, or

(ii) Diagonals of the quadrilateral bisect each other, or

(iii) A pair of opposite sides is parallel and is of equal length, or

(iv) Opposite sides are equal.

(v) Every diagonal divides the parallelogram into two congruent triangles.

$$EXERCISE$$

OBJECTIVE DPP # 11.1

~~ 1. In a parallelogram $ABCD, \angle D=105^{\circ}$, then the $\angle A$ and $\angle B$ will be.

(A) $105^{\circ}, 75^{0}$ (B) $75^{0}, 105^{0}$ (C) $105^{0}, 105^{0}$ (D) $75^{0}, 75^{0}$

~~ 2. In a parallelogram $ABCD$ diagonals $AC$ and $BD$ intersects at $O$ and $AC=12.8 cm$ and $BD=7.6 cm$, then the measure of $O C$ and $O D$ respectively equal to :

(A) $1.9 cm, 6.4 cm$ (B) $3.8 cm, 3.2 cm$ (C) $3.8 cm, 3.2 cm$ (D) $6.4 cm, 3.8 cm$

~~ 3. Two opposite angles of a parallelogram are $(3 x-2)^{0}$ and $(50-x)^{0}$ then the value of $x$ will be :

(A) $17^{0}$ (B) $16^{0}$ (C) $15^{0}$ (D) $13^{0}$

~~ 4. When the diagonals of a parallelogram are perpendicular to each other then it is called.

(A) Square (B) Rectangle (C) Rhombus (D) Parallelogram

~~ 5. In a parallelogram $A B C D, E$ is the mid-point of side $B C$. If $D E$ and $A B$ when produced meet at $F$ then :

(A) $AF=\frac{1}{2} AB$ (B) $AF=2 AB$ (C) $AF=4 AB$ (D) Data Insufficient

~~ 6. $ABCD$ is a rhombus with $\angle ABC=56^{\circ}$, then the $\angle ACD$ will be.

(A) $56^{0}$ (B) $62^{0}$ (C) $124^{0}$ (D) $34^{0}$

~~ 7. In a triangle, $P, Q$, and $R$ are the mid-points of the sides $B C, C A$ and $A B$ respectively. If $A C=16 cm, B C=20$ $cm$ and $AB=24 cm$ then the perimeter of the quadrilateral $A R P Q$ will be.

(A) $60 cm$ (B) $30 cm$ (C) $40 cm$ (D) None

~~ 8. LMNO is a trapezium with $LM || NO$. If $P$ and $Q$ are the mid-points of $LO$ and $MN$ respectively and $LM=$ $5 cm$ and $ON=10 cm$ then $PQ=$

(A) $2.5 m$ (B) $5 cm$ (C) $7.5 cm$ (D) $15 cm$

~~ 9. In a Isosceles trapezium $ABCD$ if $\angle A=45^{\circ}$ then $\angle C$ will be.

(A) $90^{\circ}$ (B) $135^{0}$ (C) $90^{\circ}$ (D) None

~~ 10. In a right angle triangle $A B C$ is right angled at $B$. Given that $A B=9 cm, A C=15 cm$ and $D, E$ are the midpoints of the sides $AB$ and $AC$ respectively, then the area of $\triangle ADE=$

(A) $67.5 cm^{2}$ (B) $13.5 cm^{2}$ (C) $27 cm^{2}$ (D) Data insufficient

~~ 11. Find the measures of all the angles of a parallelogram , if one angle is $24^{0}$ less than twice the smallest angle.

~~ 12. In the following figure, $ABCD$ is a parallelogram in which $\angle DAB=75^{\circ}$ and $\angle DBC=60^{\circ}$. Find $\angle COB$ and $\angle ADB$.

~~ 3. In the following figure, $ABCD$ is a parallelogram $\angle DAO=40^{\circ}, \angle BAO=35^{\circ}$ and $\angle COD=65^{\circ}$. Find

(i) $\angle ABO$

(ii) $\angle ODC$

(iii) $\angle ACB$

(iv) $\angle CBD$

~~ 4. In the following figure, $ABCD$ is a parallelogram in which $\angle A=65^{\circ}$. Find $\angle B, \angle C$ and $\angle D$.

~~ 5. In the following figure, $ABCD$ is a parallelogram in which $\angle A=60^{\circ}$. If the bisectors of $\angle A$ and $\angle B$ meet at $P$, prove that $\angle APB=90^{\circ}$. Also, prove that $AD=DP, PC=BC$ and $DC=2 AD$.

$\quad \ggg$
QUADRILATERAL
$\lll$

ML - 12

MID-POINT THEOREM

Statement : In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side and is half of it.

Given : A triangle $A B C$ is which $P$ is the mid-point of side $A B$ and $Q$ is the mid-point of side $A C$.

To prove : $P$ is parallel to $B C$ and is half of it i.e., $P Q || B C$ and $P Q=\frac{1}{2} B C$

Construction : Produce $P Q$ upto point $R$ such that $P Q=Q R$. Join $T$ and $C$.

Proof : In $\triangle APQ$ and $\triangle CRQ$ :-

$PQ=QR \quad \text{[By construction]}$

$AQ=QC \quad \text{[Given]}$

And, $\angle AQP=\angle CQR \quad \text{[Vertically opposite angles]}$

$\Rightarrow \triangle APQ \cong \triangle CRQ \quad \text{[By SAS]}$

$\Rightarrow AP=CR \quad \text{[By cpctc]}$

And, $\angle APQ=\angle CRQ \quad \text{[By cpctc]}$

$But, \angle APQ$ and $\angle CRQ$ are alternate angles and we know, whenever the alternate angles are equal, the lines are parallel.

$\begin{matrix} \Rightarrow & AP || CR \\ \\ \Rightarrow & AB || CR \\ \\ \Rightarrow & BP || CR\end{matrix} $

Given, $P$ is mid-point of $AB$

$\Rightarrow AP=BP$

$\Rightarrow CR=BP \quad[As, AP=CR]$

Now, $\quad B P=C R$ and $B P || C R$

$\Rightarrow$ BCRP is a parallelogram.

[When any pair of opposite sides are equal and parallel, the quadrilateral is a parallelogram]

BCRP is a parallelogram and opposite sides of a parallelogram are equal and parallel.

$\therefore \quad PR=BC$ and $PR || BC$

Since, $P Q=Q R$

$\Rightarrow PQ=\frac{1}{2} PR$

$ =\frac{1}{2} BC $

Also, $\quad PQ || B C \quad \text{[As, PR || BC]}$

$\therefore \quad PQ || BC$ and $P+\frac{1}{2} BC$

ALTERNATIVE METHOD :

Construction : Draw CR parallel to BA intersecting PQ produced at point $R$.

Proof : In $\triangle APQ$ and $\triangle CRQ$

$AQ = CQ \quad \text{[Given]}$

$\angle AQP = \angle RQC \quad \text{[Vertically opposite angles]}$

$\text{And} \quad \angle PAQ = \angle RCQ \quad \text{[Alternate angles, as AB || CR]}$

$\triangle APQ \cong \triangle CRQ \quad \text{[By ASA]}$

$\Rightarrow CR=AP$ and $QR=PQ$

Since, $\quad CR=AP$ and $AP=PB$

$\Rightarrow CR=PB$

$\text{Also, } \quad CR || PB \quad \text{[By construction]}$

$\therefore$ PBCR is a parallelogram $\quad$ [As, opposite sides PB and CR are equal and parallel]

$\Rightarrow BC || PR \text{and} BC = PR$

$\Rightarrow BC || PQ \text{and} BC = 2PQ \quad [\because PQ = QR]$

$\Rightarrow PQ || BC$ and $PQ=\frac{1}{2} BC$

Hence Proved.

(a) Converse of the Mid-Point Theorem

Statement : The line drawn through the mid-point of one side of a triangle parallel to the another side bisects the third side.

Given : A triangle $A B C$ in which $P$ is the mid-point of side $A B$ nd $P Q$ is parallel to $B C$.

To prove: $P Q$ bisects the third side $A B$ i.e., $A Q=Q C$.

Construction : Through C, draw CR parallel to BA, which meets $P Q$ produced at point $R$.

Proof : Since, PQ || BC i.e., PR || BC [Given] and CR || BA i.e., CR || BP [By construction]

$\therefore$ Opposite sides of quadrilateral $PBCR$ are parallel.

$\Rightarrow$ PBCR is a parallelogram

$\Rightarrow BP=CR$

Also, $\quad BP=AP \quad[As, P$ is mid-point of $AB]$

$\therefore \quad CR=AP$

$\therefore \quad AB || CR $ and AC is transversal,

$\angle PAQ=\angle RCQ \quad \text{[Alternate angles]}$

$\therefore \quad AB || CR$ and PR is transversal,

$\angle APQ=\angle CRQ \quad \text{[Alternate angles]}$

In $\triangle APQ$ and $\triangle CRQ$

$CR=AP, \angle PAQ=\angle RCQ$ and $\angle APQ=\angle CRQ$

$\Rightarrow \triangle APQ \cong \triangle CRQ \quad \text{[By ASA]}$

$\Rightarrow AAQ=QC$

Hence Proved.

Ex. 1 $\quad A B C D$ is a rhombus and $P, Q, R$ and $S$ are the mid-points of the sides $A B, B C, C D$ and $D A$ respectively. Prove that the quadrilateral $PQRS$ is a rectangle.

Sol. According to the given statement, the figure will be a shown alongside; using mid-point theorem :-

In $\triangle ABC, PQ || AC$ and $PQ=\frac{1}{2} AC \quad \ldots \text{(i)}$

In $\triangle ADC, SR || AC$ and $SR=\frac{1}{2} AC \quad \ldots \text{(ii)}$

$\therefore \quad P=SR$ and $PQ || SR \quad \text{[From (i) and (ii)]}$

$\Rightarrow$ PQRS is a parallelogram.

Now, PQRS will be a rectangle if any angle of the parallelogram PWRS is $90^{\circ}$

$ \begin{matrix} PQ || AC & \text { [By mid-point theorem] } \\ \\ QR=BD & \text { [By mid-point theorem] } \end{matrix} $

But, $AC \perp BD \quad$ [Diagonals of a rhombus are perpendicular to each other]

$\therefore \quad PQ \perp QR \quad$ [Angle between two lines $=$ angles between their parallels]

$\Rightarrow$ PQRS is a rectangle

Hence Proved.

Ex.2 $ \quad A B C D$ is a trapezium in which $A B || D C, B D$ is a diagonal and $E$ is the mid-point of $A D$. A line is drawn through E parallel to $A B$ intersecting $B C$ at $F$ (as shown). Prove that $F$ is the mid-point of BC.

Sol. Given line $EF$ is parallel to $AB$ and $AB || DC$

$\therefore \quad EF||AB|| DC$.

According to the converse of the mid-point theorem, is $\triangle ABD, E$ is the mid-point of $AD$.

$EP$ is parallel to $AB$

[As EF $|| AB]$

$\therefore \quad$ P is mid-point of side $BD$

[The line through the mid-point of a side of a triangle and parallel to the other side, bisects the third side]

Now, in $\triangle B C D, P$ is mid-point of $B D$

[Proved above]

And, PF is parallel to DC

[As EF $|| DC]$

$\therefore \quad F$ is mid-point of $BC$

[The line through the mid-point of a side of a triangle and parallel to the other side, bisects the third side]

Hence Proved.

REMARK :

In quadrilateral $A B C D$, if side $A D$ is parallel to side $B C ; A B C D$ is a trapezium.

Now, $P$ and $Q$ are the mid-points of the non-parallel sides of the trapezium; then $P Q=\frac{1}{2}(A D+B C)$. i.e. The length of the line segment joining the mid-points of the two non-parallel sides of a trapezium is always equal to half of the sum of the lengths of its two parallel sides.

Theorem.3: If there are three or more parallel lines and the intercepts made by them on a transversal are equal, then the corresponding intercepts on any other transversal are also equal.

Given : Three parallel lines $I, m$ and $n$ i.e., $I||m|| n$. A transversal $p$ meets these parallel lines are points A, $B$ and $C$ respectively such that $AB=BC$. Another transversal $q$ also meets parallel lines $I, m$ and $n$ at points $D, E$ and $F$ respectively.

To prove : $DE=EF$

Construction : Through point A, draw a line parallel to DEF; which meets BE at point $P$ and $CF$ and point Q.

Proof : In $\triangle ACQ, B$ is mid-point of $AC$ and $BP$ is parallel to $CQ$ and we know that the line through the midpoint of one side of the triangle and parallel to another sides bisects the third side.

$\therefore \quad AP=PQ$

When the opposite sides of a quadrilateral are parallel, it is a parallelogram and so its opposite sides are equal.

$ \therefore AP || DE \text { and } AD || PE \quad \Rightarrow \quad \text { APED is a parallelogram. } $

$ \Rightarrow \quad AP=DE \quad \ldots .(ii) $

$ \text { And } \quad PQ || EF \text { and } PE || QF \quad \Rightarrow PQFE \text { is a parallelogram } $

$ \Rightarrow \quad PQ=EF . . .(iii) $

From above equations, we get

$ DE=EF $

Hence Proved.

Ex. 3In the given figure, $E$ and $F$ are respectively, the mid-points of non-parallel sides of a trapezium $ABCD$.

Prove that

(i) EF $|| AB$

(ii) $EF=\frac{1}{2}(AB+DC)$.

Sol. Join $B E$ and produce it to intersect $CD$ produced at point P. In $\triangle AEB$ and $\triangle DEP, AB || PC$ and $BP$ is transversal

$ \begin{aligned} & \Rightarrow \angle ABE=\angle DPE \quad \text { [Alternate interior angles] } \\ \\ & \angle AEB=\angle DEP \quad \text { [Vertically opposite angles] } \\ \\ & \text { And } \quad AE=DE \quad[E \text { is mid }- \text { point of } AD] \\ \\ & \Rightarrow \Delta AEB \cong \Delta DEP \quad \text { [By ASA] } \\ \\ & \Rightarrow BE=PE \quad[\text { By cpctc }] \\ \\ & \begin{matrix} \text { And } \quad AB=DP & {[\text { By cpctc }]} \end{matrix} \end{aligned} $

Since, the line joining the mind-points of any two sides of a triangle is parallel and half of the third side, therefore, is $\triangle BPC$,

$E$ is mid-point of $BP \quad[As, BE=PE]$

and $F$ is mid-point of $BC \quad$ [Given]

$\Rightarrow EF || PC$ and $EF=\frac{1}{2} PC$

$\Rightarrow EF || DC$ and $EF=\frac{1}{2}(PD+DC)$

$\Rightarrow \quad EF || AB$ and $EF=\frac{1}{2}(AB+DC) \quad[As, DC || AB$ and $PD=AB] \quad$

Hence Proved.

$$EXERCISE$$

~~ 1. When the opposite sides of quadrilateral are parallel to each other then it is called.

(A) Square (B) Parallelogram (C) Trapezium (D) Rhombus

~~ 2. In a $\triangle A B C, D, E$ and $F$ are respectively, the mid-points of $B C, C A$ and $A B$. If the lengths of side $A B, B C$ and $C A$ are $17 cm, 18 cm$ and $19 cm$ respectively, then the perimeter of $\triangle D E F$ equal to :

(A) $54 cm$ (B) $18 cm$ (C) $27 cm$ (D) $13.5 cm$

~~ 3. When only one pair of opposite sides of a quadrilateral parallel to each other it is called.

(A) Square (B) Rhombus (C) Parallelogram (D) Trapezium

~~ 4. When the diagonals of a parallelogram are equal but not perpendicular to each other it is called a.

(A) Square (B) Rectangle (C) Rhombus (D) Parallelogram

~~ 5. When each angle of a rhombus equal to 90.0 , it is called a.

(A) Square (B) Rectangle (C) Trapezium (D) Parallelogram

~~ 6. In the adjoining figure, $AP$ and $BP$ are angle bisectors of $\angle A$ and $\angle B$ which meets at $P$ on the parallelogram $ABCD$. Then $2 \angle APB=$

(A) $\angle C+\angle D$ (B) $\angle A+\angle C$ (C) $\angle B+\angle D$ (D) $2 \angle C$

~~ 7. In a quadrilateral $ABCD, AO \And DO$ are angle bisectors of $\angle A$ and $\angle D$ and given that $\angle C=105^{\circ}, \angle B=70^{\circ}$ then the $\angle AOD$ is :

(A) $67.5^{0}$ (B) $77.5^{0}$ (C) $87.5^{0}$ (D) $99.75^{0}$

~~ 8. In a parallelogram the sum of the angle bisectors of two adjacent angle is :

(A) $30^{\circ}$ (B) $45^{0}$ (C) $60^{\circ}$ (D) $90^{\circ}$

~~ 9. In the adjoining parallelogram $ABCD$, the angles $x$ and $y$ are :

(A) $60^{\circ}, 30^{\circ}$ (B) $30^{\circ}, 60^{0}$ (C) $45^{0}, 45^{0}$ (D) $90^{0}, 90^{\circ}$

~~ 10. From the figure find the value of $\angle SQP$ and $\angle QSP$ of parallelogram $PQRS$.

(A) $60^{\circ}, 50^{\circ}$ (B) $60^{\circ}, 45^{0}$ (C) $70^{0}, 35^{0}$ (D) $35^{0}, 70^{\circ}$

SUBJECTIVE DPP 12.2

~~ 1. Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to each to the parallel sides and is equal to half of the difference of these sides.

~~ 2. $ABCD$ is a parallelogram. $P$ is a point on $AD$ such that $AP=\frac{1}{3} AD \cdot Q$ is a point on $BC$ such that $CQ=\frac{1}{3} BC$. Prove that AQCP is a parallelogram.

~~ 3. In the following figure, $AD$ is a median and $DE || AB$. Prove that $BE$ is a median.

~~ 4. Prove that “If a diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angle and then the two diagonals are perpendicular to each other.

~~ 5. Prove that the figure formed by joining the mid-points of the consecutive sides of a quadrilateral is a parallelogram.

~~ 6. In a parallelogram $A B C D$, the bisector of $\angle A$ also bisects $B C$ at $P$. Prove that $A D=2 A B$.

~~ 7. The diagonals of parallelogram $A B C D$ intersect at $O$. A line through $O$ intersects $A B$ at $X$ and $D C$ at $Y$. Prove that $OX=OY$.

~~ 8. Show that the quadrilateral formed by joining the mid points of the sides of square is also a square.

~~ 9. $\quad ABCD$ is a trapezium in which side $AB$ is parallel to side $DC$ and $E$ is the mid-point of side $AD$. If $F$ is a point on side $B C$ such that segment $E F$ is parallel to side $D C$. Prove that $E F=\frac{1}{2}(A B+D C)$.

~~ 10. In $\triangle ABC, AD$ is the median through $A$ and $E$ is the mid-point of $AD$. $BE$ produced meets $AC$ in $F$. Prove that $AF=\frac{1}{3} AC$.

$$\text{ANSWER KEY}$$

(Objective DPP # 11.1)

Qus.	1	2	3	4	5	6	7	8	9	10
Ans.	$B$	$D$	$D$	$C$	$B$	$B$	$C$	$C$	$B$	$C$

(Subjective DPP # 11.2)

~~ 1. $68^{0}, 12^{0}, 68^{0}, 112^{0}$

~~ 2. $45^{\circ} \And 60^{\circ}$

~~ 3. (i) $80^{\circ}$

(ii) $80^{\circ}$

(iii) $40^{\circ}$ (iv) $25^{0}$

~~ 4. $115^{\circ}, 65^{0}$ and $115^{0}$

(Objective DPP # 12.1)

Qus.	1	2	3	4	5	6	7	8	9	10
Ans.	B	C	D	B	A	A	C	D	A	A

$> > >$ AREA OF PARALLELOGRAMS AND TRIANGLE $< < <$

ML - 13

POLYGONAL REGION

Polygon region can be expressed as the union of a finite number of triangular regions in a plane such that if two of these intersect, their intersection is either a point or a line segment. It is the shaded portion including its sides as shown in the figure.

(a) Area Axioms :

Every polygonal region $R$ has an area, measure in square unit and denoted by $ar(R)$.

(i) Congruent area axiom : if $R_1$ and $R_2$ be two regions such that $R_1 \cong R_2$ then $ar(R_1)=ar(R_2)$.

(ii) Area monotone axiom : If $R_1 \subset R_2$, then are $(R_1) \leq ar(R_2)$.

(iii) Area addition axiom : If $R_1$ are two polygonal regions, whose intersection is a finite number of points and line segments and $R=R_1 \cup R_2$, then $ar(R)=ar(R_1)+ar(R_2)$.

(iv) Rectangular area axiom : If $AB=a$ metre and $AD=b$ metre then,

ar (Rectangular region $ABCD)=ab$ sq. $m$.

(b) Unit of Area :

There is a standard square region of side 1 metre, called a square metre, which is the unit of area measure.

The area of a polygonal region is square metres (sq. $m$ or $m^{2}$ ) is a positive real number

AREA OF A PARALLELOGRAM

(a) Base and Altitude of a Parallelogram :

(i) Base : Any side of parallelogram can be called its base.

(ii) Altitude : The length of the line segment which is perpendicular to the base from the opposite side is called the altitude or height of the parallelogram corresponding to the given base.

In the Adjoining Figure

(i) DL is the altitude of $||^{gm}$ ABCD, corresponding to the base $A B$.

(ii) DM is the altitude of $||^{gm} \quad A B C D$, corresponding to the base $B C$.

Theorem -1 A diagonal of parallelogram divides it into two triangles of equal area.

Given : A parallelogram $ABCD$ whose one of the diagonals is $BD$.

To prove : $ar(\triangle ABD)=ar(\triangle CDB)$.

Proof : In $\triangle ABD$ and $\triangle CDB$.

$AB=DC$ [Opp. sides of a $||^{gm}$ ]

$AD=BC$ [Opp. sides of a $||^{gm}$ ]

$BD=BD$ [Common side]

$\therefore \quad \triangle ABD \cong \triangle CDB$ [By SSS]

$\therefore \quad ar(\triangle ABD)=ar(\Delta CDB)$ [Congruent area axiom]

Hence Proved.

Theorem -2: Parallelograms on the same base or equal base and between the same parallels are equal in area.

Given : Two $||^{g m} \quad A B C D$ and $A B E F$ on the same base $A B$ and between the same parallels $A B$ and $F C$.

To prove : $ar(||^{gm} \quad A B C D)=ar(||^{gm} \quad A B E F)$

Proof : In $\triangle ADF$ and $\triangle BCE$, we have

$ \begin{aligned} AD=BC & \quad {[\text { Opposite sides of a } ||^{gm}] } \\ \\ AF=BE & \quad {[\text { Opposite sides of a } ||^{gm}] } \\ \\ \angle DAF=\angle CBE & {[\because AD || BC \text { and } AF || BE] } \end{aligned} $

[Angle between $AD$ and $AF=$ angle between $BC$ and $BE$ ]

$ \begin{aligned} & \therefore \quad \Delta ADF \cong \triangle BCE \quad \text { [By SAS] } \\ \\ & \therefore \quad ar(\Delta ADF)=ar(\Delta BCE) \quad \ldots .(i) \\ \\ & \therefore \quad ar(||^{gm} ABCD)=ar(ABED)+ar(\Delta BCE) \\ \\ & \quad=ar(ABED)+ar(\Delta ADF)[Using(i)] \\ \\ & =ar(||^{gm} \quad ABEF) . \end{aligned} $

Hence, $ar(||^{gm} \quad gmCD)=ar(||^{gm} ABEF)$.

Hence Proved.

NOTE : A rectangle is also parallelogram.

Theorem -3: The are of parallelogram is the product of its base and the corresponding altitude.

Given : $A ||^{gm} \quad A B C D$ in which $A B$ is the base and $A L$ is the corresponding height.

To prove : Area $(||^{gm} A B C D)=A B \times A L$.

Construction : Draw BM $\perp$ DC so that rectangle $A B M L$ is formed.

Proof : $||^{gm} \quad ABCD$ and rectangle $ABML$ are on the same base $AB$ and between the same parallel lines $AB$ and LC.

$\therefore \quad ar(||^{gm}$ ABCD $)=ar($ rectangle $ABML)=AB \times AL$.

$\therefore$ area of a $||^{gm}=$ base $\times$ height.

Hence Proved.

Theorem-4 : Parallelograms on equal bases and between the same parallels are equal in area.

Given : Two $||^{gm}$ ABCD and PQRS with equal base $AB$ and $PQ$ and between the same parallels, $AQ$ and DR.

To prove: $ar(||^{gm} \quad gmCD)=ar(||^{gm}$ PQRS $)$.

Construction : Draw AL $\perp DR$ and $PM \perp DR$.

Proof : AB $|| D R, A L \perp D R$ and $P M \perp D r$

$\therefore \quad AL=PM$.

$\therefore \quad ar(||^{gm} \quad ABCD)=AB \times AL$

$ \begin{matrix} =PQ \times PM & {[\because AB=PQ \text { and } AL=PM]} \\ \\ =a(||^{gm} PQRS) . & \end{matrix} $

Hence Proved.

ILLUSTRATIONS :

Ex. 1 In a parallelogram $A B C D, A B=8 cm$. The altitudes corresponding to sides $A B$ and $A D$ are respectively $4 m$ and $5 cm$. Find AD.

Sol. We know that, Area of a parallelogram $=$ Base $\times$ Corresponding altitude

$ \begin{matrix} \therefore & \text { Area of parallelogram } ABCD=AD \times BN=AB \times DM \\ \Rightarrow & AD \times 5=8 \times 4 \\ \Rightarrow & AD=\frac{8 \times 4}{5} \\ & =6.4 cm . \quad \text{Ans.} \end{matrix} $

Ex. 2 In figure, $ABCD$ is a parallelogram, $AE \perp DC$ and $CF \perp AD$. If $AB=16 cm, AE=8 cm$ and $CF=10 cm$ find AD.

Sol. We have $AB=16 cm, AE=8 cm CF=10 cm$.

We know that are of parallelogram $=$ Base $\times$ Height $\quad$ [Base $=CD$, height $=AE$ ]

$ABCD=CD \times AE=16 \times 8=128 cm^{2}$

Again, Area of parallelogram $=$ Base $\times$ Height $=AD \times CF \quad[$ Base $=AD$, height $=CF$ ]

$128=AD \times 10$

$\Rightarrow \quad AD=\frac{128}{10}=12.8 cm \quad$Ans.

Ex. $3 \quad ABCD$ is a quadrilateral and $BD$ is one of its diagonal as shown in the figure. Show that the quadrilateral $ABCD$ is a parallelogram and find its area.

Sol. From figure, the transversal $DB$ is intersecting a pair of lines $DC$ and $AB$ such that $\angle CDB=\angle ABD=90^{\circ}$.

Hence these angles from a pair of alternate equal angles.

$\therefore \quad DC || AB$.

Also $DC=AB=2.5$ units.

$\therefore$ Quadrilateral $ABCD$ is a parallelogram.

Now, area of parallelogram $ABCD$

$=$ Base $\times$ Corresponding altitude

$=2.5 \times 4$

$=10$ sq. unitsAns.

Ex. 4 The diagonals of a parallelogram $A B C D$ intersect in $O$. A line through $O$ meets $AB$ is $X$ and the opposite side $C D$ in $Y$. Show that ar (quadrilateral $A X Y D)=\frac{1}{2}$ far(parallelogram $.A B C D)$.

Sol. $\therefore \quad A C$ is a diagonal of the parallelogram $ABCD$.

$ ar(\triangle ACD)=\frac{1}{2} ar(ABCD) $

Now, in $\triangle$ s AOX and COY,

$AO=CO$

$\because \quad$ Diagonals of parallelogram bisect each other.

$ \begin{matrix} \angle AOX=\angle COY & \text { [Vert. opp. } \angle s] \\ \\ \angle OAX=\angle OCY & {[\text { Alt. Int. } \angle s]} \end{matrix} $

$\therefore \quad AB || DC$ and transversal $AC$ intersects them

$\therefore \quad \triangle AOX \cong \triangle COY$

$\therefore \quad ar(\Delta AOX)=ar(\Delta COY)$

Adding ar(quad. AOYD) to both sides of (ii), we get

$ar($ quad. $A O Y D)+ar(\triangle AOX)=ar($ quad. $A O Y D)+ar(\Delta C O Y)$

$\Rightarrow ar($ quad. $A X Y D)=ar(\triangle ACD)=\frac{1}{2} ar(||^{gm} ABCD) \quad($ using $(i))$

Construction : Through B, draw BD || CA intersecting PA produced in D and through C, draw CQ || BP, intersecting line $AP$ in $Q$.

Proof : We have,

BD $|| CA$ And, $\quad$ BC $|| DA$

[By construction]

[Given]

$\therefore \quad$ Quad. BCAD is a parallelogram.

Similarly, Quad. BCQP is a parallelogram.

Now, parallelogram BCQP and BCAD are on the same base BC, and between the same parallels.

$\therefore \quad ar(|| g^{g m} B C Q P)=ar(||^{gm}$ BCAD $)$

We know that the diagonals of a parallelogram divides it into two triangles of equal area.

$\therefore \quad ar(\triangle PBC=\frac{1}{2} ar(||^{gm} BCQP).$

And $ ar(\triangle ABC)=\frac{1}{2} ar(||^{gm} BCAD) $

Now, $\quad ar(|| g^{g m}.$ BCQP $)=ar(||^{gm}$ gCAD $)$

[From (i)]

$ \Rightarrow \quad \frac{1}{2} ar(||^{gm} \text { BCAD })=\frac{1}{2} ar(||^{gm} \text { BCQP }) $

Hence, $ar(\triangle ABC)=ar(\triangle PBC)$

[Using (ii) and (iii)]

Hence Proved.

Theorem-6 : The area of a trapezium is half the product of its height and the sum of the parallel sides.

Given : Trapezium $A B C D$ in which $A B || D C, A L \perp D C, C N \perp A B$ and $A L=C N=h$ (say) $AB=a, DC=b$.

To prove : $ar(trap . A B C D)=\frac{1}{2} h \times(a+b)$.

Construction : Join AC.

Proof : AC is a diagonal of quad. $A B C D$.

$\therefore \quad ar(trap . ABCD)=ar(\triangle ABC)+ar(\triangle ACD)=\frac{1}{2} h \times a+\frac{1}{2} h \times b=\frac{1}{2} h(a+b)$.

Hence Proved.

Theorem -7: Triangles having equal areas and having one side of the triangle equal to corresponding side of the other, have their corresponding altitudes equal

Given : Two triangles $A B C$ and $P Q R$ such that (i) ar $(\triangle A B C)=ar(\triangle P Q R)$ and (ii) $A B=P Q$.

$CN$ and $RT$ and the altitude corresponding to $AB$ and $PQ$ respectively of the two triangles.

To prove : CR $=$ RT.

Proof : In $\triangle ABC, CN$ is the altitude corresponding to the side $AB$.

$ar(\triangle ABC)=\frac{1}{2} AB \times CN$

Similarly, $\quad ar(\triangle PQR)=\frac{1}{2} PQ \times RT$

Since $ar(\triangle ABC)=ar(\triangle PQR) \quad \text{[Given]}$

$\therefore \quad \frac{1}{2} AB \times CN=\frac{1}{2} PQ \times RT$

Also, $AB=PQ \quad \text{[Given]}$

$ CN=RT $

Hence Proved.

Ex. 5 In figure, $E$ is any point on median $AD$ of a $\triangle ABC$. Show that $ar(ABE)=ar(ACE)$.

Sol. Construction : From A draw $AG \perp BC$ and from $E$ draw $EF \perp BC$.

Proof : $ar(\triangle ABD)=\frac{BD \times AG}{2}$

$ar(\triangle ADC)=\frac{DC \times G}{2}$

But, $\quad BD=DC \quad[\therefore D$ is the mid-point of $BC, AD$ being the median $]$

$ar(\triangle ABD)=ar(\triangle ADC) \quad \ldots \text{(i)}$

Again, $ar(\triangle EBD)=\frac{BD \times EF}{2}$

$ar(\triangle EDC)=\frac{DC \times EF}{2}$

But, $\quad BD=DC$

$\therefore \quad ar(\triangle EBD)=ar(\triangle EDC) \quad \ldots \text{(ii)}$

Subtracting (ii) from (i), we get

$ar(\triangle ABD)-ar(\triangle EBD)=ar(\triangle ADC)-ar(\Delta EDC)$

$\Rightarrow ar(\triangle ABE)=ar(\triangle ACE)$

Hence Proved.

Ex. 6 Triangles $A B C$ and $D B C$ are on the same base $B C$; with $A, D$ on opposite sides of the line $B C$, such that $ar(\triangle ABC)=ar(\triangle DBC)$. Show that $BC$ bisects $AD$.

Sol. Construction : Draw $AL \perp BC$ and $DM \perp BC$.

Proof : $ar(\triangle ABC)=ar(\triangle DBC)\quad \text{[Given]}$

$\Rightarrow \frac{BC \times AL}{2}=\frac{BC \times DM}{2}$

$\Rightarrow AL=DM$

Now in $\triangle s$ OAL and OMD

$ AL=DM $

$ \Rightarrow \angle ALO=\angle DMO $

$ \Rightarrow \angle AOL=\angle MOD $

$ \Rightarrow \angle OAL=\angle ODM $

$ \therefore \Delta OAL \cong \Delta OMD $

$ \therefore OA=OD $

Hence Proved.

Ex. 7 ABC is a triangle in which $D$ is the mid-point of $BC$ and $E$ is the mid-point of AD. Prove that the area of $\triangle BED=\frac{1}{4}$ area of $\triangle ABC$.

Sol. Given : $A \triangle ABC$ in which $D$ is the mid-point of $BC$ and $E$ is the mid-point of $AD$.

To prove: $ar(\triangle BED)=\frac{1}{4} ar(\triangle ABC)$.

Proof : $\because AD$ is a median of $\triangle ABC$.

$\therefore \quad ar(\triangle ABD)=ar(\triangle ADC)=\frac{1}{2} ar(\triangle ABC) \quad \ldots \text{(i)}$

$[\therefore$ Median of a triangle divides it into two triangles of equal area $)=\frac{1}{2} ar(\triangle ABC)$

Again,

$ \because B E \text { is a median of } \triangle ABD, $

$ \therefore ar(\triangle BEA)=ar(\triangle BED)=\frac{1}{2} ar(\triangle ABD) $

$ {\therefore \text { Median of a triangle divides it into two triangles }} $

$ \text { And } .\frac{1}{2} ar(\triangle ABD)=\frac{1}{2} \times \frac{1}{2} ar(\triangle ABC) \quad \text { From }(i)] $

$\therefore ar(\triangle BED)=\frac{1}{4} ar(\triangle ABC) $

$\text { } \therefore \text { Median of a triangle divides it into two triangles of equal area } $

Hence Proved.

Ex. 8 if the medians of a $\triangle ABC$ intersect at $G$, show that $ar(\triangle AGB)=ar(\triangle BGC)=\frac{1}{3} ar(\triangle ABC)$.

Sol. Given : $A \triangle ABC$ its medians $AD, BE$ and $CF$ intersect at $G$.

To prove : $ar(\triangle AGB)=ar(\triangle AGC)=ar(\triangle BGC)=\frac{1}{3} ar(\triangle ABC)$.

Proof : A median of triangle divides it into two triangles of equal area. In $\triangle ABC, AD$ is the median.

$\therefore \quad ar(\triangle ABD)=ar(\triangle ACD) \quad \ldots \text{(i)}$

In $\triangle GBC, GD$ is the median.

$\therefore \quad ar(\triangle GBD)=ar(\triangle GCD) \quad \ldots \text{(ii)}$

From (i) and (ii), we get

$ ar(\triangle ABD)-ar(\Delta GBD)=ar(\Delta ACD)-ar(\Delta GCD) $

$\therefore \quad a(\Delta AGB)=ar(\Delta AGC) $

Similarly,

$ ar(\Delta AGB)=ar(\Delta AGC)=ar(\Delta BGC) \quad \ldots \text{(iii)} $

But, $ar(ABC)=ar(\triangle AGB)+ar(\triangle AGC)+ar(\triangle BGC)$

$ =3 ar(\Delta AGB) \quad \text{[Using (iii)]} $

$\therefore \quad ar(\triangle AGB) \quad=\frac{1}{3} ar(\triangle ABC)$.

Hence, $ar(\triangle AGB)=ar(\triangle AGC)=ar \Delta(BGC)=\frac{1}{3} ar(\triangle ABC)$.

Hence Proved.

Ex. 9 D,E and $F$ are respectively the mid points of the sides $B C, C A$ and $A B$ of a $\triangle A B C$. Show that

(i) $BDEF$ is parallelogram

(ii) $ar(||^{gm} BDEF)=\frac{1}{2} ar(\triangle ABC)$

(iii) $ar(\triangle DEF)=\frac{1}{4} ar(\triangle ABC)$

Sol. Given : $A \triangle ABC$ in which $D, E, F$ are the mid-point of the side $BC, CA$ and $AB$ respectively.

To prove:

(i) Quadrilateral BDEF is parallelogram.

(ii) $ar(|| g^{g m} BDEF)=\frac{1}{2} ar(\triangle ABC)$.

(iii) $ar(\triangle DEF)=\frac{1}{4} ar(\triangle ABC)$.

Proof:

(i) In $\triangle ABC$,

$\therefore \quad F$ is the mid-point of side $AB$ and $E$ is the mid point of side $AC$.

$\therefore \quad EF || BD$

[ $\because$ Line joining the mid-points of any two sides of a $\Delta$ is parallel to the third side.]

Similarly,

ED $|| FB$.

Hence, BDEF is a parallelogram.

Hence Proved.

(ii) Similarly, we can prove that AFDE and FDCE are parallelograms.

$\therefore \quad$ FD is diagonals of parallelogram BDEF.

$\therefore \quad ar(\triangle FBD)=ar(\triangle DEF) \quad \ldots \text{(i)}$

Similarly,

$ ar(\Delta FAE)=ar(\Delta DEF) \quad \ldots \text{(ii)} $

And $ ar(\triangle DCE)=ar(\triangle DEF) \quad \ldots \text{(iii)} $

From above equations, we have

$ \begin{aligned} & ar(\Delta FBD)=ar(\Delta FAE)=ar(\Delta DCE)=ar(\Delta DEF) \\ \\ & \text { And } \quad ar(\Delta FBD)+ar(\Delta DCE)+ar(\Delta DEF)+ar(\Delta FAE)=ar(\Delta ABC) \\ \\ & \Rightarrow \quad 2[ar(\Delta FBD)+ar(\Delta DEF)]=ar(\Delta AC) \quad[By \text { using }(i),(ii) \text { and }(iii)] \\ \\ & \Rightarrow \quad 2[ar(||^{gm} BDEF)]=ar(\Delta ABC) \\ \\ & \Rightarrow ar(||^{gm} BDEF)=\frac{1}{2} ar(ABC) \end{aligned} $

(iii) Since, $\triangle ABC$ is divided into four non-overlapping triangles FBD, FAE, DCE and DEF.

$ \begin{aligned} & \therefore \quad ar(\triangle ABC)=ar(\Delta FBD)+ar(\Delta FAE)+ar(\Delta DCE)+ar(\Delta DEF) \\ & \Rightarrow \quad ar(\triangle ABC)=4 ar(\Delta DEF) \quad \text { [Using (i), (ii) and (iii)] } \end{aligned} $

$\Rightarrow ar(\triangle DEF)=\frac{1}{2} ar(\triangle ABC)$

Hence Proved.

Ex. 10 Prove that the area of an equilateral triangle is equal to $\frac{\sqrt{3}}{4} a^{2}$, where $a$ is the side of the triangle.

Sol. Draw $AD \perp BC$

$ \begin{matrix} \Rightarrow & \Delta ABD \cong \triangle ACD & \text { [Br R.H.S.] } \\ \\ \therefore & BD=DC & \text { [By cpctc] } \\ \\ \therefore & BC=a & \\ \\ \therefore & BD=DC=\frac{a}{2} & \end{matrix} $

In right angled $\triangle ABD$

$ \begin{aligned} AD^{2} & =AB^{2}-BD^{2}=a^{2}-\Big(\frac{a}{2}\Big)^{2}=a^{2}-\frac{a^{2}}{4}=\frac{3 a^{2}}{4} \\ \Rightarrow \quad AD & =\frac{\sqrt{3} a}{2} \end{aligned} $

Area of $\triangle ABC=\frac{1}{2} BC \times AD=\frac{1}{2} a \frac{\sqrt{3} a}{2}=\frac{\sqrt{3} a^{2}}{4}$.

Hence Proved.

Ex. 11 In figure, $P$ is a point in the interior of rectangle $ABCD$. Show that

(i) $ar(\triangle APB)+ar(\triangle PCD)=\frac{1}{2} ar($ rect. $ABCD)$

(ii) $ar(APD)+ar(PBC)=ar(APB)+ar(PCD)$

Sol. Given : A rect. ABCD and P is a point inside it. PA, PB, PC and PD have been joined.

To prove :

(i) $ar(\triangle APB)+ar(\triangle PCD)=\frac{1}{2} ar($ rect. $ABCD)$

(ii) $ar(\triangle APD)+ar(\triangle BPC)=ar(\triangle APB)+ar(\triangle CPD)$.

Construction : Draw EPF $|| A B$ and LPM $|| A D$.

Proof : (i) EPF ||AB and DA cuts them,

$\therefore \quad \angle DEP=\angle EAB=90^{\circ} \quad$ [Corresponding angles]

$\therefore \quad PE \perp AD$.

Similarly, $PR \perp BC ; PL \perp AB$ and $PM \perp DC$.

$\therefore \quad ar(\triangle APD)+ar(\triangle BPC)$

$=\Big(\frac{1}{2} \times AD \times PE\Big)+ar\Big(\frac{1}{2} \times BC \times PF\Big)=\frac{1}{2} AD \times(PE+PF) \quad[\therefore BC=AD]$

$=\frac{1}{2} \times AD \times EF=\frac{1}{2} \times AD \times AB \quad[\therefore EF=AB]$

$=\frac{1}{2} \times($ rectangle $ABCD)$.

(ii) $ar(\triangle APB)+ar(PCD)$

$ \begin{matrix} =\Big(\frac{1}{2} \times AB \times PL\Big)+\Big(\frac{1}{2} \times DC \times PM\Big)=\frac{1}{2} \times AB \times(PL+PM) & {[\therefore EF=AB]} \\ \\ =\frac{1}{2} \times AB \times LM=\frac{1}{2} \times AB \times AD & {[\because LM=AD]} \\ \\ =\frac{1}{2} \times ar(\text { rect. } ABCD) & \end{matrix} $

$ar(\triangle APD) + ar(PBC) = ar(\triangle APB)+ ar(PCD)$

Hence Proved

Ex. 12 Diagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect each other at $P$. Show that:

$ ar(APB)+ar(CPD)=ar(APD) \times ar(BPC) $

Sol. Draw perpendiculars $AF$ and $CE$ on $BD$.

$ \begin{aligned} & ar(APB) \times ar(CPD)=\Big(\frac{1}{2} \times PB \times AF\Big) \times\Big(\frac{1}{2} \times PD \times CE\Big) \quad \ldots \text{(i)} \\ \\ & ar(APD) \times ar(BPC)=(\frac{1}{2} \times PD \times AF) \times(\frac{1}{2} \times BP \times CE) \quad \ldots \text{(ii)} \end{aligned} $

From above equations, we get $ar(APB) \times ar(CPD)=ar(APD) \times ar(BPC)$

Hence Proved.

$$EXERCISE$$

OBJECTIVE DPP - 13.1

~~ 1. The sides $BA$ and $DC$ of the parallelogram $ABCD$ are produced as shown in the figure then

(A) $a+x=b+y$ (B) $a+y=b+a$ (C) $a+b=x+y$ (D) $a-**b=x-y$

~~ 2. The sum of the interior angles of polygon is three times the sum of its exterior angles. Then numbers of sides in polygon is

(A) 6 (B) 7 (C) 8 (D) 9

~~ 3. In the adjoining figure, $AP$ and $BP$ are angle bisector of $\angle A$ and $\angle B$ which meet at a point $P$ of the parallelogram $ABCD$. Then $2 \angle APB=$

(A) $\angle A+\angle B$ (B) $\angle A+\angle C$ (C) $\angle B+\angle D$ (D) $\angle C+\angle D$

~~ 4. In a parallelogram the sum of the angle bisector of two adjacent angles is

(A) $30^{\circ}$ (B) $45^{\circ}$ (C) $60^{\circ}$ (D) $90^{\circ}$

~~ 5. In a parallelogram $ABCD \angle D=60^{\circ}$ then the measurement of $\angle A$

(A) $120^{\circ}$ (B) $65^{0}$ (C) $90^{\circ}$ (D) $75^{0}$

~~ 6. In the adjoining figure $A B C D$, the angles $x$ and $y$ are

(A) $60^{\circ}, 30^{\circ}$ (B) $30^{\circ}, 60^{\circ}$ (C) $45^{\circ}, 45^{0}$ (D) $90^{\circ}, 90^{\circ}$

~~ 7. From the figure parallelogram $PQRS$, the values of $\angle SQP$ and $\angle QSP$ are are

(A) $45^{0}, 60^{\circ}$ (B) $60^{\circ}, 45^{0}$ (C) $70^{0}, 35^{0}$ (D) $35^{\circ}, 70$

~~ 8. In parallelogram $A B C D, A B=12 cm$. The altitudes corresponding to the sides $A B$ and $A D$ are respectively 9 $cm$ and $11 cm$. Find AD.

(A) $\frac{108}{11} cm$ (B) $\frac{108}{10} cm$ (C) $\frac{99}{10} cm$ (D) $\frac{108}{17} cm$

~~ 9. In $\triangle ABC, AD$ is a median and $P$ is a point is $AD$ such that $AP: PD=1: 2$ then the area of $\triangle ABP=$

(A) $\frac{1}{2} \times$ Area of $\triangle A B C$ (B) $\frac{2}{3} \times$ Area of $\triangle ABC$ (C) $\frac{1}{3} \times$ Area of $\triangle ABC$ (D) $\frac{1}{6} \times$ Area of $\triangle ABC$

~~ 10. In $\triangle ABC$ if $D$ is a point in $BC$ and divides it the ratio $3: 5$ i.e., if $BD: DC=3: 5$ then, $ar(\triangle ADC): ar(\triangle ABC)$ $=?$

(A) $3: 5$ (B) $3: 8$ (C) $5: 8$ (D) $8: 3$

SUBJECTIVE DPP - 13.2

~~ 1. If each diagonal of a quadrilateral separates into two triangles of equal area, then show that the quadrilateral is a parallelogram.

~~ 2. In the adjoining figure, $PQRS$ and $PABC$ are two parallelograms of equal area. Prove that $QC || BR$.

~~ 3. In the figure $ABCD$ is rectangle inscribed in a quadrant of a circle of radius $10 cm$. If $AD=2 \sqrt{5} cm$. Find the area of the rectangle.

~~ 4. $\quad P$ and $Q$ are any two points lying on the sides $DC$ and $AD$ respectively of parallelogram $ABCD$. Prove that $: ar(\triangle APB)=ar(\triangle BQC)$.

~~ 5. In the figure, given alongside, PQRS and ABRS are parallelograms and $X$ is any point on side BR. Prove that

(i) $ar(PQRS)=ar(ABRS)$

(ii) $ar(AXS)=\frac{1}{2} ar(PQRS)$

~~ 6. Find the area a rhombus, the lengths of whose diagonals are $16 cm$ and $24 cm$ respectively.

~~ 7. Find the area of trapezium whose parallel sides are $8 cm$ and $6 cm$ respectively and the distance between these sides is $8 cm$.

~~ 8. (i) Calculate the area of quad. $ABCD$, given in fig. (i)

(ii) Calculate the area of trap. PQRS, given in fig. (ii).

~~ 9. In figure, $A B C D$ is a trapezium in which $A B || D C ; A B=7 cm ; A D=B C=5 cm$ and the distance between $AB$ and $DC$ is $4 cm$.

Find the length of DC and hence, find the area of trap. ABCD.

~~ 10. $BD$ is one of the diagonals of quadrilateral $ABCD$. If $AL \perp BD$ and $CM \perp BD$, show that : ar(quadrilateral $ABCD)=\frac{1}{2} \times BC \times(AL+CM)$

~~ 11. In the figure, $ABCD$ is a quadrilateral in which diag. $BD=20 cm$. If $AL \perp BD$ and $CM \perp BD$, such that : $AL=10 cm$ and $CM=5 cm$, find the area of quadrilateral $ABCD$.

~~ 12. In fig. $A B C D$ is a trapezium in which $A B || D C$ and $D C=40 cm$ and $A B=60 cm$. If $X$ and $Y$ are, respectively, the mid - points of $AD$ and $BC$, prove that

(i) $X Y=50 cm$

(ii) DCYX is a trapezium

(iii) Area (trapezium DCYX) $=\frac{9}{11}$ Area (trapezium XYBA)

~~ 13. Show that a median of a triangle divides it into two triangles of equal area.

~~ 14. In the figure, given alongside, $D$ and $E$ are two points on $BC$ such that $BD=DE=EC$. Prove that : $ar(ABD)$ $=ar(ADE)=ar(AEC)$

~~ 15. In triangle $ABC$, if a point $D$ divides $BC$ in the ratio $2: 5$, show that $: ar(\triangle ABD): ar(\triangle ACD)=2: 5$.

$$\text{ANSWER KEY}$$

(Objective DPP # 13.1)

Qus.	1	2	3	4	5	6	7	8	9	10
Ans.	$C$	$C$	$D$	$D$	$A$	$A$	$A$	$A$	$D$	$C$

(Subjective DPP # 13.2)

~~ 3. $40 cm^{2}$

~~ 6. $\quad 192 cm^{2}$

~~ 7. $56 cm^{2}$

~~ 8. (i) $114 cm^{2}$ (ii) $195 cm^{2}$

~~ 9. $40 cm^{2}$

~~ 11. $\quad 150 cm^{2}$

$\ggg$

CIRCLE

$\lll$

ML - 14

DEFINITIONS

(A) Circle :

The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle.

The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle.

In figure, $O$ is the centre and the length $OP$ is the radius of the circle. So the line segment joining the centre and any point on the circle is called a radius of the circle.

(b) Interior and Exterior of a Circle :

A circle divides the plane on which it lies into three parts. They are

(i) inside the circle (or interior of the circle)

(ii) the circle nd

(iii) outside the circle (or exterior of the circle.)

The circle and its interior make up the circular region.

(c) Chord :

If we take two points $P$ and $Q$ on a circle, then the line segment PQ is called a chord of the circle.

(d) Diameter:

The chord which passes through the centre of the circle, is called a diameter of the circle.

A diameter is the longest chord and all diameter have the same length, which is equal to two times the radius. In figure, $AOB$ is a diameter of circle.

(e) Arc :

A piece of a circle between two points is called an arc. If we look at the pieces of the circle between two points $P$ and $Q$ in figure, we find that there are two pieces, one longer and the other smaller. The longer one is called the major arc PQ and the shorter one is called the minor arc PQ. The minor arc PQ is also denoted by PQ and the major arc PQ by PRQ, where $R$ is some point on the arc between $P$ and $Q$. Unless otherwise states, arc PQ or PQ stands for minor arc PQ. When $P$ and $Q$ are ends of a diameter, then both arcs are equal and each is called a semi circle.

(f) Circumference:

The length of the complete circle is called its circumference.

(g) Segment :

The region between a chord and either of its arcs is called a segment of the circular region or simply a segment of the circle. There are two types of segments also, which are the major segment and the minor segment (as in figure).

(h) Sector :

The region between an arc and the two radii, joining the centre to the end points of the arc is called a sector.

Like segments, we find that the minor arc corresponds to the minor sector and the major arc corresponds to the major sector. In figure, the region OPQ in the minor sector and the remaining part of the circular region is the major sector. When two arcs are equal, then both segments and both sectors become the same and each is known as a semicircular region.

Theorem-1 : Equal chords of a circle subtend equal angles at the centre.

Given : $A B$ and $C D$ are the two equal chords of a circle with centre $O$.

To prove : $\angle AOB=\angle COD$.

Proof : In $\triangle AOB$ and $\triangle COD$,

$ \begin{aligned} & OA=OC, & \text { [Radii of a circle] } \\ & OB=OD, & \text { [Radii of a circle] } \\ & AB=CD, & \text { [Given] } \\ & \therefore \quad \Delta AOB \cong \Delta COD, \text { [By SSS] } \\ & \therefore \quad \angle AOB=\angle COD \text { [By cpctc] } \end{aligned} $

Converse of above Theorem :

In the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.

Given : $\angle AOB$ and $\angle POQ$ are two equal angles subtended by chords $AB$ and $PQ$ of a circle at its centre $O$.

To prove : $AB=PQ$

Proof : In $\triangle AOB$ and $\triangle POQ$,

$ \begin{aligned} OA & =OP & & \text { [Radii of a circle] } \\ OB & =OQ & & \text { [Radii of a circle] } \\ \angle AOB & =\angle POQ & & \text { [Given] } \\ \therefore \Delta AOB & \approx \Delta POQ & & {[\text { By SAS }] } \end{aligned} $

$\therefore \quad AB=PQ$ [By cpctc]

Hence Proved.

Theorem-2 : The perpendicular from the centre of a circle to a chord bisects the chord.

Given : A circle with centre $O$. $AB$ is a chord of this circle. $OM \perp AB$.

To prove : $MA=MB$.

Construction : Join $OA$ and $OB$.

Proof : In right triangles OMA and OMB,

$OA = OB \quad \text{[Radii of a circle]}$

$OM = OM \quad \text{[Common]}$

$\therefore \angle OMA = \angle OMB \quad \text{[By RHS]}$

$\therefore MA = MB \quad \text{[By cpctc]}$

Hence Proved.

Converse of above Theorem :

The line drawn through the centre of a circle to bisect a chord a perpendicular to the chord.

Given : A circle with centre $O . AB$ is a chord of this circle whose mid-point is $M$.

To prove : $OM \perp AB$.

Construction : Join OA and $OB$.

Proof : In $\triangle OMA$ and $\triangle OMB$.

$MA=MB \quad \text{[Given]}$

$OM=OM \quad \text{[Common]}$

$OA=OB \quad \text{[Radii of a circle]}$

$\therefore \quad \triangle OMA \cong \triangle OMB \quad \text{[By SSS]}$

$\therefore \quad \angle AMO=\angle BMO \quad \text{[By cpctc]}$

But $\angle AMO+\angle BMO=180^{\circ} \quad \text{[Linear pair axiom]}$

$\therefore \angle AMO=\angle BMO=90^{\circ}$

[Linear pair axiom]

$\Rightarrow OM \perp AB$.

Theorem-3 : There is one and only one circle passing through three given non-collinear points.

Proof : Let us take three points A, B and C, which are not on the same line, or in other words, they are not collinear [as in figure]. Draw perpendicular bisectors of $AB$ and $BC$ say, $PQ$ and $RS$ respectively. Let these perpendicular bistros intersect at one point $O$. (Note that $PQ$ and RS will intersect because they are not parallel) [as in figure].

$\therefore \quad O$ lies on the perpendicular bisector $PQ$ of $AB$.

$\therefore OA=OB$

[ $\because$ Every point on the perpendicular bisector of a line segment is equidistant from its end points]

Similarly,

$\therefore \quad O$ lies on the perpendicular bisector $RS$ of $BC$.

$\therefore OB=OC$

[ $\because$ Every point on the perpendicular bisector of a line segment is equidistant from its end points]

So, $OA=OB=OC$

i.e., the points $A, B$ and $C$ are at equal distances from the point $O$.

So, if we draw a circle with centre $O$ and radius $OA$ it will also pass through $B$ and $C$. This shows that there is a circle passing through the three points $A, B$ and $C$. We know that two lines (perpendicular bisectors) can intersect at only one point, so we can draw only one circle with radius $OA$. In other words, there is a unique circle passing through $A, B$ and $C$.

Hence Proved.

REMARK :

If $ABC$ is a triangle, then by above theorem, there is a unique circle passing through the three vertices $A, B$ and $C$ of the triangle. This circle the circumcircle of the $\triangle A B C$. Its centre and radius are called respectively the circumcentre and the circumradius of the triangle.

Ex. 1 In figure, $A B=C B$ and $O$ is the centre of the circle. Prove that $B O$ bisects $\angle A B C$.

Sol. Given : In figure, $A B=C B$ and $O$ is the centre of the circle.

To prove : BO bisects $\angle A B C$.

Construction : Join OA and OC.

Proof : In $\triangle OAB$ and $\triangle OCB$,

$ OA =OC \text { [Radii of the same circle] } $

$AB =CB \text { [Given] } $

$OB =OB \text { [Common] } $

$\therefore \quad \Delta OAB \cong \Delta OCB \text { [By SSS] } $

$\therefore \quad \angle ABO =\angle CBO \text { [By cpctc] } $

$\Rightarrow \quad \text { BO bisects } \angle ABC .$

Hence Proved.

Ex. 2 Two circles with centres $A$ and $B$ intersect at $C$ and $D$. Prove that $\angle A C B=\angle A D B$.

Sol. Given : Two circles with centres $A$ and $B$ intersect at $C$ and $D$.

To prove : $\angle A C B=\angle A D B$.

Construction : Join $AC, AD, BC, BD$ and $AB$.

Proof : In $\triangle ACB$ an $\triangle ADB$,

$ \begin{aligned} & AC=AD \quad[\text { Radii of the same circle }] \\ & BC=BD \quad[\text { Radii of the same circle }] \\ & AB=AB \quad[\text { Common }] \\ & \therefore \quad \triangle ACB \cong \triangle ADB \quad \text { [By SSS] } \\ & \therefore \quad \angle ACB=\angle ADB . \quad \text { [By cpctc] } \end{aligned} $

Ex. 3 In figure, $AB \cong ACand O$ is the centre of the circle. Prove that $OA$ is the perpendicular bisector of $BC$.

Sol. Given : In figure, $AB \cong AC$ and $O$ is the centre of the circle.

To prove : OA is the perpendicular bisector of BC.

Construction : Join OB and OC.

Proof :

$\therefore AB \cong AC \quad \text{[Given]}$

$\therefore \quad$ chord $AB=$ chord $AC$.

[ $\because$ If two arcs of a circle are congruent, then their corresponding chords are equal.]

$\therefore \angle AOB=\angle AOC \quad$….(i) $\quad[\because$ Equal chords of a circle subtend equal angles at the centre $]$

In $\triangle OBC$ and $\triangle OCD$,

$\angle DOB=\angle DOC \quad[$ From (1)]

$OB=OC \quad \text{[Radii of the same circle]}$

$OD=OD \quad \text{[Common]}$

$\therefore \quad \triangle OBD \cong \triangle OCD \quad \text{[By SAS]}$

$\therefore \angle ODB=\angle ODC$

And $\quad BD=CD \quad \text{….(ii) [By cpctc]}$

But $\angle BDC=180^{\circ}$

$\therefore \quad \angle ODB+\angle ODC=180^{\circ}$

$\Rightarrow \angle ODB+\angle ODB=180^{\circ} \quad \text{[From equation (ii)]}$

$\Rightarrow 2 \angle ODB=180^{\circ}$

$\Rightarrow \angle ODB=90^{\circ}$

$\therefore \quad \angle ODB=\angle ODC=90^{\circ} \quad \ldots \text{(iv)} \quad \text{[From (ii)]}$

So, by (iii) and (iv), OA is the perpendicular bisector of BC.

Ex. 4 Prove that the line joining the mid-points of the two parallel chords of a circle passes through the centre of the circle.

Sol. Let $A B$ and $C D$ be two parallel chords of a circle whose centre is $O$.

Let 1 and $M$ be the mid-points of the chords $AB$ and $CD$ respectively. Join PL and $OM$.

Draw $\quad$ OX $|| AB$ or $CD$.

$\therefore \quad L$ is the mid-point of the chord $AB$ and $O$ is the centre of the circle

$\therefore \angle OLB=90^{\circ}$

$[\because$ The perpendicular drawn from the centre of a circle to chord bisects the chord]

But, OX $|| AB$

$\therefore \angle LOX=90^{\circ} \quad \ldots \text{(i)}$.

[ $\because$ Sum of the consecutive interior angles on the same side of a transversal is $180^{\circ}$ ]

$\therefore \quad M$ is the mid-point of the chord $CD$ and $O$ is the centre of the circle.

$\therefore \angle OMD=90^{\circ}$

$[\because$ The perpendicular drawn from the centre of a circle to a chord bisects the chord]

But OX $|| CD \quad \ldots \text{(ii)}$

[ $\because$ Sum of the consecutive interior angles on the same side of a transversal is $180^{\circ}$ ]

$\therefore \angle MOX=90^{\circ}$

From above equations, we get

$ \begin{aligned} & \angle LOX+\angle MOX=90^{\circ}+90^{\circ}=180^{\circ} \\ \Rightarrow & \angle LOM=180^{\circ} \end{aligned} $

$\Rightarrow$ LM is a straight line passing through the centre of the circle.

Hence Proved.

Ex.5 $ \quad \ell$ is a line which intersects two concentric circle (i.e., circles with the same centre) with common centre $O$ at $A, B, C$ and $D$ (as in figure). Prove that $AB=CD$.

Sol. Given : $\ell$ is a line which intersects two concentric circles (i.e., circles with the same centre) with common centre $O$ at $A, B, C$ and $D$.

To prove : $A B=C D$.

Construction : Draw OE $\perp \ell$

Proof :

$\therefore$ The perpendicular drawn from the centre of a circle to a chord bisects the chord

$\therefore \quad AE=ED \quad \ldots \text{(i)}$

And $\quad BE=EC \quad \ldots \text{(i)}$

Subtracting (ii) from (i), we get

$AE-BE=ED-EC$

$\Rightarrow AB=CD$.

Hence Proved.

Ex. 6 PQ and RS are two parallel chords of a circle whose centre is $O$ and radius is $10 cm$. If $PQ=16 cm$ and $RS=$ $12 cm$, find the distance between $P Q$ and RS, if they lie.

(i) on the same side of the centre $O$.

(ii) on opposite sides of the centre $O$.

Sol. (i) Draw the perpendicular bisectors $OL$ and $OM$ of $PQ$ and RS respectively.

$\therefore \quad$ PQ $|| RS$

$\therefore \quad OL$ and $OM$ are in the same line.

$\Rightarrow O, L$ and $M$ are collinear.

Join OP and OR.

In right triangle OLP,

$OP^{2}=OL^{2}+PL^{2} \quad \text{[[By Pythagoras Theorem]]}$

$\Rightarrow(10)^{2}=OL^{2}+\Big(\frac{1}{2} \times pq\Big)^{2}$

[ $\therefore$ The perpendicular drawn from the centre of a circle to a chord bisects the chord]

$\Rightarrow 100=OL^{2}+\Big(\frac{1}{2} \times 16\Big)^{2}$

$\Rightarrow \quad 100=OL^{2}+(8)^{2}$

$\Rightarrow 100=OL^{2}+64$

$\Rightarrow OL^{2}=100-64$

$\Rightarrow OL^{2}=36=(6)^{2}$

$\Rightarrow OL=6 cm$

In right triangle $OMR$,

$OR^{2}=OM^{2}+RM^{2} \quad \text{[By Pythagoras Theorem]}$

$\Rightarrow OR^{2}=OM^{2}+\Big(\frac{1}{2} \times RS\Big)^{2}$

$[\because$ The perpendicular drawn from the centre of a circle to a chord bisects the chord]

$\Rightarrow(10)^{2}=OM^{2}+\Big(\frac{1}{2} \times 12\Big)^{2}$

$\Rightarrow(10)^{2}=OM^{2}+(6)^{2}$

$\Rightarrow OM^{2}=(10)^{2}-(6)^{2}=(10-6)(10+6)=(4)(16)=64=(8)^{2}$

$\Rightarrow OM=8 cm$

$\therefore \quad LM=OM-OL=8-6=2 cm$

Hence, the distance between $PQ$ and $RS$, if they lie on he same side of the centre $O$, is $2 cm$.

(ii) Draw the perpendicular bisectors $OL$ and $OM$ of $PQ$ and RS respectively.

$\therefore \quad PQ || RS$

$\therefore \quad OL$ and $OM$ are in the same line

$\Rightarrow L, O$ and $M$ are collinear.

Join OP nd OR.

In right triangle $O L P$,

$OP^{2}=OL^{2}+PL^{2}$

$\Rightarrow OP^{2}=OL^{2}+(\frac{1}{2} \times pQ)^{2} \quad \text{[By Pythagoras Theorem]}$

$[\because$ The perpendicular drawn from the centre of a circle to a chord bisects the chord]

$\Rightarrow(10))^{2}=OL^{2}+(\frac{1}{2} \times 16)^{2}$

$\Rightarrow 100=OL^{2}+(8)^{2}$

$\Rightarrow 100=OL^{2}+64$

$\Rightarrow OL^{2}=100-64$

$\Rightarrow OL^{2}=36=(6)^{2}$

$\Rightarrow OL=6 cm$

In right triangle $OMR$,

$OR^{2}=OM^{2}+RM^{2} \quad[$ By Pythagoras Theorem $]$

$\Rightarrow OR^{2}=OM^{2}+(\frac{1}{2} \times 12)^{2}$

$[\because$ The perpendicular drawn from the centre of a circle to a chord bisects the chord]

$\Rightarrow(10)^{2}=OM^{2}+(\frac{1}{2} \times RS)^{2}$

$\Rightarrow(10)^{2}=OM^{2}+(6)^{2}$

$\Rightarrow OM^{2}=(10)^{2}-(6)^{2}=(10-6)(10+6)=(4)(16)=64=(8)^{2}$

$\Rightarrow OM=8 cm$

$\therefore LM=OL+OM=6+8=14 cm$

Hence, the distance between PQ and RS, if they lie on the opposite side of the centre $O$, is $14 cm$.

Theorem-4 : Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

Given : A circle have two equal chords $AB \And CD$. e. $AB=CD$ and $OM \perp AB, ON \perp CD$

To prove : $OM=ON$

Construction : Join OB \And OD

Proof : $AB=CD($ Given)

[ $\because$ The perpendicular drawn from the centre of a circle to bisect the chord.]

$\therefore \quad \frac{1}{2} AB=\frac{1}{2} CD$

$\Rightarrow BM=DN$

In $\triangle OMB \And \Delta OND$

$\angle OMB=\angle OND=90^{\circ} \quad \text{[Given]}$

$OB=OD \quad \text{[Radii of same circle]}$

Side $BM=$ Side $DN \quad \text{[Proved above]}$

$\therefore \quad \triangle OMB \cong \triangle OND \quad \text{[By R.H.S.]}$

$\therefore OM=ON \quad \text{[By cpctc]}$

Hence Proved.

REMARK :

Chords equidistant from the centre of a circle are equal in length.

Ex. $7 \quad A B$ and $C D$ are equal chords of a circle whose centre is $O$. When produced, these chords meet at E. Prove that $EB=ED$.

Sol. Given : $AB$ and $CD$ are equal chords of a circle whose centre is $O$. When produced, these chords meet at $E$.

To prove : EB = ED.

Construction : From $O$ draw $OP \perp AB$ and $OQ \perp CD$. Join $OE$.

Proof : $\therefore AB=CD$ [Given]

$\therefore \quad OP=OQ$

[ $\therefore$ Equal chords of a circle are equidistant from the centre]

Now in right tingles $OPE$ and $OQE$,

$OE=OE \quad \text{[Common]}$

Side $OP=$ side $OQ \quad \text{[Proved above]}$

$\therefore \quad \triangle OPE \cong \triangle OQE \quad \text{[By RHS]}$

$\therefore OE=QE \quad \text{[By cpctc]}$

$\Rightarrow PE-\frac{1}{2} AB=QE-\frac{1}{2} CD \quad [\because AB = CD \text{(Given)}]$

$\Rightarrow \quad PE-PB=QE-QD$

$\Rightarrow EB=ED$.

Ex. 8 Bisector $AD$ of $\angle BAC$ of $\triangle ABC$ passed through the centre $O$ of the circumcircle of $\triangle ABC$. Prove that $AB=$ AC.

Sol. Given : Bisector $AD$ of $\angle BAC$ of $\triangle ABC$ passed through the centre $O$ of the circumcircle of $\triangle ABC$,

To prove : $A B=A C$.

Construction : Draw $OP \perp AB$ and $OQ \perp AC$.

Proof :

In $\triangle APO$ and $\triangle AQO$

$ \angle OPA=\angle OQA \quad[\text { Each }=90^{\circ} \text { (by construction) }] $

$ \begin{matrix} \angle OAP=\angle OAQ & {[\text { Given }]} \\ OA=OA & {[\text { Common }]} \end{matrix} $

$\therefore \quad \triangle APO \cong \triangle AQO \quad$ [By ASS cong. prog.]

$\therefore \quad OP=OQ \quad$ [By cpctc]

$\therefore \quad AB=AC . \quad[\because$ Chords equidistant from the centre are equal $] \quad$Hence Proved.

Ex. 9 $\quad AB$ and $CD$ are the chords of a circle whose centre is $O$. They intersect each other at $P$. If $PO$ be the bisector of $\angle APD$, prove that $AB=CD$.

In the given figure, $O$ is the centre of the circle and $PO$ bisect the angle $APD$. prove that $AB=CD$.

Sol. Given : $A B$ and $C D$ are the chords of a circle whose centre is $O$. They interest each other at P. PO is the bisector of $\angle A P D$.

To prove : $A B=C D$.

Construction : Draw $OR \perp AB$ and $OQ \perp CD$.

Proof : In $\triangle OPR$ and $\triangle OPQ$,

$ \angle OPR=\angle OPQ {[\text { Given }]} $\

$ OP=OP {[\text { Common }]} $

And $\quad \angle ORP=\angle OQP[.$ Each $.=90^{\circ}]$

$ \therefore \quad \Delta ORP \cong \triangle OPQ \quad \text { [By AAS] } $

$ \begin{matrix} \therefore & OR=OQ & {[\text { By cpctc }]} \\ \\ \therefore & AB=CD & {[\because \text { Chords of a circle which are equidistant from the centre are equal }]} \end{matrix} $

REMARK :

Angle Subtended by an Arc of a Circle :

In figure, the angle subtended by the minor arc $PQ$ at $O$ is $\angle POQ$ and the angle subtended by the major arc $PQ$ at $O$ is reflex angle $\angle POQ$.

$$EXERCISE$$

OBJECTIVE DPP # 14.1

~~ 1. If two circular wheels rotate on a horizontal road then locus of their centres will be

(A) Circles (B) Rectangle (C) Two straight line (D) Parallelogram

~~ 2. In a plane locus of a centre of circle of radius $r$, which passes through a fixed point

(A) rectangle (B) A circle (C) A straight line (D) Two straight line

~~ 3. In a circle of radius $10 cm$, the length of chord whose distance is $6 cm$ from the centre is

(A) $4 cm$ (B) $5 cm$ (C) $8 cm$ (D) $16 cm$

~~ 4. If a chord a length $8 cm$ is situated at a distance of $3 cm$ form centre, then the diameter of circle is :

(A) $11 cm$ (B) $10 m$ (C) $12 cm$ (D) $15 cm$

~~ 5. In a circle the lengths of chords which are situated at a equal distance from centre are :

(A) double (B) four times (C) equal (D) three times

SUBJECTIVE DPP # 14.2

~~ 1. The radius of a circle is $13 cm$ and the length of one of its chords is $10 cm$. Find the distance of the chord from the centre.

~~ 2. Show is the figure, $O$ is the centre of the circle of radius $5 cm$. $OP \perp AB, OQ \Re CD, AB || CD, AB=6 cm$ and $C D=8 cm$. Determine PQ.

~~ 3. $AB$ and $CD$ are two parallel chords of a circle such that $AB=10 cm$ and $CD 24 cm$. If the chords are on the opposite side of the centre and the distance between is $17 cm$, Find the radius of the circle.

~~ 4. In a circle of radius $5 cm, A B$ and $A C$ are two chords such that $A B=A C=6 cm$. Find the length of the chord BC.

~~ 5. $\quad A B$ and $C D$ are two parallel chords of a circle whose diameter is $A C$. Prove that $A B=C D$.

~~ 6. Two circles of radii $10 cm$ and $8 cm$ interest and the length of the common chord is $12 cm$. Find the distance between their centries.

~~ 7. Two circles with centre $A$ and $B$ and of radii $5 cm$ and $3 cm$ touch each other internally. If the perpendicular bisector of segment $A B$ meet the bibber circle is $P$ and $Q$, find the length of $P Q$.

$> > >$
CIRCLE
$< < <$

ML - 15

SOME IMPORTANT THEOREMS

Theorem-1 : Equal chords of a circle subtend equal angles at the centre.

Given : A circle with centre $O$ in which chord $PQ=$ chord $RS$.

To prove : $\angle POQ=\angle ROS$.

Proof : In $\triangle POQ$ and $\triangle ROS$,

$ \begin{aligned} & OP=OR \quad[\text { Radii of the same circle }] \\ & OQ=OS \quad[\text { Radii of the same circle }] \\ & PW=RS \quad \text { [Given] } \\ & \Rightarrow \Delta POQ=\Delta ROS \quad[By SSS] \\ & \Rightarrow \angle POQ=\angle ROS \quad[\text { By cpctc }] \end{aligned} $

Hence Proved.

Theorem-2 : If the angles subtended by the chords at the centre (of a circle) are equal then the chords are equal.

Given : A circle with centre $O$. Chords $PQ$ and RS subtend equal angles at the enter of the circle.

i.e. $\angle POQ=\angle ROS$

To prove : Chord $PQ=$ chord RS.

Proof : In $\triangle POQ$ and $\triangle ROS$,

$ \begin{matrix} \angle POQ=\angle ROS & {[\text { Given }]} \\ OP=OR & {[\text { Radii of the same circle }]} \\ OQ=OS & {[\text { Radii of the same circle }]} \end{matrix} $

$\Rightarrow \quad \Delta POQ \cong \Delta ROS \quad[By SSS]$

$\Rightarrow$ chord $PQ=$ chord $RS \quad[$ By cpctc]

Corollary-1 : Two arc of a circle are congruent, if the angles subtended by them at the centre are equal.

Corollary 2 : If two arcs of a circle are equal, they subtend equal angles at the centre.

Corollary 3 : If two arc of a circle are congruent (equal), their corresponding chords are equal.

Corollary 4: If two chords of a circle are equal, their corresponding arc are also equal.

$ \angle A O B=\angle C O D $

$\therefore$ Chord $AB=$ Chord $CD$

$\therefore \quad$ Arc APB $=$ Arc COD.

Hence Proved.

Theorem-3 : The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Given : An arc PQ of a circle subtending angles POQ at the centre $O$ and PAQ at a point $A$ on the remaining part of the circle.

To prove : $\angle POQ=2 \angle PAQ$.

Construction : Join $AO$ and extend it to a point $B$.

(A)

(B)

(C)

Proof: There arises three cases :

(A) are $PQ$ is minor

(B) arc PQ s a semi - circle

In all the cases,

$ \angle BOQ=\angle OAQ+\angle AQO \quad \ldots \text{(i)} $

[ $\because$ An exterior angle of triangle is equal to the sum of the two interior opposite angles]

In $OAQ$,

$ \begin{aligned} & OA=OQ \quad \text{[Radii of a circle]}\\ \therefore & \angle OAQ=\angle OQA \quad \ldots \text{(ii)} \text{[Angles opposite equal of a triangle are equal]} \end{aligned} $

(i) and (ii), give,

$ \angle BOQ=2 \angle OAQ \quad \ldots \text{(iii)} $

Similarly,

$ \angle BOP=2 \angle OAP \quad \ldots \text{(iv)} $

Adding (iii) and (iv), we get

$ \begin{aligned} & \angle BOP+\angle BOQ=2(\angle OAP+\angle OAQ) \\ \Rightarrow & \angle POQ=2 \angle PA \quad \ldots \text{(v)} \end{aligned} $

NOTE : For the case $(C)$, where $PQ$ is the major arc, $(v)$ is replaced by reflex angles.

Thus, $\angle POQ=2 \angle PAQ$.

Theorem- 4 : Angles in the same segment of a circle are equal.

Proof : Let $P$ and $Q$ be any two points on a circle to form a chord $PQ, A$ and $C$ any other points on the remaining part of the circle and $O$ be the centre of the circle. Then,

$ \angle POQ=2 \angle PAQ \quad \ldots \text{(i)} $

And $\quad \angle POQ=2 \angle PCQ \quad \ldots \text{(ii)}$

From above equations, we get

$2 \angle PAQ=2 \angle PCQ$

$\Rightarrow \angle PAQ=\angle PCQ$

Hence Proved

Theorem-5 : Angle in the semicircle is a right angle.

Proof : $\angle PAQ$ is an angle in the segment, which is a semicircle.

$\therefore \angle PAQ=\frac{1}{2} \angle PAO=\frac{1}{2} \times 180^{\circ}=90^{\circ}$

[ $\therefore \angle PQR$ is straight line angle or $\angle PQR=180^{\circ}$ ]

If we take any other point $C$ on the semicircle, then again we get

$ \angle PCQ=\frac{1}{2} \angle POQ=\frac{1}{2} \times 180^{\circ}=90^{\circ} $

Hence Proved.

Theorem-6: If a line segment joining two points subtend equal angles at two other points lying on the same side of the lien containing the line segment the four points lie on a circle (i.e., they are concyclic).

Given : $AB$ is a line segment, which subtends equal angles at two points $C$ and $D$. i.e., $\angle ACB=\angle ADB$.

To prove : The points A, B, C and D lie on a circle.

Proof : Let us draw a circle through the points A, C and B.

Suppose it does not pass through the point $D$.

Then it will intersect $A D$ (or extended $A D$ ) at a point, say $E$ (or $E^{\prime}$ ).

If points $A, C, E$ and $B$ lie on a circle,

But it is given that $ \angle ACD=\angle AEB$

$\quad[\therefore \text { Angles in the same segment of circle are equal }]$

Therefore, $\angle ACB=\angle ADB$

$\angle AEB=\angle ADB$

This is possible only when $E$ coincides with $D$. [As otherwise $\angle AEB>\angle ADB$ ]

Similarly, E’ should also coincide with D. So A, B, C and D are concyclic

Hence Proved.

CYCLIC QUADRILATERAL

A quadrilateral $ABCD$ is called cyclic if all the four vertices of it lie on a circle.

Theorem-7 : The sum of either pair of opposite angles of a cyclic quadrilateral is $180^{\circ}$

Given : A cyclic quadrilateral $A B C D$.

To prove : $\angle A+\angle C=\angle B+\angle D=180^{\circ}$

Construction : Join $AC$ and $BD$.

Proof : $\angle ACB=\angle ADB \quad \text{[Angles of same segment]}$

$ \text { And } \quad \angle BAC=\angle BDC \quad \text { [Angles of same segment] } $

$\therefore \quad \angle ACB+\angle BAC=\angle ADB+\angle BDC=\angle ADC$.

Adding $\angle ABC$ to both sides, we get

$ \angle ACB+\angle BAC+\angle ABC=\angle ADC+\angle ABC $

The left side being the sum of three angles of $\triangle ABC$ is equal to $180^{\circ}$.

$\therefore \quad \angle ADC+\angle ABC=180^{\circ}$

i.e., $\angle D+\angle B=180^{\circ}$

$\therefore \quad \angle A+\angle C=360^{\circ} \quad-(\angle B+\angle D)=180^{\circ} \quad[\therefore \angle A+\angle B+\angle C+\angle D=360^{\circ}] \quad$Hence Proved.

Corollary : If the sum of a pair of opposite angles of a quadrilateral is $180^{\circ}$, then quadrilateral is cyclic.

Ex. 1 In figure, $\angle ABC=69^{\circ}, \angle ACB=31^{\circ}$, find $\angle BDC$.

Sol. In $\triangle ABC$.

$ \angle BAC+\angle ABC+\angle ACB=180^{\circ} $

[Sum of all the angles of a triangle is $180^{\circ}$ ]

$ \Rightarrow \angle B A C+69^{0}+31^{0}=180^{\circ} $

$ \Rightarrow \angle B A C+100^{\circ}=180^{\circ} $

$\Rightarrow \angle B A C=180^{\circ}-100^{\circ}=80^{\circ}$

Now, $\angle B D C=\angle B A C=80^{\circ}$Ans. [Angles in the same segment of a circle are equal]

Ex. 2 $\quad ABCD$ is a cyclic quadrilateral whose diagonals intersect at a point $E$. If $\angle DBC=70^{\circ}, \angle BAC$ is $30^{\circ}$, find $\angle B C D$. Further, if $B=B C$, find $\angle E C D$.

Sol. $ \angle CDB =\angle BAC=30^{\circ} \quad \text{…(i) [Angles in the same segment of a circle are equal]} $

$ \angle DBC =70^{\circ} \quad \ldots \text{(ii)}$

In $\triangle BCD$,

$ \begin{aligned} & \angle BCD+\angle DBC+\angle CDB=180^{\circ} \quad[\text { Sum of all he angles of a triangle is } 180^{\circ}] \\ & \Rightarrow \quad \angle BCD+70^{\circ}+30.0=180^{\circ} \quad[Using(i) \text { and (ii) } \\ & \Rightarrow \angle B C D+100^{\circ}=180^{\circ} \\ & \Rightarrow \quad \angle BCD=180^{\circ}-100^{\circ} \\ & \Rightarrow \quad \angle BCD=80^{\circ} \quad \ldots \text{(iii)} \end{aligned} $

In $\triangle ABC,$

AB = BC

$\therefore \angle BCA = \angle BAC = 30 \degree \quad \text{(iv)} \quad \text{[Angles opposite to equal sides of a triangle are equal]}$

Now, $\angle BCD = 80 \degree \quad \text{[From (iii)]}$

$\Rightarrow \angle BCA + \angle ECD = 80 \degree$

$\Rightarrow 30 \degree + \angle ECD = 80 \degree$

$\Rightarrow \angle ECD = 80 \degree - 30 \degree$

$\Rightarrow \angle ECD = 50 \degree$

Ex. 3 If the nonparallel side of a trapezium are equal, prove that it is cyclic.

Sol. Given : ABCD is a trapezium whose two non-parallel sides $A B$ and $B C$ are equal.

To prove : Trapezium $A B C D$ is a cyclic.

Construction : Draw BE $|| A D$. Proof : $\therefore AB || DE \quad \text{[Given]}$

$AD || BE \quad \text{[By construction]}$

$\therefore$ Quadrilateral $ABCD$ is a parallelogram.

$\therefore \quad \angle BAD=\angle BED \quad \text{….(i) [Opp. angles of a $||^{gm}$]}$

And, $AD=BE \quad \text{….(ii) [Opp. sides of a $|| s m]$}$

But $AD=BC \quad \text{…(iii) [Given]}$

From (ii) and (iii),

$ \begin{aligned} & BE=BC \\ \\ & \therefore \quad \angle BEC=\angle BCE \quad \text {….(iv) [Angles opposite to equal sides] } \\ \\ & \angle BEC+\angle BED=180^{\circ} \quad[\text { Linear Pair Axiom }] \\ \\ & \Rightarrow \angle B C E+\angle B A D=180^{\circ} \quad[\text { From (iv) and (i)] } \end{aligned} $

[ $\therefore$ If a pair of opposite angles of a quadrilateral $180^{\circ}$, then the quadrilateral is cyclic]Hence Proved.

Ex. 4 Prove that a cyclic parallelogram is a rectangle.

Sol. Given : ABCD is a cyclic parallelogram.

To prove : $A B C D$ is a rectangle.

Proof : $\therefore$ ABCD is a cyclic quadrilateral

$\therefore \quad \angle 1+\angle 2=180^{\circ}$

$[\therefore$ Opposite angles of a cyclic quadrilateral are supplementary]

$\therefore \quad ABCD$ is a parallelogram

$\therefore \quad \angle 1=\angle 2 \quad \text{…(ii) [Opp. angles of a $||^{gm}$]}$

From (i) and (ii),

$ \angle 1=\angle 2=90^{\circ} $

$\therefore \quad ||^{gm} ABCD$ is a rectangle.

Hence Proved.

Ex. 5 Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.

Prove that the angles of the triangle DEF are $90^{\circ}-\frac{1}{2} A, 90^{\circ}-\frac{1}{2} B$ and $90^{\circ}-\frac{\angle C}{2}$.

Sol. Given : Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.

To prove : The angles of the $\triangle DEF$ are $90^{\circ}-\frac{\angle A}{2}, 90^{\circ}-\frac{\angle B}{2}$ and $90^{\circ}-\frac{C}{2}$ respectively.

Construction : Join DE, EF and FD.

Proof: $\angle FDE = \angle FDA + \angle EDA = \angle FCA + \angle EBA \quad [\because \text{Angles in the same segment are equal}]$

$ =\frac{1}{2} \angle C+\frac{1}{2} \angle B $

$ \Rightarrow \quad \angle D =\frac{\angle C+\angle B}{2}=\frac{180^{\circ}-\angle A}{2} {[\therefore \text { In } \triangle ABC, \angle A+\angle B+\angle C=180^{\circ}]} $

$\Rightarrow \angle D = 90 \degree - \frac{\angle A}{2}$

Similarly, we can show that

$ \angle E=90^{\circ}-\frac{\angle B}{2} $

And $\angle F=90^{\circ}-\frac{\angle C}{2}$.

Hence Proved.

Ex. 6 Find the area of a triangle, the radius of whose circumcircle is $3 cm$ and the length of the altitude drawn from the opposite vertex to the hypotenuse is $2 cm$.

Sol. We know that the hypotenuse of a right angled triangle is the diameter of its circumcircle.

$\therefore \quad BC=2 OB=2 \times 3=6 cm$

Let, $AD \perp BC$

$AD=2 cm \quad \text{[Given]}$

$\therefore \quad$ Area of $\triangle ABC=\frac{1}{2}(BC)(AD)$

$ \begin{aligned} & =\frac{1}{2}(6)(2) \\ & =6 cm^{2} \end{aligned} $

Ans.

Ex. 7 In figure, $PQ$ is a diameter of a circle with centre O. IF $\angle PQR=65^{\circ}, \angle SPR=40^{\circ}, \angle PQ M=50^{\circ}$, find $\angle QPR$, $\angle PRS$ and $\angle QPM$.

Sol. (i) $\angle QPR$

$\therefore \quad PQ$ is a diameter

$\therefore \angle PRQ = 90 \degree \quad \text{[Angle in a semi-circle is 90 \degree]}$

In $\triangle PQR$,

$ \angle QPR+\angle PRQ+\angle PQR=180^{\circ} \quad \text { [Angle Sum Property of a triangle] } $

$\Rightarrow \quad \angle QPR+90^{\circ}+65^{\circ}=180^{\circ}$

$\Rightarrow \angle QPR+155^{\circ}=180^{\circ}$

$\Rightarrow \angle QPR=180^{\circ}-155^{\circ}$

$\Rightarrow \angle Q P R=25^{\circ}$.

(ii) $\angle PRS$

$\therefore \quad$ PQRS is a cyclic quadrilateral

$\therefore \angle PSR + \angle PQR = 180 \degree \quad [\because \text{Opposite angles of a cyclic quadrilateral are supplementary}]$

$\Rightarrow \angle PSR + 65 \degree = 180 \degree$

$\Rightarrow \angle PSR = 180 \degree - 65 \degree$

$\Rightarrow \angle PSR = 115 \degree$

In $\triangle P S R$,

$\angle PSR+\angle SPR+\angle PRS=180 \circ \quad [\text { Angles Sum Property of a triangle}] $

$\Rightarrow 115^{\circ}+40^{\circ}+\angle PRS=180^{\circ}$

$\Rightarrow 115^{\circ}+\angle PRS=180^{\circ}$

$\Rightarrow \angle PRS=180^{\circ}-155^{\circ}$

$\Rightarrow \angle PRS=25^{\circ}$

(iii) $\angle QPM$

$\therefore \quad PQ$ is a diameter

$\therefore \angle PMQ=90^{\circ} \quad[\because.$ Angle in a semi - circle is $.90^{\circ}]$

In $\triangle PMQ$,

$\angle PMQ+\angle PQM+\angle QPM=180^{\circ} \quad$ [Angle sum Property of a triangle]

$\Rightarrow 90^{\circ}+50^{\circ}+\angle QPM=180^{\circ}$

$\Rightarrow 140^{\circ}+\angle QPM=180^{\circ}$

$\Rightarrow \angle QPM=180^{\circ}-140^{\circ}$

$\Rightarrow \angle QPM=40^{\circ}$.

Ex. 8 In figure, $O$ is the centre of the circle. Prove that

$\angle x+\angle y=\angle z$.

Sol. $\angle EBF=\frac{1}{2} \angle EOF=\frac{1}{2} \angle z \quad[\because$ Angle subtended by an arc of a circle at the centre in twice the angle subtended by it at any point of the remaining part of the circle] $\therefore \quad \angle ABF=180^{\circ}-\frac{1}{2} \angle z$ [Linear Pair Axiom] $\angle EDF=\frac{1}{2} \angle EOF=\frac{1}{2} \angle z$

[ $\because$ Angle subtend by any arc of a circle at the centre is twice the angle subtended by it at any point of the remaining part of the circle]

$ \begin{aligned} & \therefore \quad \angle ADE=180^{\circ}-\frac{1}{2} \angle z \\ & \angle BCD=\angle ECF=\angle y \\ & \angle BAD=\angle x \end{aligned} $

In quadrilateral $ABCD$

$ \begin{aligned} & \angle ABC+\angle BCD+\angle CDA+\angle BAD=2 \times 180^{\circ}[\text { Angle Sum Property of a quadrilateral] } \\ \Rightarrow \quad & 180^{\circ}-\frac{1}{2} \angle z+\angle y+180^{\circ}-\frac{1}{2} \angle z+\angle x=2 \times 180^{\circ} \end{aligned} $

$ \Rightarrow \quad \angle x+\angle y=\angle z $

Hence Proved.

Ex. $9 \quad AB$ is a diameter of the circle with centre $O$ and chord $CD$ is equal to radius $OC, AC$ and $BD$ produced meet at $P$. Prove that $\angle CPD=60^{\circ}$.

Sol. Given : $A B$ is a diameter of the circle with centre $O$ and chord $C D$ is equal to radius $O C$. $A C$ and $B D$ produced meet at $P$.

To prove : $\angle CPD=60^{\circ}$

Construction : Join AD.

Proof : In $\triangle O C D$,

$ \begin{aligned} & O C=O D \\ & O C=C D \end{aligned} $

From (i) and (ii),

$ OC=OD=CD $

[Radii of the same circle] [Given]

$\therefore \quad \triangle OCD$ is equilateral

$\therefore \quad \angle C O D=60^{\circ}$

$\therefore \quad \angle CAD=\frac{1}{2} \angle COD=\frac{1}{2} \angle(60^{\circ})=30^{\circ}$

[ $\because$ Angle subtended by any arc of a circle at the centre is twice the angle subtended by it at any point of the reaming part of the circle]

$\Rightarrow \angle PAD=30^{\circ} \quad \ldots \text{(iii)}$

And, $\angle ADB=90^{\circ} \quad \ldots \text{(iv)} \quad \text{[Angle in a semi-circle]}$

$\Rightarrow \angle ADB+\angle ADP=180^{0} \quad \text{[Linear Pair Axiom]}$

$\Rightarrow 90^{\circ}+\angle ADP=180^{\circ} \quad \text{[From (iv)]}$

$\Rightarrow \angle ADP=90^{\circ} \quad \ldots \text{(v)}$

In $\triangle ADP$,

$ \begin{aligned} & \angle ADP+\angle PAD+\angle ADP=180^{\circ} \quad[\because \text { The sum of the three angles of a triangles is } 180^{\circ}] \\ \Rightarrow & \angle APD+30^{\circ}+90^{\circ}=180^{\circ} \quad[\text { From (iii) and (v) }] \\ \Rightarrow & \angle APD+120^{\circ}=180^{\circ} \\ \Rightarrow & \angle APD=180^{\circ}-120^{\circ}=60^{\circ} \\ \Rightarrow & \angle CPD=60^{\circ} . \end{aligned} $

Ex. 10 Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.

Sol. Given : $A B C D$ is a cyclic quadrilateral. Its angle bisectors from a quadrilateral PQRS.

To prove : $PQRS$ is a cyclic quadrilateral.

$ \text {Proof :} \angle 1+\angle 2+\angle 3=180^{\circ} \quad \text {…(i) } \quad[\because \text { Sum of the angles of a } \triangle \text { is } 180^{\circ}] $

$ \angle 4+\angle 5+\angle 6=180^{\circ} \quad \text {…(ii) } $

$ \therefore \quad \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6=360 \text {.. ….(iii) \newline [Adding (i) and (ii)] } $

$ \text { But } \quad \angle 2+\angle 3+\angle 6+\angle 5=\frac{1}{2}[\angle A+\angle B+\angle C+\angle D] $

$ =\frac{1}{2} \cdot 360^{\circ}=180^{\circ} \quad[\because \text { Sum of the angles of quadrilateral is } 360^{\circ}] $

$ \therefore \quad \angle 1+\angle 4=360^{\circ}-(\angle 2+\angle 3+\angle 6+\angle 5) $

[ $\because$ If the sum of any pair of opposite angles of a quadrilateral is $180^{\circ}$, then the quadrilateral is a cyclic]Hence Proved.

Ex. 11 Prove that the angle bisectors of the angles formed by producing opposite sides of a cyclic quadrilateral (Provided they are not parallel) intersect a right angle.

Sol. Given : $A B C D$ is a cyclic quadrilateral. Its opposite sides $D A$ and $C B$ are produced to meet at $P$ and opposite sides $AB$ and $DC$ are produced to meet at $Q$. The bisectors of $\angle P$ and $\angle Q$ meet is $F$.

To prove : $\angle PFQ=90^{\circ}$.

Construction : Produce PF to meet DC is G.

Proof : In $\triangle PEB$,

$ \angle 5=\angle 2+\angle 6 \quad \ldots \text{(i)} $

[ $\because$ Exterior angle of a triangle is equal to the sum of interior opposite angles]

But $\angle 2=\angle 1$

And, $\angle 6=\angle D$ $\quad$ [ $ \because$ In a cyclic quadrilateral, exterior angle $=$ interior opposite angle] [From (i)]

$\therefore \quad \angle 5=\angle 1+\angle D \quad \ldots \text{(ii)} \text{[From (i)]}$

Now in $\triangle PDG$,

$ \angle 7=\angle 1+\angle D \quad \ldots \text{(iii)} $

[ $\because$ Exterior angle of a triangle is equal to the sum of interior opposite angles]

Frim (ii) and (iii), we have

$ \angle 5=\angle 7 $

Now, in $\Delta$ QEF and $\Delta$ QGF, $\quad \text{[Proved above]}$

$\angle 5=\angle 7 \quad \text{[Common side]}$

$QF=QF \quad \text{[Given]}$

$\angle 3=\angle 4 \quad \text{[AAS criterion]}$

$\therefore \quad \Delta QEF \cong \Delta QGE \quad \text{[By cpctc]}$

$\therefore \quad \angle 8=\angle 9$

But $\angle 8+\angle 9=180^{\circ}$

$\therefore \angle 8=\angle 9=90^{\circ} \quad \text{[Linear Pair Axiom]}$

$\therefore \angle PFQ=90^{\circ}$

Hence Proved.

Ex. 12 Two concentric circles with centre O have A, B, C, D as the points of intersection with the line $\ell$ as shown in the figure. If $A D=12 cm$ and $B C=8 cm$, find the length of $A B, C D, A C$ and $B D$.

Sol. Since $OM \perp BC$, a chord of the circle,

$\therefore$ is bisects $B C$.

$\therefore \quad BM=CM=\frac{1}{2}(BC)=\frac{1}{2}(8)=4 cm$

Since, $OM \perp AD$, a chord of the circle,

$\therefore$ it bisects AD.

$\therefore \quad AM=AD=\frac{1}{2} AD=\frac{1}{2}(8)=4 cm$

Since, $OM \perp CD$, a chord of the circle,

$\therefore$ it bisects AD.

$\therefore \quad AM=MD=\frac{1}{2} AD=\frac{1}{2}(12)=6 cm$

Now, $\quad A B=A M-B M=6-4=2 cm$

$CD=MD-MD=6-4=2 cm$

$AC=AM+MC=6+4=10 cm$

$BD=BM+MD=4+6=10 cm$

Ex. 13 $OABC$ is a rhombus whose three vertices, $A B$ and $C$ lie on a circle with centre $O$. If the radius of the circle is $10 cm$. Find the area of the rhombus.

Sol. Since $O A B C$ is a rhombus

$\therefore \quad OA=AB=BC=OC=10 cm$

Now, $OD \perp BC \Rightarrow CD=\frac{1}{2} BC=\frac{1}{2}(10)=5 cm$

$\therefore \quad$ By Pythagoras theorem,

$OC^{2}=OD^{2}+DC^{2}$

$\Rightarrow OD^{2}=OC^{2}-DC^{2}=(10)^{2}-(5)^{2}=100-25=75$

$\Rightarrow OD=\sqrt{75}=5 \sqrt{3}$

$\therefore \quad$ Area $(\triangle OBC)=\frac{1}{2} BC \times OD=\frac{1}{2}(10) \times 5 \sqrt{3}=25 \sqrt{3} sq . cm$.

Ex. 14 Chords $AB$ and $CD$ of a circle with centre $O$, intersect at a point $E$. If $OE$ objects $\angle AED$. Prove that $AB=CD$.

Sol. In $\triangle OLE$ and $\triangle OME$

$\angle OLE=\angle OME$ $\quad [90^{0}.$ each $]$

$\angle LEO=\angle MEO \quad \text{[Given]}$

And $OE=OE \quad \text{[[Common]]}$

$\therefore \quad \triangle OLE \cong \triangle OME \quad \text{[By AAS Criteria]}$

$\Rightarrow OL=OM \quad \text{[By cpctc]}$

This chords $AB$ and $CD$ are equidistant from centre. But we know that only equal chords are equidistant from centre.

$\Rightarrow AB=DC$

Ex. 15 In the given figure. $A B$ is the chord of a circle with centre $O . A B$ is produced to $C$ such that $B C=O B$. $C O$ is joined and produced to meet the circle in $D$. If $\angle ACD=y^{0}$ and $\angle AOD=x^{0}$, prove that $x^{0}=3 y^{0}$.

Sol. Since $BC=OB$

[Given]

$ \begin{aligned} \therefore \quad \angle OCB & =\angle BOC=y^{0} \quad[\because \text { Angles opposite to equal sides are equal }] \\ \\ \angle OBA & =\angle BOC+\angle OCB=y^{0}+y^{0}=2 y^{0} . \end{aligned} $

[ $\because$ Exterior angle of a $\Delta$ is equal to the sum of the opposite interior angles]

$ \begin{matrix} \text { Also } & \begin{matrix} OA=OB & \text { [Radii of the same circle] } \\ \\ \angle OAB=\angle OBA=2 y^{0} & \text { [Angles opposite to equal sides of a triangle are equal] } \\ \\ \angle AOD=\angle OAC+\angle OCA & \\ \\ =2 y^{0}+y^{0}=3 y^{0} & \end{matrix} \end{matrix} $

[ $\because$ Exterior angle of a $\Delta$ is equal to the sum of the opposite interior angles]

Hence $x^{0}=3 y^{0}$

Hence Proved.

Ex. 16 In the given figure, the chord $ED$ is parallel to the diameter $AC$. Find $\angle CED$.

Sol.

$\angle CBE = \angle 1 \quad$ $\text{[ $\angle$ s in the same segment]}$

$\angle 1 = 50 \degree \quad \ldots \text{(i)} \quad \text{[$\because$ $\angle$ CBE = 50 \degree]}$

$\angle AEC = 90 \degree \quad \ldots \text{(ii)} \quad \text{[Semicircle Angle is right angle ]}$

Now, in $\triangle AEC$,

$ \begin{aligned} & \angle 1+\angle AEC+\angle 2=180^{\circ} \quad \text{[$\because$ Sum of angles of a $\triangle$ = 180 $\degree$]}\\ & \therefore \quad 50^{\circ}+90^{\circ}+\angle 2=180^{\circ} \\ & \Rightarrow \quad \angle 2=180^{\circ}-140^{\circ}=40^{\circ} \end{aligned} $

Thus $\angle 2 = 40 \degree \quad \ldots \text{(iii)}$

Also, $ED || AC \quad \text{[Given]}$

$\therefore \angle @ = \angle 3 \quad \text{[Alternate angles]}$

$\therefore 40 \degree = \angle i.e., \angle 3 = 40 \degree$

Hence $\angle CED = 40 \degree$Ans

Ex. 17 ABCD is a parallelogram. The circle through A, B, C intersects CD (produced if necessary) at E. Prove that $AD=AE$.

Sol. Given : $A B C D$ is a parallelogram. The circle through $A, B, C$ intersects $C D$, when produced in $E$.

To prove : $A E=A D$.

Proof : Since $A B C E$ is a cyclic quadrilateral

$\therefore \quad \angle 1+\angle 2=180^{\circ} \quad$….(i) [opposite angles of a cyclio quadrilateral are supplementary]

Also $\angle 3+\angle 4=180^{\circ}$ [linear pair] $\quad \ldots \text{(ii)}$

From (i) and (ii), we get $\angle 1+\angle 2=\angle 3+\angle 4 \quad \ldots \text{(iii)}$

But $\angle 2=\angle 3 \quad \ldots \text{(iv)}$

$\therefore \quad$ From (iii) and (iv), we get $\angle 1=\angle 4$

Now in $\triangle ADE$, since $\angle 1=\angle 4$

$AD=AE \quad \text{[Sides opp. to equal angles of a triangle are equal]}$

Hence Proved.

$$EXERCISE$$

OBJECTIVE DPP # 15.1

~~ 1. I the given circle $ABCD, O$ is the centre and $\angle BDE=42^{\circ}$. The $\angle ACB$ is equal to :

(A) $48^{0}$

(B) $45^{0}$

~~ 2. In the diagram, $O$ is the centre of the circle. The angles $CBD$ is equal to :

(A) $25^{0}$

(B) $50^{\circ}$

(D) $130^{\circ}$

~~ 3. In the given figure, $\angle CAB=80^{\circ}, \angle ABC=40^{\circ}$. The sum of $\angle DAB+\angle ABD$ is equal to :

(A) $80^{\circ}$

(B) $100^{\circ}$

(D) $140^{\circ}$

~~ 4. In the given figure, if $C$ is the centre of the circle and $\angle PC=25^{\circ}$ and $\angle PRC=15^{\circ}$, then $\angle QCR$ is equal to :

(A) $40^{0}$

(B) $60^{0}$

(D) $120^{\circ}$

~~ 5. In a cyclic quadrilateral if $\angle B-\angle D=60^{\circ}$, then the smaller of the angles $B$ and $D$ is :

(A) $30^{\circ}$ (B) $45^{0}$ (C) $60^{\circ}$ (D) $75^{0}$

~~ 6. Three wires of length $\ell_1, \ell_2, \ell_3$ from a triangle surmounted by another circular wire, If $\ell_3$ is the diameter and $\ell_3=2 \ell_1$, then the angle between $\ell_1$ and $\ell_3$ will be

(A) $30^{0}$ (B) $60^{0}$ (C) $45^{0}$ (D) $90^{\circ}$

~~ 7. In a circle with centre $O, OD \perp$ chord $AB$. If $BC$ is the diameter, then :

(A) $AC=BC$ (B) $OD=BC$ (C) $AC=2 OD$ (D) None of these

~~ 8. In the diagram two equal circles of radius $4 cm$ intersect each other such that each passes through the centre of the other. Find the length of the common chord.

(A) $2 \sqrt{3} cm$

(B) $4 \sqrt{3} cm$

(D) $8 cm$

~~ 9. The sides $AB$ and $DC$ of cyclic quadrilateral $ABCD$ are produced to meet at $P$, the sides $AD$ and $BC$ are produced to meet at $Q$. If $\angle ADC=85^{\circ}$ and $\angle BPC=40^{\circ}$, then $\angle CQD$ equals.

(A) $30^{\circ}$ (B) $45^{0}$ (C) $60^{\circ}$ (D) $75^{0}$

~~ 10. In the given figure, if $\angle ACB=40^{\circ}, \angle DPB=120^{\circ}$, then will be :

(A) $40^{\circ}$

(B) $20^{\circ}$

(D) $60^{\circ}$

~~ 11. Any cyclic parallelogram is a.

(A) rectangle (B) rhombus (C) trapezium (D) square

~~ 12. The locus of the centre of all circles of given radius $r$, in the same planes, passing through a fixed point is :

(A) A point (B) A circle (C) A straight line (D) Two straight lines

~~ 13. In a cyclic quadrilateral if $\angle A-\angle C=70^{\circ}$, then the greater of the angles $A$ and $C$ is equal to :

(A) $95^{\circ}$ (B) $105^{\circ}$ (C) $125^{0}$ (D) $115^{0}$

~~ 14. The length of a chord a circle is equal to the radius of the circle. The angle which this chord subtends on the longer segment of the circle is equal to :

(A) $30^{\circ}$

(B) $45^{\circ}$

(D) $90^{\circ}$

~~ 15. If a trapezium is cyclic then,

(A) Its parallel sides are equal.

(B) Its non-parallel sides are equal.

(D) None of these above

SUBJECTIVE DPP - 15.2

~~ 1. In the given figure, $BC$ is diameter bisecting $\angle ACD$, find the values of $a, b$ (o is centre of circle).

~~ 2. In the given figure, find the value of $a \And b$.

~~ 3. Find the value of .

~~ 4. Find the value of $a \And b$.

~~ 5. Prove that $a+2 b=90^{\circ}$

~~ 6. $ABCD$ is a cyclic quadrilateral in which $\angle A=(x+y+10)^{0}, \angle B=(y+20)^{0}, \angle C=(x+y-30)^{0}$ and $\angle D=(x+$ $y)^{0}$. Find $x$ and $y$.

~~ 7. Find the value of $a$ and $b$, if $b=2 a$.

~~ 8. Find the value of a if BC $||$ EA

~~ 9. In the adjoining fig., $O$ is centre of the circle, chord $AC$ and $BD$ are perpendicular to each other, $\angle OAB=a$ and $\angle DBC=b$. Show that $a=b$.

~~ 10. In the fig. given below, $AB$ is diameter of the circle whose centre is $O$. Given that : $\angle ECD=\angle EDC=32^{\circ}$. Show that $\angle COF=\angle CEF$.

~~ 11. In the given fig., $AC$ is the diameter of circle centre $O$. Chord $BD$ is perpendicular to $AB$. Write down the angles $p, q \And r$ in terms of $x$.

~~ 12. Prove that the line segment joining the mid-point of hypotenuse of a right triangle to its opposite vertex is half of the hypotenuse.

(Objective DPP # 14.1)

Qus.	1	2	3	4	5
Ans.	C	B	D	B	C

(Subjective DPP # 14.2)

~~ 1. $12 cm$

~~ 2. $\quad 7 cm$

~~ 3. $13 cm$

~~ 4. $\quad 9.6 cm$

~~ 5. $\quad 10 cm$

~~ 6. $\quad 13.29 cm$

~~ 7. $4 \sqrt{6} cm$

(Objective DPP # 15.1)

Qus.	1	2	3	4	5	6	7	8	9	10
Ans.	A	A	C	C	C	B	C	B	A	B
A	11	12	13	14	15
Ans.	A	B	C	A	B

(Subjective DPP # 15.2)

~~ 1. $b=90^{\circ}, a=45^{\circ}$

~~ 2. $a=5^{0}, b=170^{0}$

~~ 3. $a=140^{\circ}, b=70^{\circ}$

~~ 4. $a=40^{\circ}, b=90^{\circ}$

~~ 5. $x=40, y=60$

~~ 6. $a=40^{\circ}, b=80^{\circ}$

~~ 7. $\quad a=108^{0}$

~~ 8. $p=90^{\circ}-\frac{x}{2}, q=\frac{x}{2}$, and $r=90-\frac{x}{2}$

$> > >$
CONSTRCUTIONS
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ML - 16

TO CONSTRUCT THE BISECTOR OF A LINE SEGMENT

Ex. 1Draw a line segment of length $7.8 cm$ draw the perpendicular bisector of this line segment.

Sol. Given the given the segment be $AB=7.8 cm$.

STEPS :

(i) Draw the line segment $AB=7.8 cm$.

(ii) With point $A$ as centre and a suitable radius, more than half the length of $A B$, draw arcs on both the sides of $AB$.

(iii) With point $B$ as centre and with the same radius draw arcs on both the sides of AB. Let these arc cut at points $P \And Q$ as shown on in the figure.

(iv) Draw a line through the points $P$ and $Q$. The line so obtained is the required perpendicular bisector of given line segment $A B$.

Line $P Q$ is perpendicular bisector of $A B$.

(A) $P Q$ bisects $A B$ i.e., $O A=O B$.

(B) $PQ$ is perpendicular to $AB$ i.e., $\angle PAO=\angle POB=90^{\circ}$.

Proof : In $\triangle APQ$ and $\triangle BPQ$ :

$ \begin{aligned} & AP=BP \quad \text{[By construction]}\\ & A Q=B Q \quad \text{[By construction]}\\ & P Q=P Q \quad \text{[Common]}\\ & \Rightarrow \quad \triangle APQ=\angle BPQ \quad \text{[By SSS]}\\ & \Rightarrow \quad \angle APQ=\angle BPQ \quad \text{[By cpctc]} \end{aligned} $

Now, in $\triangle APO \And \triangle BPO$

$AP=BP \quad \text{[By construction]}$

$OP=OP \quad \text{[Common side]}$

$\angle APO=\angle BPO \quad \text{[Proved above ]}$

$\Rightarrow \triangle APO \cong \triangle BPO \quad \text{[By SAS]}$

And, $\angle POA=\angle POB$

$ =\frac{180^{\circ}}{2}=90^{\circ} \quad[\because \angle POA+\angle POB=180^{\circ}] $

$\Rightarrow PQ$ is perpendicular bisector of $AB$.

TO CONSTRCUT THE BISECTOR OF A GIVEN ANGLE

Let $A B C$ be the given angle to be bisected.

STEPS :

(i) With $B$ as centre and a suitable radius, draw an arc which cuts ray $BA$ at point $D$ and ray $BC$ at point $E$.

(ii) Taking $D$ and $E$ as centres and with equal radii draw arcs which intersect each other at point $F$. In this step, each equal radius must be more than half the length $DE$.

(iii) Join $B$ and $F$ and produce to get the ray $BF$.

Ray $B F$ is the required bisector of the given angle $A B C$.

Proof : Join DF and EF.

In $\Delta BDF$ and $\Delta BEF$ :

$BD = BE \quad \text{[Radii of the same arc]}$

$DF = EF \quad \text{[Radii of the equal arcs]}$

$BF = BF \quad \text{[Common]}$

$\Rightarrow \triangle BDF \cong \triangle BEF \quad \text{[By SSS]}$

$\Rightarrow \angle DBF = \angle EBF \quad \text{[By cpctc]}$

$i.e., \angle ABF = \angle CBF$

$\Rightarrow BF \text{bisects} \angle ABC$

Hence Proved.

TO CONSTRUCT THE REQUIRED ANGLE

(a) To Construct the Required Angle of $60^{\circ}$ :

STEPS :

(i) Draw a line BC of any suitable length.

(ii) With $B$ as centre and any suitable radius, draw an arc which cuts $B C$ at point $D$.

(iii) With $D$ as centre and radius same, as taken in step (ii), draw one more arc which cuts previous arc at point $E$.

(iv) Join $BE$ and produce upto any point $A$.

Then, $\angle ABC=60^{\circ}$

(b) To Construct an Angle of $\mathbf{1 2 0}^{\circ}$ :

STEPS :

(i) Draw a line $BC$ of any suitable length.

(ii) Taking $B$ as centre and with any suitable radius, draw an arc which cuts $BC$ at point $D$.

(iii) Taking D as centre, draw an arc of the same radius, as taken in step (ii), which cuts the first arc at point E.

(iv) Taking E as centre and radius same, as taken in step (ii), draw one more arc which cuts the first arc at point $F$.

(v) Join BF and produce upto any suitable point A.

Then, $\angle ABC=120^{\circ}$

(c) To Construct and Angle of $30^{\circ}$ :

STEPS :

(i) Construct angle $ABC=60^{\circ}$ by compass.

(ii) Draw $BD$, the bisector of angle $ABC$.

The, $\angle DBC=30^{\circ}$

(d) To Construct an Angle of $90^{\circ}$ :

STEPS

(i) Construct angle $ABC=120.0$ by using compass.

(ii) Draw PB, the bisector of angle EBG.

Then, $\angle PBC=90^{\circ}$

Alternative Method :

(i) Draw a line segment $BC$ of any suitable length.

(ii) Produce $CB$ upto a arbitrary point $O$.

(iii) Taking $B$ as centre, draw as arc which cuts $OC$ at points $D$ and $E$.

(iv) Taking $D$ and $E$ as centres and with equal radii draw arcs with cut each other at point $P$.

[The radii in this step must be of length more than half of DE.]

(v) Join BP and produce.

Then, $\angle PBC=90^{\circ}$

(d) To Construct an Angle of $45^{\circ}$

STEPS :

(i) Draw $\angle PBC=90^{\circ}$

(ii) Draw $A B$ which bisects angle $P B C$,

Then, $\angle ABC=45^{\circ}$

Alternative Method :

STEPS :

(i) Construct $\angle ABC=60^{\circ}$

(ii) Draw $BD$, the bisector of angle $ABC$.

(iii) Draw $BE$, the bisector of angle $ABD$.

Then, $\angle EBC=45^{\circ}$

(e) To Construct an Angle of $105^{\circ}$ :

STEPS :

(i) Construct $\angle ABC=120^{\circ}$ and $\angle PBC=90^{\circ}$

(ii) Draw $BO$, the bisector of $\angle ABP$.

Then, $\angle OBC=105^{\circ}$

(f) To Construct an Angle of $150^{\circ}$.

STEPS :

(i) Draw line segment $BC$ of any suitable length. Produce $CB$ upto any point $O$.

(ii) With $B$ as centre, draw an arc (with any suitable radius) which buts $OC$ at points $D$ and $E$.

(iii) With $D$ as centre, draw an arc of the same radius, as taken in step 2, which cuts the first arc at point $F$.

(iv) With $F$ as centre, draw one more arc of the same radius, staken in step 2, which cuts the first arc at point G.

(v) Draw PB, the bisector of angle EBG.

Now $\angle FBD=\angle GBF=\angle EBG=60^{\circ}$

Then, $\angle PBC=150^{\circ}$

(g) To Construct an Angle of $135^{\circ}$.

STEPS :

(i) Construct $\angle PBC=150^{\circ}$ and $\angle GBC=120^{\circ}$

(ii) Construct $BQ$, the bisector of angle PBG.

Then, $\angle QBC=135^{\circ}$

TO CONSTRUCT A TRIANGLE

Case (i) To construct an equilateral triangle when its one side is given.

Ex. 2 Draw an equilateral triangle having each side of $2.5 cm$.

Sol. Given one side of the equilateral triangle be $2.5 cm$.

STEPS :

(i) Draw a line segment $BC=2.5 cm$.

(ii) Through $B$, construct ray $BP$ making angle $60^{\circ}$ with $BC$.

i.e. $\quad \angle PBC=60^{\circ}$

(iii) Through $C$, construct $C Q$ making angle $60^{\circ}$ with $B C$

i.e., $\angle Q C B=60^{\circ}$

(iv) Let $B P$ and $C Q$ intersect each other at point $A$.

The $n, \triangle ABC$ is the require equilateral triangle.

Proof: Since, $\angle A B C=\angle A C B=60^{\circ}$

$\therefore \quad \angle B A C=180^{\circ}-(60^{\circ}+60^{\circ})=60^{\circ}$

$\Rightarrow \quad$ All the angles of the $\triangle ABC$ drawn are equal.

$\Rightarrow \quad$ All the sides of the $\triangle ABC$ drawn are equal.

$\Rightarrow \quad \triangle ABC$ is the required equilateral triangle.

Hence Proved.

Alternate method:

If one side is $2.5 cm$, then each side of the required equilateral triangle is $2.5 cm$.

STEPS :

(i) Draw $BC=2.5 cm$

(ii) With $B$ as centre, draw an arc of radius $2.5 cm$

(iii) With $C$ as centre, draw an arc of radius $2.5 cm$

(iv) Let the two arc intersect each other at point $A$. Join $AB$ and $AC$.

Then, $ABC$ is the required equilateral triangle.

Case (ii) When the base of the triangle, one base angle and the sum of other two sides are given.

Ex. 3 Construct a triangle with $3 cm$ base and sum of other two sides is $8 cm$ and one base angle is $60^{\circ}$.

Sol. Given the base $BC$ of the triangle $ABC$ be $3 cm$, one base angle $\angle B=60^{\circ}$ and the sum of the other two sides be $8 cm$ i.e, $AB+AC=8 cm$.

STEPS :

(i) Draw $BC=3 cm$

(ii) At point $B$, draw $PB$ so that $\angle PBC=60^{\circ}$

(iii) From $B P$, cut $B C=8 cm$.

(iv) Join D and C.

(v) Draw perpendicular bisector of $C D$, which meets $B D$ at point $A$.

(vi) Join A and C.

Thus, $ABC$ is the required triangle.

Proof : Since, $OA$ is perpendicular bisector of $CD$

$ \Rightarrow \begin{gathered} OC=OD \\ \angle AOC=\angle AOD=90^{\circ} \end{gathered} $

Also, $\quad OA=OA \quad \text{[Common]}$

$\therefore \quad \triangle AOC \cong \triangle AOD \quad \text{[[By SAS]]}$

$\Rightarrow AC=AD$

$\therefore \quad BD=BA+AD$

$=BA+AC$

$=$ Given sum of the other two sides

Thus, base $BC$ and $\angle B$ are draw as given and $BD=AC$.Hence Proved.

Ex. 4 Construct a right triangle, when one side is $3.8 cm$ and the sum of the other side and hypotenuse is $6 cm$.

Sol. Here, if we consider the required triangle to be $\triangle ABC$, as shown alongside.

Clearly, $AB=3.8 cm, \angle B=90^{\circ}$ and $BC+AC=6 cm$.

STEPS :

(i) Draw $AB=3.8 cm$

(ii) Through $B$, draw line $BP$ so that $\angle ABP=90^{\circ}$

(iii) From BP, cut $B D=6 cm$

(iv) Join A and D.

(v) Draw perpendicular bisector of $A D$, which meets $B D$ at point $C$.

Thus, $ABC$ is the required triangle.

Case (iii) When the base of the triangle, one base angle and the difference of the other two sides are given.

Ex. 5 Construct a triangle with base of $8 cm$ and difference between the length of other two sides is $3 cm$ and one base angle is $60^{\circ}$

Sol. Given the base $B C$ of the required triangle $A B C$ be $8 cm$ i.e., $B C=8 m$, base angle $B=60^{\circ}$ ant the difference between the lengths of other two sides $A B$ and $A C$ be $3 cm$.

i.e., $AB-AC=3 cm$ or $AC-AB=3 cm$.

(a) When AB-AC=3 cm i.e., AB>AC :

STEPS :

(i) Draw $BC=8 cm$

(ii) Through point $B$, draw $BP$ so that $\angle PBC=60^{\circ}$

(iii) From $BP$ cut $BD=3 cm$.

(iv) Join $D$ and $C$.

(v) Draw perpendicular bisector of $DC$; which meets $BP$ at point $A$.

(vi) Join A and C.

Thus, $\triangle B C$ is the required triangle.

Proof : Since OA is perpendicular bisector of $C D$

$\Rightarrow \quad OD=OC$

$ \angle AOD=\angle AOC=90^{\circ} $

And, $\quad OA=OA \quad \text{[Common]}$

$\therefore \quad \triangle AOD \cong \triangle AOC \quad \text{[By SAS]}$

$\Rightarrow \quad AD=AC \quad \text{[By cpctc]}$

Now, $\quad B D=B A-A D$

$=BA-AD$

$=BA-AC$

$=$ Given difference of other two sides.

Thus, the base $BC$ and $\angle B$ are drawn as given and $BD=BA-AC$.

(b) When A C-A B=3 cm i.e, A B<A C :

STEPS :

(i) Draw $BC=8 cm$

(ii) Through B, draw line BP so that angle $PBC=60^{\circ}$.

(iii) Produce BP backward upto a suitable point $Q$.

(iv) Fro $B Q$, cut $B D=3 cm$.

(v) Join D and C.

(vi) Draw perpendicular bisector of $DC$, which meets $BP$ at point $A$.

(vii) Join A and C.

Thus, $\triangle ABC$ is the required triangle.

Proof : Since, $O A$ is perpendicular bisector of $C D$

$\Rightarrow OD=OC$

$ \angle AOD=\angle AOC=90^{\circ} $

And $\quad OA=OA \quad \text{[Common]}$

$\therefore \quad \triangle AOD \cong \triangle AOC \quad \text{[By SAS]}$

$\Rightarrow AD=AC \quad \text{[By cpctc]}$

Now, $\quad B D=A D-A B$

$=AC-AB \quad [\because AD=AC]$

$=$ Given difference of other two sides.

Thus, the base $BC$ and $\angle B$ are drawn as given and $BD=AC-AB$.

Hence Proved.

Case (iv) When the perimeter of the triangle and both the base angles are

Given :

Ex. 6 Contruct a triangle $A B C$ with $A B+B C+C A=12 cm \angle B=45^{\circ}$ and $\angle C=60^{\circ}$

Sol. Given the perimeter of the triangle $A B C$ be $12 cm$ i.e., $A B+B C+C A=12 cm$ and both the base angles be $45^{\circ}$ and $60^{\circ}$ i.e., $\angle B=45^{\circ}$ and $\angle C=60^{\circ}$

STEPS :

(i) Draw a line segment $P Q=12 cm$

(ii) At $P$, construct line $PR$ so that $\angle RPO=45^{\circ}$ and at $Q$, construct a line $QS$ so that $\angle SQP=60^{\circ}$

(iii) Draw bisector of angles RPQ and SQP which meet each other at point $A$.

(iv) Draw perpendicular bisector of $AP$, which meets $PQ$ at point $B$.

(v) Draw perpendicular bisector of $A Q$, which meets $P Q$ at point $C$.

(vi) Join $A B$ and $A C$.

Thus, $ABC$ is the required triangle.

Proof : Since, $M B$ is perpendicular bisector of $A P$

$ \begin{aligned} \Rightarrow \quad \Delta QNC & \cong \Delta ANC & {[\text { By SAS }] } \\ PB & =AC & \end{aligned} $

Similarly, $N C$ is perpendicular bisector of $A Q$.

$ \begin{aligned} & \Rightarrow \quad \Delta QNC \cong \triangle ANC \quad \text { [By SAS] }\\ & \Rightarrow \quad C Q=A C \end{aligned} $

Now, $\quad PQ=PB+BC+CQ$

$=AB+BC+AC$

$=$ Given perimeter of the $\triangle ABC$ drawn.

$ \text { Also, } \quad \angle BPA=\angle BAP \quad[As \triangle PMB \cong \triangle AMB] $

$\therefore \quad \angle ABC=\angle BPA+\angle BAP[Ext$. angle of a triangle $=$ sum of two interior opposite angles $]$

$\angle ABC=\angle BPA+\angle BAP=2 \angle BPA=\angle RPB=\angle ACB$ [Given]

$\angle ACB=\angle CQA+\angle CQA$

$ \begin{aligned} & =2 \angle CQA \quad[\because \Delta QNC \cong \triangle ANC \therefore \angle CQA=\angle CAQ] \\ & =\angle SQC=\text { Given base angle } ACB . \end{aligned} $

Thus, given perimeter $=$ perimeter of $\triangle ABC$.

given one base angle $=$ angle $A B C$

and, given other base angle $=$ angle $ACB$.

Ex. 7 Construct and equilateral triangle if its altitude is $3.2 cm$.

Sol. Given In an equilateral $\triangle ABC$ an altitude $AD=3.2 cm$ Required to Construct an equilateral triangle $ABC$ from the given data.

STEPS :

(i) Draw a line $P Q$ and mark and point $D$ on it.

(ii) Construct a ray $D E$ perpendicular to $P Q$.

(iii) Cut off $DA=3.2 cm$ from $DE$.

(iv) Construct $\angle DAR=30^{\circ}=\Big(\frac{1}{2} \times 60^{0}\Big)$.

The ray $A R$ intersects $P Q$ at $B$.

(v) Cut off line segment $DC=BD$.

(vi) Join $A$ and $C$. We get the required $\triangle ABC$.

Ex. 7 Construct a right angled triangle whose hypotenuse measures $8 cm$ and one side is $6 cm$.

Sol. Given Hypotenuse $A C$ of a $\triangle A B C=8 cm$ and one side $A B=6 cm$.

Required To construct a right angled $\triangle ABC$ from the given data.

STEP.

(i) Draw a line segment $AC=8 cm$.

(ii) Mark the mid point $O$ of $AC$.

(iii) With $O$ as centre and radius $OA$, draw a semicircle on $AC$.

(iv) With $A$ as centre and radius equal to $6 cm$, draw an arc, cutting the semicircle a $B$.

(v) Jon $A$ and $B ; B$ and $C$. We get the required right angled triangle $ABC$.

Ex. 8 Construct a $\triangle ABC$ in which $BC=6.4 cm$, altitude from $A$ is $3.2 cm$ and the median bisecting $BC$ is $4 cm$.

Sol. Given : one side $BC=6.4 cm$, Altitude $AD=3.2$ am and the median $AL=4 cm$.

Required : To construct a $\triangle ABC$ form the given data.

STEP :

(i) Draw $BC=6.4 cm$

(ii) Bisect $BC$ at $L$.

(iii) Draw EF $|| BC$ at a distance $3.2 cm$ for $BC$.

(iv) With $L$ as centre and radius equal to $4 cm$, draw an arc, cutting $EF$ at $A$.

(v) Join $A$ and $B ; A$ and $C$. We get the required triangle $ABC$.

Ex. 9 Construct a $\triangle ABC$ in which $\angle B=30^{\circ}$ and $\angle C=60^{\circ}$ and the perpendicular from the vertex $A$ to the base $BC$ is $4.8 cm$.

Sol. Given : $\angle B=30^{\circ} \angle C=60^{\circ}$, length of perpendicular from vertex $A$ to be base $B C=4.8 cm$.

Required : To construct a $\triangle ABC$ from the given data. STEP :

(i) Draw any ray line $PQ$.

(ii) Take a point $B$ on line $PQ$ and construct $\angle QBR=30^{\circ}$

(iii) Draw a line $EF || PQ$ a distance of $4.8 cm$ from $PQ$,

cutting BR at A.

(iv) Construct an angle $\angle F A C=60^{\circ}$, cutting $P Q C$.

(v) Join $A$ and $C$. We get the required triangle $A B C$.

Ex. 10 Construct a triangle $A B C$, the lengths of whose medians are $6 cm, 7 cm$ and $6 cm$.

Sol. Given : Median $AD=6 cm$ Median $BE=7 cm$ Median $CF=6 cm$.

Required : To construct a $\triangle ABC$ from the given data.

STEP :

(i) Construct a $\triangle APQ$ with $AP=6 cm, PQ=7 cm$ and $AQ=6 cm$.

(ii) Draw the medians $AE$ and $PF$ of $\triangle APQ$ intersecting each other at $G$.

(iii) Produce $AE$ to $B$ such that $GE=EB$

(iv) Join $B$ and $Q$ and produce it to $C$, such that $B Q=Q C$

(v) Join $A$ and $C$. We get the required triangle $ABC$.

$$EXERCISE$$

SUBJECTIVE DPP # 16.1

For each angle, given below, make a separate construction. Draw a ray BC and an another ray BA so that the $\angle ABC$ is equal to :

~~ 1. $15^{0}$

~~ 2. $22 \frac{1^{0}}{2}$

~~ 3. $\quad 75^{0}$

~~ 4. $52 \frac{1^{0}}{2}$

~~ 5. $\quad 67 \frac{1^{0}}{2}$

~~ 6. $\quad 165^{0}$

~~ 7. $135^{0}$

~~ 8. Construct and equilateral triangle with side :

(i) $5 cm$ (ii) $\quad 5.4 cm$ (iii) $6.2 cm$

~~ 9. Construct a triangle $ABC$, in which :

(i) base $AB=5.4 cm, \angle B=45^{\circ}$ and $AC+BC=9 cm$.

(ii) base $BC=6 cm, \angle B=60^{\circ}$ and $AB+AC=9.6 cm$.

(iii) base $AC=5 cm, \angle C=90^{\circ}$ and $AB+BC=10.6 cm$.

~~ 10. Construct a right triangle, with base $=4 cm$ and the sum of the other side and hypotenuse $=9.4 cm$.

~~ 11. Construct a triangle $ABC$, in which :

(i) $BC=4.8 cm, \angle B=45^{\circ}$ and $AB-**AC=2.4 cm$.

(ii) $BC=4.8 cm, \angle B=45^{0}$ and $AC-AB=2.4 cm$.

(iii) $AB=5.3 cm, \angle A=60^{\circ}$ and $AC-BC=2 cm$.

(iv) $AB=5.3 cm, \angle A=60^{\circ}$ and $BC-AC=2 cm$.

~~ 12. Construct a triangle $A B C$, with :

(i) perimeter $=12 cm, \angle B=45^{\circ}$ and $\angle C=60^{\circ}$.

(ii) perimeter $=11.6 cm, \angle B=60^{\circ}$ and $\angle C=90^{\circ}$

(iii) perimeter $=11 cm, \angle A=60^{\circ}$ and $\angle C=45^{\circ}$.

(iv) perimeter $=10 cm, \angle B=\angle C=60^{\circ}$

~~ 13. Construct as equilateral triangle with perimeter $15.6 cm$.

~~ 14. Without finding the length of each side of the equilateral triangle construct it. If its perimeter is $16 cm$.

~~ 15. Construct an equilateral triangle whose altitude is $4.8 cm$.

~~ 16. Construct a $\triangle PQR$ in which base $QR=4 cm, \angle R=30^{\circ}$ and $PR-PQ=1.1 cm$.

~~ 17. Construct a $\triangle XYZ$ with perimeter $9.6 cm$ and base angle $30^{\circ}$ and $60^{\circ}$

~~ 18. Construct a $\triangle PQR$ in which $PQ=3.7 cm, QR=3.6 cm$ and median $PA=3.1 cm$.

~~ 19. Construct a $\triangle DEF$, the lengths of whose medians are $6 cm, 7 cm$ and $8 cm$.

~~ 20. Construct on equilateral triangle, one of whose altitudes measures $6.4 cm$.

$\ggg$
HERON’S FORMULA
$\lll$

ML-17

MENSURTION

A branch of mathematics which concerns itself measurement of lengths, areas and volumes of plane and solid figure is called Monsuration.

(a) Perimeter :

The perimeter of a plane figure is the length of its boundary. In case of a triangle or a polygon, the perimeter is the sum of the lengths of its sides.

(b) Units of Perimeter :

The unit of perimeter is the same as the unit of length i.e. centimetre $(cm)$, metre $(m)$, kilometere $(k m)$ etc.

$ \begin{aligned} 1 \text { centimetre }(cm) & =10 \text { milimetre }(mm) \\ 1 \text { decimetre }(dm) & =10 \text { centrimetre } \\ 1 \text { metre }(m) & =10 \text { decimetre } \\ & =100 \text { centimetre } \\ & =1000 \text { milimetre } \\ 1 \text { decamete }(\text { dam }) & =10 \text { metre } \\ & =1000 \text { centimetre } \\ 1 \text { hectometre }(hm) & =10 \text { decametre } \\ & =100 \text { metre } \\ 1 \text { kilometre }(km)= & 1000 \text { metre } \\ & =100 \text { decametre } \\ & =10 \text { hectometre } \end{aligned} $

AREA

The area of a plane figure is the measure of the surface enclosed by its boundary.

The area of a triangle or a polygon is the measure of the surface enclosed by its sides.

(a) Units of Area :

The various units of measuring area are, squire centimetre $(cm^{2})$, square metre $(m^{2}), 1$ hectare etc.

1 square centrimetre $(cm^{2})=1 cm \times 1 cm$.

$ =10 mm \times 10 mm=100 sq . mm . $

1 square decimetre $(cm^{2}) \quad=1 dm \times dm$

$ =10 cm \times 10 cm=100 sq . cm . $

1 square metre $(m^{2})$

$=1 m \times 1 m$

$=10 dm \times 10 dm=100 sq . dm$.

1 square decametre $(.$ dam $.^{2})=1$ dam $\times 1$ dam

$=10 m \times 10 m=100 sq . m$.

1 square hectometre $(hm^{2}) \quad=10$ dam $\times 10$ dam $=100 sq$. dam

(or 1 hectare) $\quad=1000 sq . m$.

1 square kilometre $(km^{2}) \quad=1 km \times 1 km$

$=10 hm \times 10 hm=100 sq . hm$.

(b) Heron’s formula :

In $\triangle A B C$ if sides of triangle $B C, C A, \And A B$ are $a, b, c$ respectively then perimeter $=2 s=a+b+c$

Area $=\sqrt{s(s-a)(s-b)(s-c)}$

(c) Perimeter and Area of a Triangle :

(i) Right - angled triangle :

For an right-angled triangle, let $b$ be the base, $h$ be the perpendicular and $d$ be the hypotenuse. Then

(A) Perimeter $=b+h+d$

(B) Area $=\frac{1}{2}($ Base $\times$ Height $)=\frac{1}{2} bh$

(ii) Isosceles right-angled triangle

For an isosceles right-angled triangle, let a bet the equal sides, then

(A) Hypotenuse $=\sqrt{a^{2}+a^{2}}=\sqrt{2} a$

(B) Perimeter $=2 a+\sqrt{2} a$

(iii) Equilateral triangle

For an equilateral triangle, let each side be $a$, and the height of the triangle is $h$, then

(A) $\angle A=\angle B=\angle C=60^{\circ}$

(B) $\angle BAD=\angle CAD=30^{\circ}$

(D) $BD=DC=\frac{a}{2}$

(E) $\Big(\frac{a}{2}\Big)^{2}+h^{2}=a^{2} \Rightarrow h^{2}=\frac{3 a^{2}}{4}$

$\therefore \quad$ Height $(h)=\frac{\sqrt{3}}{2} a$

(F) Area $=\frac{1}{2}($ Base $\times$ Height $)=\frac{1}{2} \times a \times \frac{\sqrt{3}}{2} a=\frac{\sqrt{3}}{4} a^{2}$

(G) Perimeter $=a+a+a=3 a$.

Ex. 1 The area of a triangle is $30 cm^{2}$. Find the base if the altitude exceeds the base by $7 cm$.

Sol. Let base $B C=x cm$ then altitude $=(x+7) cm$

Area of $\triangle ABC=\frac{1}{2} \times$ base $\times$ height

$\Rightarrow 30=\frac{1}{2}(x)(x+7)$

$\Rightarrow 60=x^{2}+7 x$

$\Rightarrow x^{2}+7 x-60=0$

$\Rightarrow x^{2}+12 x-5 x-60=0$

$\Rightarrow x(x+12)-5(x+12)=0$

$\Rightarrow(x-5)(x+12)=0$

$\Rightarrow x=5$ or $x=-12$

$\Rightarrow x=5 \quad[\because x \neq-12$

$\therefore \quad$ Base $(x)=5 cm$ and Altitude $=x+7=5+7=12 cm$.

Ans.

Ex. 2 The cost of turfing a triangle field at the rate of Rs. 45 per $100 m^{2}$ is Rs. 900 . Find the height, if double the base of the triangle is 5 times the height.

Sol. Let the height of triangular field be $h$ metres.

It is given that $2 \times$ (base) $=5 \times$ (Height)

$\therefore \quad$ Base $=\frac{5}{2} h$

Area $=\frac{1}{2} \times$ Base $\times$ Height

Area $=\frac{1}{2} \times \frac{5}{2} h \times h=\frac{5}{4} h^{2} \quad \ldots \text{(i)}$

$\therefore$ Cost of turfing the field is Rs. 45 per $100 m^{2}$

$\therefore \quad$ Area $=\frac{\text { Total cost }}{\text { Rate per sq. }}$

$=\frac{900}{45 / 100}$

$=\frac{9000}{45}$

$=2000 m^{2} \quad \ldots \text{(ii)}$

From (i) and (ii)

$ \begin{aligned} & \frac{5}{4} h^{2}=2000 \\ \Rightarrow & 5 h^{2}=8000 \\ \Rightarrow & h^{2}=1600 \\ \Rightarrow & h=40 m \end{aligned} $

$\therefore$ Height of the triangular field is $40 cm$.

Ans.

Ex. 3 From a point in the interior of an equilateral triangle, perpendicular drawn to the three sides are $8 cm, 10$ $cm$ and $11 cm$ respectively. Find the area of the triangle.

Sol. Let each side of the equilateral $\triangle ABC=x cm$,

From an interior point $O, OD, OE$ and $OF$ be drawn perpendicular to $BC, AC$ and $AB$ respectively. It is given that $OD=11 cm, OE=8 cm$ and $OF=10 cm$. Join $OA, OB$ and $OC$.

Area of $\triangle ABC=$ Area of $\triangle OBC+$ Area of $\triangle OCA+$ Area of $\triangle OAB$

$ \begin{aligned} & =\frac{1}{2} \cdot x \cdot 11+\frac{1}{2} \cdot x \cdot 8+\frac{1}{2} \cdot x \cdot 10 \\ & =\frac{29}{2} x^{2} \text { cm}^{2} \end{aligned} $

But, area of an equilateral triangle, whose ease side is $x$

$ =\frac{\sqrt{3}}{4} x^{2} cm^{2} $

Therefore, $\frac{\sqrt{3}}{4} x^{2}=\frac{29}{2} x$

$\therefore \quad x=\frac{4 \times 29}{2 \times \sqrt{3}}=\frac{58}{\sqrt{3}} cm$

$\therefore \quad$ Area of $\triangle ABC=\frac{29}{2} \times \frac{58}{\sqrt{3}} cm^{2}=\frac{841}{1.73} cm^{2}$

$\therefore \quad \text{Area of} \triangle ABC=486.1 cm^{2}$

Ans.

Ex. 4 The difference between the sides at right angles in a right-angled triangle is $14 cm$. The area of the triangle is $120 cm^{2}$. Calculate the perimeter of the triangle.

Sol. Let the sides containing the right angle be $x cm$ and $(x-14) cm$.

The, its area $=\Big[\frac{1}{2} \cdot x \cdot(x-14)\Big] cm^{2}$.

But, area $=120 cm^{2} \quad \text{[Given]}$

$\therefore \quad \frac{1}{2} x(x-14)=120$

$\Rightarrow x^{2}-14 x-240=0$

$\Rightarrow x^{2}-24 x+10 x-240=0$

$\Rightarrow x(x-24)+10(x-24)=0$

$\Rightarrow(x-24)(x+10)=0$

$\Rightarrow x=24 \quad$ [Neglecting $x=-10$ ]

$\therefore \quad$ one side $=24 cm$, other side $=(24-14) cm=10 cm$

$ \begin{aligned} \text { Hypotenuse } & =\sqrt{(24)^{2}+(10)^{2}} cm \\ & =\sqrt{576+100} cm \\ & =\sqrt{576} cm \\ & =26 cm . \end{aligned} $

$\therefore \quad$ Perimeter of the triangle $=(24+10+26) c m=60 cm$.

Ans.

Ex. 5 Find the percentage increase in the area of a triangle if its each side is doubled.

Sol. Let $a, b, c$ be the sides of the given triangle and $s$ be its semi-perimeter

$ \therefore \quad s=\frac{1}{2}(a+b+c) \quad \ldots \text{(i)} $

The sides of the new triangle are $2 a, 2 b$ and $2 c$.

Let s’ be its semi-perimeter.

$\therefore \quad s^{\prime} \quad=\frac{1}{2}(2 a+2 b+2 c)=a+b+c=2 s \quad \text{[Using (i)]}$

Let $\Delta=$ Area of given triangle

$ \Delta=\sqrt{s(s-a)(s-b)(s-c)} \quad \ldots \text{(ii)} $

And, $\Delta^{\prime}=$ Area of new triangle

$ \begin{aligned} \Delta^{\prime} & =\sqrt{s^{\prime}(s^{\prime}-2 a)(s^{\prime}-2 b(s^{\prime}-2 c..} \\ & =\sqrt{2 s(2 s-2)(2 s-2 b)(2 s-2 c)} \quad \text{[Using (i)]} \\ & =\sqrt{16 s(s-a)(s-b)(s-c} \\ \Delta^{\prime} & =4 \Delta \end{aligned} $

$\therefore \quad$ Increase in the area of the triangle $=\Delta^{\prime}-\Delta=4 \Delta-\Delta=3 \Delta$

$\therefore \quad %$ increase in area $=\Big(\frac{3 \Delta}{\Delta} \times 100\Big) %=300 % %$

Ans.

Ex. 6 An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring $20 cm, 50 cm$ and $50 cm$. How much cloth of each colour is required for the umbrella ?

Sol. The sides of a triangular piece are

$20 cm, 50 cm$ and $50 cm$

$ s=\frac{a+b+c}{2}=\frac{20+50+50}{2}=60 cm=60 cm $

Area of one triangular piece

$=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{60(60-20)(60-50)(60-50)}$

$=\sqrt{60 \times 40 \times 10 \times 10}=\sqrt{24000}$

$=200 \sqrt{6} cm^{2}$

Area of cloth of each colour for five triangular pieces $=5 \times 200 \sqrt{6}=1000 \sqrt{6} cm^{2}$

Ans.

$$EXERCISE$$

OBJECTIVE DPP 17.1

~~ 1. The area of the field ABGFEA is :

(A) $7225 m^{2}$

(B) $7230 m^{2}$

(D) $7240 m^{2}$

~~ 2. Area of shaded portion as shown in the figure :

(A) $12 m^{2}$

(B) $13 m^{2}$

(D) $15 m^{2}$

~~ 3. The lengths of four sides and a diagonal of the given quadrilateral are indicated in the diagram. If a denotes the area of quadrilateral, then $A$ is

(A) $12 \sqrt{6}$

(B) $\sqrt{6}$

(D) $\sqrt{6}$,

~~ 4. In the sides of a triangle are doubled, then its area :

(A) Remains the same

(B) Becomes doubled

(D) Becomes four times

~~ 5. Inside a triangular garden there is a flower bed in the form of a similar triangle. Around the flower bed runs a uniform path of such a width that the side of the garden are double of the corresponding sides of the flower bed. The areas of the path and the flower bed are in the ratio :

(A) $1:1$

(B) $1: 2$

(D) $3: 1$

SUBJECTIVE DPP - 17.2

~~ 1. In the given figure, $\triangle ABC$ is a equilateral triangle the length of whose side is equal to $10 cm$ and $\triangle DBC$ is right-angled at $D$ and $BD=8 cm$. Find the area of the shaded region. Take $\sqrt{3}=1.732$.

~~ 2. Calculate the area of the triangle whose sides are $18 cm, 24 cm$ and $30 cm$ in length. Also, find the length of the altitude corresponding to the smallest side of the triangle.

~~ 3. The sides of a triangle are $10 cm, 24 cm$ and $26 cm$. Find its area and the longest altitude.

~~ 4. Two sides of a triangular field are $85 m$ and $154 m$ in length, and its perimeter is $324 cm$. Find (i) the area of the field, and (ii) the length of the perpendicular from the opposite vertex on the side measuring $154 cm$.

~~ 5. The sides of a triangular field are $165 cm, 143 cm$ and $154 cm$. Find the cost of ploughing it at 12 paise per sq. $m$.

~~ 6. The base of an isosceles triangle measures $80 cm$ and its area is $360 cm^{2}$. Find the perimeter of the triangle.

~~ 7. The perimeter of an isosceles triangle is $42 cm$ and its base is $1 \frac{1}{2}$ times each of the equal sides. Find (i) the length of each side of the triangle, (ii) the area of the triangle, and (iii) the height of the triangle.

~~ 8. The perimeter of a right angle triangle is $40 cm$. Its hypotenuse is $17 cm$. Find the sides containing the right angle. Also find the area of the triangle.

~~ 9. Find the area and perimeter of an isosceles right-angled triangle, each of whose equal sides measures $10 cm$. Take $\sqrt{2}=1.414$.

~~ 10. The area of a square field in 8 hectares. How long would a man take to cross its diagonal by walking at the rate of $4 km$ per hour?

~~ 11. A rhombus shaped field has green for 18 cows to graze. If each side of the rhombus is $30 m$ and its longer diagonal is $48 m$, how much area of grass field will each cow be getting ?

$$EXERCISE$$

(Objective DPP # 17.1)

Qus.	1	2	3	4	5
Ans.	$A$	$A$	$A$	$D$	$D$

(Subjective DPP # 17.2)

~~ 1. $\quad 19.3 cm^{2}$

~~ 2. $216 cm, 24 cm$

~~ 3. $\quad 120 cm^{2}, 24 cm$

~~ 4. () $2772 cm^{2}$ (ii) $36 cm^{2}$

~~ 5. Rs. 1219.68

~~ 6. $\quad 162 cm$

~~ 7. (i) $12 cm, 12 cm, 18 cm$ (ii) $71.42 cm^{2}$ (iii) $7.94 cm$

~~ 8. $8 cm, 15 cm \And 60 cm^{2}$

~~ 9. $50 cm^{2}, 34.14 cm$

~~ 10. 6 minutes

~~ 11. $48 m^{2}$

$\ggg$
SURFACE AREA AND VOLUME $\lll«$

ML - 18

SOLID FIGURES

If any figure such as cuboids, which has three dimensions length, width and height are known as three dimensional figures. Where are rectangle has only two dimensions i.e. length and width. Three dimensional figures have volume in addition to areas of surface from which these solid figures are formed.

(a) Cuboids :

There are six faces (rectangular), eight vertices and twelve edges in a cuboids.

Total Surface Area (T.S.A.) : The area of surface from which cuboids is formed.

(i) Total Surface Area (T.S.A.) $=2[\ell \times b+b \times h+h \times \ell]$

(ii) Lateral Surface Area (L.S.A.) $=2[b \times h+h \times \ell]$

(iii) (or Area of 4 walls)

$=2 h[\ell+b]$

iii) Volume of Cuboids $=($ Area of base $) \times$ height

(iv) Length of diagonal $=\sqrt{\ell^{2}+b^{2}+h^{2}}$

(b) Cube :

Cube ahs six faces. Each face is a square.

(i) T.S.A. $=2[x \cdot x+x \cdot x+x \cdot x]$ $=2[x^{2}+x^{2}+x^{2}]=2(33^{2})=6 x^{2}$

(ii) L.S.A. $=2[x^{2}+x^{2}]=4 x^{2}$

(iii) Volume $=$ (Area of base $) \times$ Height $=(x^{2}) \times x=x^{3}$

(iv) Length of diagonal $=x \sqrt{3}$

(c) Cylinder :

Curved surface area of cylinder (C.S.A.) : It is the area of surface from which the cylinder is formed. When we cut this cylinder, we will find a rectangle with length $2 \pi r$ are height $h$ units.

(i) C.S.A. of cylinder $=(2 \pi r) \times h=2 \pi r h$.

(ii)

T.S.A = C.S.A. + circular top $\And$ bottom

$ =2 \pi r h+(\pi r^{2})+(\pi r^{2}) $

$ =2 \pi r h+2 \pi r^{2} $

$ =2 \pi r(h+r) \text { sq.units }$

$ =(\pi r^{2}) \times h $

$ =\pi r^{2} h \text { cubic units } $

Hollow cylinder:

(i) C.S.A. of hollow cylinder $=2 \pi(R+r) h$ sq. units

(ii) T.S.A. of hollow cylinder $=2 \pi(R+r) h+\pi(R^{2}-r^{2})$

$ =\pi(R+r)[2 h+R-r] \text { sq. units } $

(iii) Volume of hollow cylinder $=\pi(R^{2}-r^{2}) h \quad$ cubic units

Where, $r=$ inner radius of cylinder

$ \begin{aligned} & R=\text { outer radius of cylinder } \\ & h=\text { height of the cylinder } \end{aligned} $

(d) Cone :

(i) C.S.A. of cone $\quad=\pi \ell$

(ii) T.S.A . of cone = C.S.A. + Base area

$=\pi r \ell+\pi r^{2}$

$=\pi r(\ell+r)$

(iii) Volume of cone

$=\frac{1}{3} \pi r^{2} h$

Where, $h=$ height

$ \begin{aligned} & r=\text { radius of base } \\ & \ell=\text { slant height } \end{aligned} $

(e) Sphere :

(i) T.S.A. of sphere $\quad=4 \pi r^{2}$

(ii) Volume of sphere $=\frac{4}{3} \pi r^{3}$

(f) Hemisphere :

(i) C.S.A.

$ =2 \pi r^{2} $

(ii) T.S.A. $\quad=$ C.S.A. + other area

$=2 \pi r^{2}+\pi r^{2}$

$=3 \pi r^{2}$

$ -3 \pi $

(iii) Volume $=\frac{2}{3} \pi r^{3}$

(g) Hollow Hemisphere :

(i) C.S.A. $\quad=2 \pi(R^{2}+r^{2})$

(ii) T.S.A $\quad=2 \pi(R^{2}+r^{2})+\pi(R^{2}-r^{2})$

(iii) Volume $\quad=\frac{2}{3} \pi(R^{3}-r^{3})$

ILLUSTRATIONS :

Ex. 1 Three equal cubes are placed adjacently in a row. Find the ratio of the total surface area of the new cuboids to that of the sum of the surface areas of three cubes.

Sol. Let the side of each of the three equal cubes be a $cm$.

Then surface area of one cube $=6 a^{2} cm^{2}$

$\therefore \quad$ Sum of the surface areas of three cubes $=3 \times 6 a^{2}=18 a^{2} cm^{2}$

For new cuboids

$ \begin{aligned} & \text { length }(\ell)=3 a cm \\ & \text { breadth }(b)=a cm \\ & \text { height }(h)=a cm \end{aligned} $

$\therefore$ Total surface are of the new cuboids $\quad=2(\ell \times b+b \times h+h=\ell)$

$ \begin{aligned} & =2[3 a \times a+a \times a+a \times 3 a] \\ & =2[3 a^{2}+a^{2}+3 a^{2}]=14 a^{2} cm .2 \end{aligned} $

$\therefore$ Required ratio $\quad=\frac{\text { Total surface area of the new cuboid }}{\text { Sum of the surace areas of three cubes }}$

$ =\frac{14 a^{2}}{18 a^{2}}=\frac{7}{9}=7: 9 $

Ans.

Ex. 2 A class room is $7 m$ long, $6.5 m$ wide and $4 m$ high. It has one door $3 m \times 1.4 m$ and three windows each measuring $2 m \times 1 m$ The interior walls are to be colour-washed. The contractor charges Rs. 15 per sq. $m$. Find the cost of colour washing.

Sol. $\quad \ell=7 m, n=6.5 m$ and $h=4 m$

$\therefore \quad$ Area of the room $=2(\ell+b) h=2(7+6.5) 4=108 m^{2}$

Area of door $=3 \times 1.4=4.2 m^{2}$

Area of one window $=3 \times 2=6 m^{2}$

$\therefore \quad$ Area of 3 windows $=3 \times 2=6 m^{2}$

$\therefore \quad$ Area of the walls of the room to be colour washed $=108-(4.2+6)$

$ =108-10.2=97.8 m^{2} $

$\therefore \quad$ Cost of colour washing $@$ Rs. 15 per square metre $=$ Rs. $97.8 \times 15=$ Rs. ~~ 1467. Ans.

Ex. 3 A cylindrical vessel, without lid, has to be tin coated including both of its sides. If the radius of its base is $\frac{1}{2}$ $m$ and its height is $1.4 m$, calculate the cost of tin-coating at the rate of Rs. 50 per $1000 cm^{2}$.

Sol. Radius of the base $(r)=\frac{1}{2} m$

$ =\frac{1}{2} \times 100 cm=50 cm $

Height $(h)=1.4 m$

$ \begin{aligned} & =1.4 \times 100 cm \\ & =140 cm . \end{aligned} $

Surface area of to tin-coated $=2(2 \pi r+\pi r^{2})$

$ \begin{aligned} & =2[2 \times 3.14 \times 50 \times 140+3.14 \times(50)^{2}] \\ & =2[43960+7850]=2(51810)=103620 cm^{2} \end{aligned} $

$\therefore \quad$ Cost of tin-coating at the rate or Rs. 50 per $1000 cm^{2}$

$ =\frac{50}{1000} \times 103620=\text { Rs } 5181 $

Ans.

Ex. 4 The diameter of a roller $120 cm$ long is $84 cm$. If its takes 500 complete revolutions to level a playground determine the cost of leveling at the rate of Rs. 25 per square metre. $(.$ Use $\pi=\frac{22}{7}$ )

Sol. $\quad 2 r=84 cm$

$\therefore \quad r=\frac{84}{2} cm=42 cm$

$h=120 cm$

Area of the playground leveled in one complete revolution $=2 \pi rh$

$ =2 \times \frac{22}{7} \times 42 \times 120 \times 31680 cm^{2} $

$\therefore \quad$ Area of the playground $=31680 \times 500 cm^{2}$

$ =\frac{31680 \times 500}{100 \times 100} m^{2}=1584 m^{2} $

$\therefore \quad$ Cost of leveling $@$ Rs 25 per square metre $=$ Rs $1584 \times 25=39600 . \quad$Ans.

Ex. 5 How many metres of cloth of $1.1 m$ width will be required to make a conical tent whose vertical height is 12 $m$ and base radius is $16 m$ ? Find also the cost of the cloth used at the rate of Rs 14 per metre.

Sol. $\quad h=12 m$

$r=16 m$

$\therefore \quad \ell=\sqrt{r^{2}+h^{2}}$

$=\sqrt{(16)^{2}+(12)^{2}}=\sqrt{256+144}$

$=\sqrt{400}=20 m$

$\therefore \quad$ Curved surface area $=\pi r \ell=\frac{22}{7} \times 16 \times 20=\frac{7040}{7} m^{2}$

Width of cloth $=1.1 m$

$\therefore \quad$ Length of cloth $=\frac{7040 / 7}{1.1}=\frac{70400}{77}=\frac{6400}{7} m$

$\therefore \quad$ Cost of the cloth used $@$ Rs 14 per metre $=$ Rs $\frac{6400}{7} \times 14=$ Rs 12800Ans.

Ex. 6 The surface area of a sphere of radius $5 cm$ is five times the area of the curved surface of cone of radius 4 $cm$. Find the height of the cone.

Sol. Surface area of sphere of radius $4 cm=\pi(4) \ell cm^{2}$ when $\ell cm$ is the slant height of the cone.

According to the question,

$4 \pi(5)^{2}=5[\pi(4) \ell] $

$ \Rightarrow \ell=5 cm \Rightarrow \sqrt{r^{2}+h^{2}}=5 $

$ \Rightarrow r^{2}+h^{2}=25 \Rightarrow (4)^{2}+h^{2}=25 $

$ \Rightarrow 16+h^{2}=25 \Rightarrow h^{2}=9 $

$ \Rightarrow h=3 $

Hence the height of the cone is $3 cm$.

Ans.

Ex. 7 The dimensions of a cinema hall are $100 m, 50 m$ and $18 m$. How many persons can sit in the hall, if each required $150 m^{3}$ of air?

Sol. $\quad \ell=100 m$

$b=50 m$

$h=18 m$

$\therefore \quad$ Volume of the cinema hall $=\ell bh$

$ =100 \times 50 \times 18=90000 m^{3} $

Volume occupied by 1 person $=150 m^{3}$

$\therefore \quad \text{Number of persons who can sit in the hall} =\frac{\text { Volume of the ball }}{\text { Volume occupied by } 1 \text { person }}$

$ =\frac{90000}{150}=600 $

Hence 600 persons can sit in the hall.

Ans.

Ex. 8 The outer measurements of a closed wooden box are $42 cm, 30 m$ and $27 cm$. If the box is made of $1 cm$ thick wood, determine the capacity of the box.

Sol. Outer dimensions

$ \begin{aligned} \ell & =42 cm \\ b & =30 cm \\ h & =27 cm \end{aligned} $

Thickness of wood $=1 cm$

Inner dimensions

$ \begin{aligned} \ell & =42-(1+1)=40 cm \\ b & =30-1(1+1)=28 cm \\ h & =27-(1+1)=25 cm \end{aligned} $

$\therefore \quad$ Capacity of the box $\ell \times b \times h$

$ =40 \times 28 \times 25=28000 cm^{2} . \quad \text{Ans.} $

Ex. 9 If $v$ is the volume of a cuboids of dimensions $a, b$, and $c$ and $s$ is its surface area, then prove that

$ \frac{1}{v}=\frac{2}{a}\Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big) $

Sol. L.H.S. $=\frac{1}{v}=\frac{1}{a b c} \quad \ldots \text{(i)}$

R.H.S. $=\frac{2}{s}\Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big)$

$=\frac{2}{2\Big(a b+b c+c a\Big)}\Big(\frac{b c+c a+a b}{a b c}\Big)$

$=\frac{1}{abc} \quad \ldots \text{(ii)}$

from (i) and (ii) $\frac{1}{v}=\frac{2}{s}\Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big)$.

Hence Proved.

Ex. 10 The ratio of the volumes of the two cones is $4: 5$ and the ratio of the radii of their bases is $2: 3$. Find the ratio of their vertical heights.

Sol. Let the radii of bases, vertically heights and volumes of the two cones be $r_1, h_1, v_1$ and $r_2, h_2, v_2$ respectively. According to the question,

$ \frac{v_1}{v_2}=\frac{4}{5} \quad \text {…(i) } \quad \frac{r_1}{r}=\frac{2}{3} \quad \ldots \text{(i)} $

From (i), we have $\frac{\frac{1}{2} \pi r_1^{2} h_1}{\frac{1}{3} \pi r_2^{2} h_2}=\frac{4}{5}$

$\Rightarrow \frac{r_1^{2} h_1}{r_2^{2} h_2}=\frac{4}{5}$

$\Rightarrow\Big(\frac{r_1}{r_2}\Big)^{2} \frac{h_1}{h_2}=\frac{4}{5}$

$\Rightarrow\Big(\frac{2}{3}\Big)^{2}=\frac{h_1}{h_2}=\frac{4}{5}$

$\Rightarrow \frac{h_1}{h_2}=\frac{4}{5}\Big(\frac{3}{2}\Big)^{2} \quad[$ Using (ii)]

$\Rightarrow \frac{h_1}{h_2}=\frac{9}{5}$

Hence the ratio of their vertical height is $9: 5$.

Ans.

Ex. 11 If $h, c$ and $v$ be the height, curved surface and volume of a cone, show that $3 \pi vh^{3}-c^{2} h^{2}+9 v^{2}=0$.

Sol. Let the radius of the base and slant height of the cone be $r$ and $\ell$ respectively. The $n$;

$ \begin{aligned} & c=\text { curved surface }= \pi r \ell=\pi r \sqrt{r^{2}+h^{2}} \quad \ldots \text{(i)}\\ & v=\text { volume }=\frac{1}{3} \pi r^{2} h \quad \ldots \text{(ii)}\\ & \therefore \quad 3 \pi vh^{3}-c^{2} h^{2}+9 v^{2}=3 \pi(\frac{1}{3} \pi r^{2} h) h^{2}-\pi^{2} r^{2}(r^{2}+h^{2}) h^{2}+9\Big(\frac{1}{3} \pi r^{2} h\Big)^{2} \\ &=^{2} r^{2} h^{4}-\pi^{2} r^{2} h^{2}-\pi^{2} r^{2} h^{4}+\pi^{2} r^{4} h^{2}=0 . \end{aligned} $

[Using (i) and (ii)]

Hence Proved.

Ex. 12 How many balls, each of radius $1 cm$, can be made from a solid sphere of lead of radius $8 cm$ ?

Sol. Volume of the spherical ball of radius $8 cm=\frac{4}{3} \pi \times 8^{3} c^{3}$

Also, volume of each smaller spherical ball of radius $1 cm=\frac{4}{3 \pi \times 1^{3}} cm^{3}$.

Let $n$ be the number of smaller balls that can be made. Then, the volume of the larger ball is equal to the sum of all the volumes of $n$ smaller balls.

Hence, $\frac{4}{3} \pi \times n=\frac{4}{3} \pi \times 8^{3}$

$\Rightarrow \quad n=8^{3}=512$

Hence, the required number of balls $=512$.

Ans.

Ex. 13 By melting a solid cylindrical metal, a few conical materials are to be made. If three times the radius of the cone is equal to twice the radius of the cylinder an the ratio of the height of the cylinder and the height of the cone is $4: 3$, find the number of cones which can be made.

Sol. Let $R$ be the radius and $H$ be the height of the cylinder and let $r$ and $h$ be the radius and height of the cone respectively. Then,

$ 3 r=2 R $

And $H ; h=4: 3 \quad \ldots \text{(i)}$

$\Rightarrow \frac{H}{h}=\frac{4}{3}$

$\Rightarrow 3 H=4 h \quad \ldots \text{(ii)}$

Let $n$ be the required number of cones which can be made from the materials of the cylinder. Then, the volume of the cylinder will be equal to the sum of the volumes of $n$ cones. Hence, we have

$ \begin{aligned} & \pi R^{2} H=\frac{n}{3} \pi r^{2} h \\ \\ \Rightarrow & 3 R^{2} H=n r^{2} h \\ \\ \Rightarrow \quad & n=\frac{3 R^{2} H}{r^{2} h}=\frac{3 \times \frac{9 r^{2}}{4} \times \frac{4 h}{3}}{r^{2} h} \quad[\because \text { From (i) and (ii), } R=\frac{3 r}{2} \text { and } H=\frac{4 h}{3}] \\ \\ & =\frac{3 \times 9 \times 4}{3 \times 4}=9 \end{aligned} $

Hence, the required number of cones is 9 .

Ans.

Ex. 14 Water flows at the rate of 10 per minute through a cylindrical pipe having its diameter as $5 mm$. How much time will it take to fill a conical vessel whose diameter of the base is $40 cm$ and depth $24 cm$ ?

Sol. Diameter of the pipe $=5 mm=\frac{5}{10} cm=\frac{1}{2} cm$.

$\therefore$ Radius of the pipe $=\frac{1}{2} \times \frac{1}{2} cm=\frac{1}{4} cm$.

In 1 minute, the length of the water column in the cylindrical pipe $=10 m=1000 cm$.

$\therefore$ Volume of water that flows out of the pipe in 1 minute $=\pi \times \frac{1}{4} \times \frac{1}{4} \times 1000 cm^{3}$.

Also, volume of the cone $=\times \frac{1}{3} \times \pi \times 20 \times 20 \times 24 cm^{3}$.

Hence, the time needed to fill up this conical vessel $=\Big(\frac{1}{3} \pi \times 20 \times 20 \times 24 \div \pi \times \frac{1}{4} \times \frac{1}{4} \times 1000\Big)$ minutes $=\Big(\frac{20 \times 20 \times 24}{3} \times \frac{4 \times 4}{100}\Big)=\frac{4 \times 24 \times 16}{30}$ minutes $=\frac{256}{5}$ minutes $=51.2$ minutes.

Hence, the required time is 51.2 minutes.

Ans.

$$EXERCISE$$

OBJECTIVE DPP # 18.1

~~ 1. The height of a conical tent at the centre is $5 m$. The distance of any point on its circular base from the top of the tent is $13 m$. The area of the slant surface is :

(A) $144 \pi sq m$

(B) $130 \pi sq m$

(D) $169 \pi sq m$

~~ 2. A rectangular sheet of paper $22 m$ long and $12 cm$ broad can be curved to form the lateral surface of a right circular cylinder in two ways. Taking $\pi=\frac{22}{7}$, the difference between the volumes of the two cylinders thus formed is :

(A) 200 c.c.

(B) 210 c.c.

(D) 252 c.c.

~~ 3. The percentage increase in the surface area of a cube when each side is increased to $\frac{3}{2}$ times the original length is

(A) 225

(B) 200

(D) 125

~~ 4. A cord in the form of a square enclose the area ’ $S^{\prime} cm^{2}$. if the same cord is bent into the form of a circle, then the area of the circle is

(A) $\frac{\pi S^{2}}{4}$

(B) $4 \pi S^{2}$

(D) $\frac{4 S}{\pi}$

~~ 5. If ’ $I$ ‘, ’ $b$ ’ and ’ $h$ ’ if a cuboids are increased, decreased and increased by $1 %, 3 %$ and $2 %$ respectively, then the volume of the cuboids

(A) increase

(B) decrease

(D) can’t be calculated with given data

~~ 6. The radius and height of a cone are each increased by $20 %$, then the volume of the cone is increased by

(A) $20 %$

(B) $40 %$

(D) $72.8 %$

~~ 7. There is a cylinder circumscribing the hemisphere such that their bases are common. The ratio of their volume is

(A) $1: 3$

(B) $1: 2$

(D) $3: 4$

~~ 8. Consider a hollow cylinder of inner radius $r$ and thickness of wall $t$ and length $\ell$. The volume of the above cylinder is given by

(A) $2 \pi \ell(r^{2}-\ell^{2})$

(B) $2 \pi r \ell t\Big(\frac{t}{2 r}+1\Big)$

(D) $2 \pi r \ell(r+t)$

~~ 9. A cone and a cylinder have the same base area. They also have the same curved surface area. If the height of the cylinder is $3 m$, then the slant height of the cone (in $m$ ) is

(A) 3

(B) 4

(D) 7

~~ 10. A sphere of radius $3 cm$ is dropped into a cylindrical vessel of radius $4 cm$. If the sphere is submerged completely, then the height (in $cm$ ) to which the water rises, is

(A) 2.35

(B) 2.30

(D) 2.15

~~ 11. The whole surface of a rectangular lock is $846 cm^{2}$. Find the length, breadth and height, if these dimensions are in the ratio $5: 4: 3$.

~~ 12. An open box is made of wood $3 cm$ thick. its external length, breadth and height are $1.48 m, 1.16 m$ and 8.3 $dm$. Find the cost of painting the inner surface at Rs 5 per $m^{2}$.

~~ 13. A room $8 m$ long $6 m$ board and $3 m$ high has two windows $1 \frac{1}{2} m \times 1 m$ and a door $2 m \times 1 \frac{1}{2} m$. Find the cost of papering the walls will paper $50 cm$ wide at Rs. 40 per metre.

~~ 14. 50 circular plates, each of radius $7 cm$ and thickness $\frac{1}{2} cm$, are placed one above the other to form a solid right circular cylinder. Find the total surface area.

~~ 15. A tent in the shape of a right circular cylinder surmounted by a right circular cone. The heights of the cylindrical and the conical parts are $40 m$ and $21 m$ respectively. If the base diameter of the tent is $56 m$, find the area of the required canvas to make this tent if $20 %$ of the area is consumed in folding and sewing.

~~ 16. A toy is in the form of a right circular cylinder closed at one end and with a hemisphere on the other end. The height and the radius of the base are $15 cm$ and $6 cm$ respectively. The radius of the hemisphere are cylinder are same. Calculate the total surface area and the volume of the toy. if the toy is painted at the rate of Rs. 2.50 per $10 cm^{2}$, find the cost of painting the toy.

~~ 17. An iron pillar has some portion in the form of a right circular cylinder an remaining in the form of a right circular cone. The radius of the base of each of the cone and the cylinder is $8 cm$. The cylindrical portion is $240 cm$ high and the conical part is $36 cm$ high. Find the weight of the pillar, if one cubic $cm$ of iron weights $7.8 g$.

~~ 18. A solid metallic sphere of diameter 28 is melted and recasted into a number of smaller cones, each of diameter $4 \frac{2}{3} cm$ and height $3 cm$. Find the number of cones so formed.

$$\text{ANSWER KEY}$$

(Objective DPP # 18.1)

Qus.	1	2	3	4	5	6	7	8	9	10
Ans.	$C$	$B$	$D$	$D$	$B$	$D$	$C$	$B$	$C$	$C$

(Subjective DPP # 18.2)

~~ 1. $15 cm, 12 cm, 9 cm$

~~ 2. Rs. 27.97

~~ 3. Rs. 62.40

~~ 4. $\quad 1408 cm^{2}$

~~ 5. Total surface area $=12144 cm^{2}$

~~ 6. Surface area $\approx 678.86 cm^{2}$, Volume $\approx 1470.86 cm^{3}$, Cost $\approx$ Rs. 170

~~ 7. $\quad 395.37 kg$.

~~ 8. 672 cones.

$\ggg$
STATISTICS
$\lll$

ML - 19

INTRODUCTION

The branch of science known as Statistics has been used in India from ancient times. Statistics deals with collection of numerical facts i.e., data, their classification \And tabulation and their interpretation. In statistics we shall try to study, in detail about collection, classification and tabulation of such data.

(a) Importance of Data :

Expressing facts with the helps of data is of great importance in our day-today life. For example, instead of saying that India has a large population it is more appropriate to say that the population of India, based on the census of 2000 is more than one billion.

(b) Collection of Data :

On the basis of methods of collection, data can be divided into two categories :

(i) Primary data : Data which are collected for the first time by the statistical investigator or with help of his workers is called primary data. As example if an investigator wants to study the condition of the workers working in a factory then fro this he collects some data like their monthly income, expenditure, number or brother, sisters, etc.

(ii) Secondary data : These are the data already collected by a person or a society and these may be in published or unpublished form. These data should be carefully used. These are generally obtained from the following two sources.

(A) Published sources

(B) unpublished sources

(c) Classification of Data :

When the data is complied in the same form and order in which it is collected, it is known as Raw Data, It is alsoCrude Data. For example, the marks obtained by 20 students of class $X$ in English out of 10 marks are as follows :

| 7, | 4, | 9, | 5, | 8, | 9, | ~~

6.	7,	9,	2,
0	3,	7,	6,	2,	1,	9,	8,	3,	8,

(i) Geographical basis : Here, the data is classified on the basis of place or region. For example the production of food grains of different state is shown in the following table :

S.No.	State	Production (in Tons)
1	Andhdra Pradesh	9690
2	Bihar	8074
3	Haryana	10065
4	Pubjab	17065
5	Uttar Pradesh	28095

(ii) Chronological classification : If data’s classification is based on hour, day, week and month or year, then it is called chronological classification, For example, the population of India in different year is shown in following table :

S.No	Year	Production (in Crores)
1	1951	46.1
2	1961	53.9
3	1971	61.8
4	1981	68.5
5	1991	88.4
6	2001	100.01

(iii) Qualitative basis : When the data are classified into different groups on the basis of their descriptive qualities and properties, such a classification is known as descriptive or qualitative classification. Since the attributes can not be measured directly, they are counted on the basis of presence or absence of qualities. For example intelligence, literacy, unemployment, honesty etc. The following table shows classification on the basis of sex and employment.

alt text

(iv) Quantitative basis : if facts are such that they can be measured physically e.g. marks obtained height, weight, age, income, expenditure etc. Such facts are known as variable values. If such facts are kept into classes then it is called classification according to quantitative or class intervals.

Marks obtained	$10-20$	$20-30$	$30-40$	$40-50$
No. of students	7	9	15	6

DEFINITIONS

(i) Variate : The numerical quantify whose value varies in objective is called a variate, generally a variate is represented by $x$. There are two types of variate.

(A) Discrete variate : its magnitude is fixed. For example, the number of teacher in different branches of a institute are 30, 35, 40 etc.

(B) Continuous variate : is magnitude is not fixed. It is expressed in groups like $10-20,20$ - 30,…. etc.

(ii) Rage : The difference of the maximum and the minimum values of the variable $x$ is called range.

(iii) Class frequency : In each class the number of times a data is repeated in known as its class frequency.

(iv) Class Interval $=\frac{\text { Range }}{\text { Number of classes }}$

It is generally denoted by $h$ or $i$.

(v) Class limits : The lowest and the highest value of the class are known as lower and upper limited restively of that class.

(vi) Class mark : The average of the lower and the upper limits of a class is called the mid value or the class mark of that class. It is generally denoted by $x$.

If $x$ be the mid value and $h$ be the class interval, then the class limits are $\Big(x-\frac{h}{2}, x+\frac{4}{2}\Big)$.

Ex. 1 The mid values of a distribution are 54, 64, 74, 84 and ~~ 94. Find the class interval and class limits.

Sol. The class interval is the difference of two consecutive class marks, therefore class interval $(h)=64-54=10$.

Here the mid values are given and the class interval is 10 .

So class limits are

For $1^{\text {st }}$ class $\quad 54-\frac{10}{2}$ to $54+\frac{10}{2}$ or $\quad 49$ to 59

For $2^{\text {nd }}$ class $\quad 64-\frac{10}{2}$ to $64+\frac{10}{2}$ or $\quad 59$ to 69

For 3 rd class $\quad 74-\frac{10}{2}$ to $74+\frac{10}{2}$ or $\quad 69$ to 79

For $4^{\text {th }}$ class $\quad 84-\frac{10}{2}$ to $84+\frac{10}{2}$ or $\quad 79$ to 89

For $5^{\text {th }}$ class $\quad 94-\frac{10}{2}$ to $94+\frac{10}{2}$ or 89 to 99

Therefore class limits are 49 - 59, 59 - 69, 79 - 89, and 89 - 99 .

FREQUENCY DISTRIBUTION

The marks scored by 30 students of IX class, of a school in the first test of Mathematics our of 50 marks are as follows :

6	32	10	17	22	28	0	48	6	22
32	6	36	26	48	10	32	48	28	22
22	22	28	26	17	36	10	22	28	0

The number of times a mark is repeated is called its frequency. It is denoted by $f$.

Marks obtained	Taly mark	Frequency	Marks obtained	Tally mark	Frequency
0	II	2	26	II	2
6	III	3	28	IIII	4
10	III	3	32	III	3
17	II	2	36	II	2
22	IIII I	6	48	III	3

Above type of frequency distribution is calledungrouped frequency distribution. Although this representation of data is shorter than representation of raw data, but from the angle of comparison and analysis it is quite bit. So to reduce the frequency distribution, it can be classified into groups is following ways and it is calledgrouped frequency distribution.

Class	Frequency
$0-10$	8
$11-20$	2
$21-30$	12
$31-40$	5
$41-50$	3

(a) Kinds of Frequency Distribution :

Statistical methods like comparison, decision taken etc. depends of frequency distribution. Frequency distribution are of three types.

(i) Individual frequency distribution : Here each item or original price of unit is written separately. In $n$ this category, frequency of each variable is one.

Ex. 2 Total marks obtained by 10 students in a class.

S.No.	1	2	3	4	5	6	7	8	9	10
Marks obtained	46	18	79	12	97	80	5	27	67	54

(ii) Discrete frequency distribution : When number of terms is large and variable are discrete, i.e., variate can accept some particular values only under finite limits and is repeated then its called discrete frequency distribution. For example the wages of employees and their numbers is shown in following table.

Monthly wages	No. Of employees
4000	10
6000	8
8000	5
11000	7
20000	2
25000	1

The above table shows ungrouped frequency distribution the same facts can be written in grouped frequency as follows :

NOTE :

Monthly wages	No. of employees
$0-10,000$	23
$11,000-20,000$	9
$21,000-30,000$	1

If variable is repeated in individual distribution then it can be converted into discrete frequency distribution.

(iii) Continuous frequency distribution : When number of terms is large and variate is continuous. i.e., variate can accept all values under finite limits and they are repeated then it is called continuous frequency distribution. For example age of students in a school is shown in the following table :

Age (in year)	Class	No. of students
Less than 5 year	$0-5$	72
Between 5 and 10 y ear	$5-10$	103
Between 10 and 15 year	$10-15$	50
Between 15 and 20 year	$15-20$	25

NOTE :

Continuous frequency distribution is generally represented in form of grouped frequency distribution and variate is continuous in $i$, so $0-5,6-10,11$ - 15, 16- 20 types of classes can not be made here. If such classes are made in the table then students of age 5 to 6 year or 10 to 11 year or 15 to 16 years can not be classified. if such type of classes are given then they should be made continuous by following methods. Half of the difference between classes should be added to the upper limit of lower class and subtracted from lower limit o upper class. Thus the classes $0-5.5,5.5-10.5,10.5-15.5,15.5$ - 19.5 are obtained which are continuous.

Classes can be made mainly by two methods :

(i) Exclusive series : In this method upper limit of the previous class and lower limit of the next class is same. In this method the term of upper limit in a class is not considered in the same class, it is considered in the next class.

(ii) Inclusive series : In this method value of upper and lower limit are both contained in same class. In this method the upper limit of class and lower limit of other class are not same. Some time the value is not a whole number, it is a fraction or in decimals and lies in between the two intervals then in such situation the class interval can be constructed as follows

A		B
Class	Frequency	Or Class	Frequency
$0-9$	4	$0-9.99$	4
$10-19$	7	$10-19.99$	7
$20-29$	6	$20-29.99$	6
$30-39$	3	$30-39.99$	3
$40-49$	3	$40-49.99$	3

CUMULIVE FREQUENCY

(i) Discrete frequency distribution : From the table of discrete frequency distribution, it can be identified that number of employees whose monthly income is 4000 or how many employees of monthly income 1100 are there. But if we want to know how many employees whose monthly income is upto 11000 , then we should add $10+8+57$ i.e., number of employees whose monthly income is upto 11000 is 30 . Here we add all previous frequency and get cumulative frequency. If will be more clear from the following table

Class	Frequency (f)	Cumulative frequency (cf)	Explanation
4000	10	10	$10=10$
6000	8	18	$10+8$
8000	5	23	$18+5$
11000	7	30	$23+7$
20000	2	32	$30+2$
25000	1	33	$32+1$

(ii) Continuous frequency distribution : In the previous page we obtained cumulative frequency for discrete series. Similarly cumulative frequency table can be made from continuous frequency distribution also. For example, for table :

Monthly income	No. of employee	Cumulative	Explanation
Variate $(x)$	Frequency $(f)$	Frequency $(cf)$
$0-5$	72	72	$72=72$
$5-10$	103	175	$72+103=175$
$10-15$	50	225	$175+50=225$
$15-20$	25	250	$225+25=250$

Above table can also be written as follows :

Clas	Cumulative Frequency
Less than 5	72
Less than 10	175
Less than 15	225
Less than 20	250

From this table the number of students of age less than the upper limit of a class, i.e., number of student whose age is less than $5,10,15,20$ year can determined by merely seeing the table but if we need the number students whose age is more than zero, more than 5 , more than 10 or more than 15 , then table should be constructed as follows :

Class	Frequency	Age Cumulative frequency	Explanation
$0-5$	72	0 and more 50	$250=250$
$5-10$	103	5 and more 78	$250-72=178$
$10-15$	50	10 and more 75	$178-103=75$
$15-20$	25	15 and more 25	$75-50=25$

GRAPHICAL REPRESANTATION OF DATA

(i) Bar graphs

(ii) Histograms

(iii) Frequency polygons

(iv) Frequency curves

(v) Cumulative frequency curves or Ogives.

(vi) Pie Diagrams

(a) Bar Graphs :

Ex. 3 A family with monthly income of Rs. 20,000 had planned the following expenditure per month under various heads: Draw bar graph for the data given below :

Heads	Expenditure (in Rs. 1000)
Grocery	4
Rent	5
Education of children	5
Medicine	2
Fuel	2
Entertainment	1
Miscellaneous	1

Sol.

Histogram : Histogram is rectangular representation of grouped and continuous frequency distribution in which class intervals are taken as base and height of rectangles are proportional to corresponding frequencies. To raw the histogram class intervals are marked along $x$-axis on a suitable scale. Frequencies are marked along $y$-axis on a suitable scale, such that theareas of drawn rectangles are proportional to corresponding frequencies.

Now we shall study construction of histograms related with four different kinds of frequency distributions.

(i) When frequency distribution is grouped and continuous and class intervals are also equal.

(ii) When frequency distribution is grouped and continuous but class interval are not equal.

(iii) When frequency distribution is grouped but not continuous.

(iv) When frequency distribution is ungrouped and middle points of the distribution are given.

Now we try to make the above facts clear with some examples.

Ex. 4 Draw a histogram of the following frequency distribution.

Clas (Age in year)	$0-5$	$5-10$	$10-15$	$15-20$
No. of students	72	103	50	25

Sol. Here frequency distribution is grouped and continuous and class intervals are also equal. So mark the class intervals on the $x$-axis i.e., age in year (scale $1 cm=5$ year). Mark frequency i.e., number of students (scale 1 $cm=25$ students) on they y-axis.

Now, since the number of students in class interval 0 - 5 is 72 , so draw a parallel line to $x$-axis in front of frequency to construct a rectangle on class interval 0 - 5 . Repeating this procedure construct rectangle $A, B$, $C$ and $D$.

Ex. 5 The weekly wages of workers of a factory are given in the following table. Draw histogram for it.

Weekly wages	$1000-2000$	$2000-2500$	$2500-3000$	$3000-5000$	$5000-5500$
No. of workers	26	30	20	16	1

Sol. Here frequency distribution is grouped and continuous but class intervals are not same. Under such circumstances the following method is used to find heights of rectangle so that heights are proportional to frequencies.

(i) Write interval (h) of the least interval, here $h=500$.

(ii) Redefine the frequencies of classes by the using the following formula.

Redefined frequency of class $=\frac{h}{\text { clssinterval }} \times$ frequency of class interval.

So here the redefined frequency table is obtained as follows :

Weekly wages (in Rs.)	No. of workers	Redefined of workers
$1000-2000$	26	$\frac{500}{1000} \times 26=13$
$2000-2500$	30	$\frac{500}{500} \times 30=30$
$2500-3000$	20	$\frac{500}{500} \times 20=20$
$3000-5000$	16	$\frac{500}{2000} \times 16=4$
$5000-5500$	1	$\frac{500}{500} \times 1=1$

Now mark class interval on x-axis (scale $1 cm=500$ ) and no. of workers on $y$-axis $($ scale $1 cm=5)$. On the basis of redefined frequency distribution construct rectangle $A, B, C D$ and $E$.

This is the required histogram of the given frequency distribution

(a) Difference Between Bar Graph and Histogram

(i) In histogram there is no gap in between consecutive rectangle as in bar graph.

(ii) The width of the bar is significant in histogram. In bar graph, width is not important at all.

(iii) In histogram the areas of rectangles are proportional to the frequency, however if the class size of the frequencies are equal then height of the rectangle are proportional to the frequencies.

Frequency polygon : A frequency polygon is also a form a graphical representation of frequency distribution. Frequency polygon can be constructed in two ways :

(i) With the help of histogram

(ii) Without the help of histogram

(A) Following procedure is useful to draw a frequency polygon with the help of histogram.

(a) Construct the histogram for the given frequency distribution.

(b) Find the middle point of each upper horizontal line of the rectangle.

(d) Join the middle point of the initial rectangle with the middle point of the previous expected class interval on the x-axis.

Ex. 6 For the following frequency distribution, draw a histogram and construct a frequency polygon with it.

Class	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$
Frequency	8	12	17	9	4

Sol. The given frequency distribution is grouped and continuous, so we construct a histogram by the method given earlier. Join the middle points $P, Q, R, S, T$ of upper horizontal line of each rectangles $A, B, C, D, E$ by straight lines.

Ex. 7 Draw a frequency polygon of the following frequency distribution table.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$	$90-100$
Frequency	8	10	6	7	9	8	8	6	3	4

Sol. Given frequency distribution is grouped and continuous. So we construct a histogram by using earlier method. Join the middle points of $P, Q, R, S, T, U, V, W, X, Y$ of upper horizontal lines of each rectangle $A, B, C, D, E, F, G, H, I, J$ by straight line in successions.

Ex. 8 Draw a frequency polygon of the following frequency distribution.

Age (in years)	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Frequency	15	12	10	4	11	14

Sol. Here frequency distribution is grouped and continuous so here we obtain following table on the basis of class.

Age (in years)	$0-10$	$10-20$	$20-20$	$30-40$	$40-50$	$50-60$
Class mark	5	15	25	35	45	55
Frequency	15	12	10	4	11	14

Now taking suitable scale on graph mark the points $(5,15),(15,12),(25,10)(35,4),(45,11),(55,14)$. Since age can not be negative so instead of joining corner $(5,15)$ with middle point of zero frequency of earlier assumed class, we draw vertical line from the lower limit of this class i.e., 0 and point of half frequency of this lie i.e., $(0,7.5)$ is joined by the end point. Joint the last point $(55,14)$ with the points of zero frequency of the next assumed class i.e, with $(65,0)$.

MEASURES OF CENTRAL TENDENCY

The commonly used measure of central tendency are -

(i) Mean

(ii) Median

(iii) Mode

(a) Mean :

The mean of a number of observation is the sum of the values of all the observations divided by the total number of observations. It is denoted by the symbol $\bar{x}$, read as $x$ bar.

(i) properties of mean :

(a) If a constant real number ’ $a$ ’ is added to each of the observation than new mean will be $\bar{x}+a$.

(b) If a constant real number ’ $a$ ’ is subtracted from each of the observation then new mean will be $\bar{x}-a$

(d) If each of the observation is dived by a constant no ’ $a$ ’ then new mean will be $\frac{\bar{x}}{a}$.

(ii) Mean of ungrouped data: If $x_1, x_2, x_3, \ldots . ., x_n$ are then $n$ values (or observations) then A.M. (Arithmetic mean) is

$ \begin{aligned} & \bar{x}=\frac{x_1+x_1+\ldots . .+x_n}{n}=\frac{\sum_{i-1}^{n} x_i}{n} \\ & n \bar{x}=\text { Sum of observation }==\sum_{i-1}^{n} x_i \end{aligned} $

i.e. product of means \And no. of items given sum of observation.

Ex. 9 Find the mean of the factors of 10

Sol. factors of 10 are $1,2,5 \And 10$.

$ \bar{x}=\frac{1+2+5+10}{4}=\frac{18}{4}=4.5 $

Ex. 10 If the mean of $6,4,7 P$ and 10 is 8 find $P$.

Sol. $\quad 8=\frac{6+4+7+P+10}{5} \Rightarrow P=13 \Rightarrow P=13$

(iii) Method for Mean of ungrouped frequency distribution.

$\mathbf{x} _{\mathbf{i}}$	$\mathbf{f} _{\mathbf{i}}$	$\mathbf{f} _{\mathbf{i}} \mathbf{x} _{\mathbf{i}}$
$x_1$	$f_1$	$f_1 x_1$
$x_2$	$f_2$	$f_2 x_2$
$x_3$	$f_3$	$f_3 f_3$
$\cdot$	$\cdot$	$\cdot$
$\cdot$	$\cdot$	$\cdot$
$\cdot$	$\cdot$	$\cdot$
$x_n$	$f_n$	$f_n x_n$

	$\sum f_i=$	$\sum f_i x_i=$

Then mean $#### ILLUSTRATIONS :=\frac{\sum f_i x_i}{\sum f_i}$

(iv) Method for Mean of grouped frequency distribution.

Ex. 11 (1) Direct Method : for finding mean

Marks	No. of students fi	mid values xi	fixi
$10-20$	6	15	90
$20-30$	8	25	200
$30-40$	13	35	455
$40-50$	7	45	315
$50-60$	3	55	165
$60-70$	2	65	130
$70-80$	1	75	$\sum f_i x_i$
	$\sum f_1=40$		$\sum f_i$
		$\sum f_i x_i=40$

(v) Combined Mean :

$ \text{Combined Mean}:=\frac{n_1 \overline{x}_1+n_2+\overline{x}_2+\ldots}{n_1+n_2+\ldots} $

(vi) Uses of Arithmetic Mean

(A) It is used for calculating average marks obtained by a student.

(B) It is extensively used in practical statistics.

(D) It is used by businessman to find out profit per unit article, output per machine, average monthly income and expenditure etc.

(b) Median :

Median of a distribution is the value of the variable which divides the distribution into two equal parts.

(i) Median or ungrouped data

(A) Arrange the data in ascending order.

(B) Count the no. of observations (Let there be ’ $n$ ’ observations)

(D) If $n$ is even the median $=$ value of mean of $\Big(\frac{n}{2}\Big)^{\text {th }}$ observation and $\Big(\frac{n}{2}+1\Big)^{\text {th }}$ observation.

Ex. 12 Find the median of the following values :

$37,31,42,43,46,25,39,45,32$

Sol. Arranging the data in ascending order, we have

$25,31,32,37,39,42,43,45,46$

Here the number of observations $n=9$ (odd)

$ \therefore \text{Median} =\text{Value of}\Big(\frac{9+1}{2}\Big)^{\text {th }} \text { observation } $

$ =\text { Value of } 5^{\text {th }} \text { observation } $

$ =39 $

Ex. 13 The median of the observation 11, 12, 14, 18, $x+2, x+4,30,32,35,41$ arranged in ascending order is 24 . Find the value of $x$.

Sol. Here, the number of observations $n=10$. Since $n$ is even, therefore

Median $=\frac{\Big(\frac{n}{2}\Big)^{th} \text { conservation }+\Big(\frac{n}{2}+1\Big)^{th} \text { observation }}{2}$

$\Rightarrow \quad 24=\frac{5^{\text {th }} \text { observation }+6^{\text {th }} \text { observation }}{2}$

$\Rightarrow \quad 24=\frac{(x+2)+(x+4)}{2}$

$\Rightarrow 24=\frac{2 x+6}{2} \Rightarrow 24=x+3 \Rightarrow x=21$.

Hence, $x=21$

(ii) Uses of Median :

(A) Median is the only average to be used while dealing with qualitative data which cannot be measured quantitatively but can be arranged in ascending or descending order or magnitude.

(B) It is used for determining the typical value in problems concerning wages, distribution of wealth etc.

(c) Mode :

(i) Mode or ungrouped data (By inspection only) : Arrange the data in an array and then count the frequencies of each variate. The variate having maximum frequency is the mode.

Ex. 13 Find the mode of the following array of an individual series of scores $7,10,12,12,12,11,13,13$, 17 .

Number	7	10	11	12	13	17
Frequency	2	1	1	3	2	1

$\therefore \quad$ Mode is 12

(ii) Uses of Mode : Mode is the average to be used to find the ideal size, e.g., in business forecasting, in manufacture of ready-made garments, shoes etc.

(c) Empirical Relation between Mode, Median \And Mean :

Mode $=3$ Median -2 Mean

RANGE

The range is the difference between the highest and lowest scores of a distribution. It is the simplest measure of dispersion. It gives a rough idea of dispersion. This measure is useful for ungrouped data.

(a) Coefficient of the Range :

If $\ell$ and $h$ are the lowest and highest scores in a distribution then the coefficient of the Range $=\frac{h-\ell}{h+\ell}$

Ex. 14 Find the range of the following distribution : 1, 3, 4, 7, 9, 10, 12, 13, 14, 16 and ~~ **19. ** Sol. $\quad \ell=1, h=19$

$\therefore \quad$ Range $=h-\ell=19-1=18$Ans.

Ex. 15 Find the range of the following frequency distribution :

Class - Interval	Frequency
$0-5$	6
$5-10$	8
$10-15$	12
$15-20$	5
$20-25$	4

Sol. The range is the difference between the mid value of the least class-interval and the greatest class interval.

Mid value of least class interval $=\frac{0+5}{2}=2.5$

Mid value of greatest class interval $=\frac{20+25}{2}=22.5$

$\therefore \quad$ Range $=22.5-2.5=20 \quad$Ans.

$$EXERCISE$$

OBJECTIVE DPP # 19.1

~~ 1. The median of following series is $520,20,340,190,35,800,1210,50,80$

(A) 1210

(B) 520

(D) 35

~~ 2. If the arithmetic mean of $5,7,9, x$ is 9 then the value of $x$ is

(A) 11

(B) 15

(D) 16

~~ 3. The mode of the distribution $3,5,7,4,2,1,4,3,4$ is

(A) 7

(B) 4

(D) 1

~~ 4. If the mean and median of a set of numbers are 8.9 and 9 respectively, then the mode will be

(A) 7.2

(B) 8.2

(D) 10.2

~~ 5. A student got marks in 5 subjects in a monthly test is given below :

(A) 2,3,4,5,6, in these obtained marks, 4 is the

(A) Mean and median

(B) Median but no mean(C) Mean but no median

(D) Mode

~~ 6. What is the mode from the following table :

Marks obtained	3	1	23	33	43
Frequency (f)	7	11	15	8	3

(A) 13

(B) 43

(D) 23

~~ 7. If the class intervals in a frequency distribution are (72 - 73.9), (74 - 75.9), (76 - 77.9), (78 - 79.9) etc., the midpoint of the class $(74-75.9)$ is

(A) 74.50

(B) 74.90

(D) 75.00

~~ 8. Which one of the following is not correct -

(A) Statistics is liable to be misused

(B) The data collected by the investigator to be used by himself are called primary data

(D) Statistics do not take into account of individual cases

~~ 9. If the first five elements of a se replaced by $(x_1+5)$, where $i=1,2,3, \ldots{ }^{5}$ and the next five elements are replaced by $(x_i-5)$, where $=6 \ldots . .10$ then the mean will change by

(A) 25

(B) 10

(D) 0

~~ 10. The following numbers are given $61,62,63,61,63,64,64,60,65,63,64,65,66,64$. The difference between their mean and median is

(A) 0.4

(B) 0.3

(D) 0.1

~~ 11. The value of $\sum_{i=1}^{n}(x_i-\bar{x})$ where $\bar{x}$ is the arithmetic mean of $x_1$ is

(A) 1

(B) $n \bar{x}$

(D) None of these

~~ 12. The average of 15 numbers is 18 . The average of first 8 is 19 and that last 8 is 17 , then the 8 th number is

(A) 15

(B) 16

(D) 20

~~ 13. In an examination, 10 students scores the following marks in Mathematics 35, 19, 28, 32, 63, 02, 47, 31, 13, ~~ 98. It rage is

(A) 96

(B) 02

(D) 50

Direction : question 15 is based on the histogram given in the adjacent figure.

~~ 14. The percentage of students in science faculty in 1990-91 is :

(A) $26.9 %$

(B) $27.8 %$

(D) $30.2 %$

~~ 15. For the scores $8,6,10,12,1,5,6$ and 6 the Arithmetic mean is

(A) 6.85

(B) 6.75

(D) 7

Direction : Each question from 16 to 18 is based on the histogram given in the adjacent figure.

~~ 16. What is the number of worker earning Rs. 300 to 350 ?

(A) 50

(B) 40

(D) 130

~~ 17. In which class interval of wages there is the least number of workers ?

(A) 400 - 450

(B) 350 - 400

(D) 200 - 250

~~ 18. What is the upper limit of the class-interval 200-250

(A) 200

(B) 250

(D) None of these

SUBJECTIVE DPP # 19.2

~~ 1. Find the mean of following data $13,17,16,14,11,13, 10,16,11,18,12,17$.

~~ 2. Find the median of following data $38,70,48,34,42,55,63,46,54,44$.

~~ 3. Find the mode of following data 2,2,6,5,4,3,4,5,7,9, 4,5,3,1,10,~~ **4. ** ~~ 4. Find the median of :

(i) $5,30,15,6,18,22,26,32,6,9,18$

(ii) $92,88,62,53,55,59,60,61,85,89$

(iii) $66,69,108,72,78,82,98,99,102,101$

~~ 5. Find the value of pm if the median of following observations is 48 .

$14,17,33,35, p-5, p+7,57,63,69,80$. The above observation are in ascending order.]

~~ 6. Find the missing frequencies of the following distribution if it is known that mean of the distribution is 50 .

$\mathrm{x:} \quad \begin{matrix} 10 & 30 & 50 & 70 & 90 & \text { Total }\end{matrix} $

~~ 7. Find the mean for following data.

Age (Years)	$25-30$	$30-35$	$35-40$	$40-45$	$45-50$	$50-55$
No. of teachers	30	23	20	14	10	3

~~ 8. Calculate the mean of the following frequency distribution :

Marks	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$	$70-80$	$80-90$
No. of Students	3	6	13	15	14	5	4

~~ 9. The mean of a certain group of observations is ~~ 78. Find the resulting mean, if the value of each observation is :

(i) increased by 2

(ii) decreased by 3

(iii) multiplied by 1.5

(iv) divided by 2

(v) increased by $30 %$

(iv) diminished by $25 %$

~~ 10. Draw a histogram to represent the following data :

Class- Interval	$40-60$	$60-80$	$80-100$	$100-120$	$120-140$	$140-160$	$160-180$	$180-200$
Frequency	20	40	30	50	30	20	10	40

~~ 11. Draw a bar-graph to represent the following

A	B	C	D	E	F
60	70	55	40	90	50

$$\text{ANSWER KEY}$$

(Objective DPP 19.1)

Qus.	1	2	3	4	5	6	7	8	9	10
Ans.	$C$	$B$	$B$	$C$	$A$	$D$	$C$	$C$	$D$	$B$
Qus.	11	12	13	14	15	16	17	18
Ans.	$C$	$C$	$A$	$C$	$B$	$A$	$D$	$B$

~~ 1. 14

~~ 2. $\quad 47$

~~ 3. 4

~~ 4. (i) 18

$8 . \quad 55.33$

~~ 5. $\quad P=47$

~~ 6. $\quad 28.24$

~~ 7. $\quad 35.5$

~~ 8. 55.33

~~ 9. (i) 80 (ii) 75 (iii) 117 (iv) 39 (v) 101.4 (iv) 58.5

$\ggg$
PROBABILITY
$\lll$

ML - 20

PROBABILITY

Theory of probability deals with measurement of uncertainty of the occurrence of same event or incident in terms of percentage or ratio.

(i) Sample Space : Set of possible out comes.

(ii) Trial : Trial is an action which results in one of several outcomes.

(iii) An experiment : An experiment is any kind of activity such as throwing a die, tossing a coin, drawing a card. outcome of an experiment. The different possibilities which can occur during an experiment. e.g. on throwing a dice, 1 dot, 2 dots, 3 dots, 4 dots, 5 dots, 6 dots can occur.

(iv) An event : getting a ‘six’, in a throw of dice, getting a head, in a toss of a coin.

(v) A random experiment : Whenever we do some experiment at once.

(vi) Equally likely outcomes : there are equal uncertainty in getting 1 dot, 2 dots, 3 dots, 4 dots, 5 dots, 6 dots when we throw a single dice.

(vii) Probability of an event A: Written as $P**(A)$ in a random experiment and is defined as -**

$ P(A)=\frac{\text { Number of outcomes in favour of } A}{\text { Total number of possible outcomes }} $

(a) Important Properties :

(i) $0 \leq P(A) \leq 1$

(ii) $P($ not happening of $(A)+P($ happening of $A)=1$

$ \begin{matrix} \text { or } & P(\bar{A})=P(A)=1 \\ \therefore & P(\bar{A})=1-P(A) \end{matrix} $

Probability of the happening of $A=\frac{\text { Number of favourable outcomes }}{\text { Total number possible outcomes }}$

$ \frac{m}{m+n} $

Probability of not happening of $A($ falling of $A)=\frac{n}{m+n}$

where is for an event $A$ can happen in $m$ ways and fail in $n$ ways all these ways being equally likely to occur.

(b) Problems of Die :

(i) A die is thrown once. What is the probability of -

(A) Getting an even number in the throwing of a die, the total number of outcomes is 6 .

Let be the event of getting an even number then there are three even numbers $2,4,6$.

$\therefore \quad$ number of favourable outcomes $=3$.

$\therefore \quad P(A)=\frac{\text { no. of faourable outcomes }}{\text { total no. of outcomes }}=\frac{3}{6}=\frac{1}{2}$.

(B) Getting an odd number (A) total outcomes $=6$, favourable outcomes $=3(1,3,5)$

$\therefore \quad P(A)=\frac{3}{6}=\frac{1}{2}$

(D) Getting a number which is multiple of 2 and $3=\Big(\frac{\text { Fabourable cases }}{6}\Big)$

(E) Getting a number $\geq 3(3,4,5,6)$

$ P(A)=\frac{4}{6}=\frac{2}{3} $

(F) Getting a number 5 or $6(5$ or 6$) P(A)=\frac{2}{6}=\frac{1}{3}$

(G) Getting a number $\leq 5 P(A)=\frac{5}{6}(1,2,3,4,5)$

(c) Problems Concerning Drawing a Card :

(i) A pack of 52 cards

(ii) Face cards (King, Queen, Jack)

Ex. 1 A card is drawn from a well shuffled deck of 52 cards. Find the probability of (i) A king.

(ii) A heart.

(iii) A seven of heart.

(iv) A jack, queen or a king.

(v) A two of heart or a two of diamond.

(vi) A face card.

(vii) A black card.

(viii) Neither a heart nor a king.

(ix) Neither an ace nor a king.

Sol. Total no. of outcomes $=52$

(i) A king.

No. of kings $=4$ (favorable cases) $ P(A)=\frac{4}{42}=\frac{1}{13} $

(ii) A heart $P(A)=\frac{13}{52}=\frac{1}{4}$

(iii) A seven of heart $P(A)=\frac{1}{52}$

(iv) A jack, queen or a king $ P(A)=\frac{12}{52}=\frac{3}{13} $

(v) A two of heat or a two of diamond. $P(A)=\frac{2}{52}=\frac{1}{26}$

(vi) A face card $ P(A)=\frac{12}{52}=\frac{3}{13} $

(vii) A black card $ P(A)=\frac{26}{52}=\frac{1}{2} $

(viii) Neither a heart nor a king (13 heart +4 king, but 1 common)

$ P(A)=1-\frac{16}{52}=\frac{52-16}{52}=\frac{36}{52}=\frac{9}{13} $

(ix) Neither an ace nor a king. $ P(A)=\frac{44}{52}=\frac{11}{13} $

Ex. 2 Two coins are tossed simultaneously. Find the probability of getting (i) two heads (ii) at least one head (iii) no head

$\therefore$ On tossing two coins simultaneously, all the possible outcomes are

$ \text { HH, HT, TH, TT. } $

(i) The probability of getting two heads $=P(HH)$

$ =\frac{\text { Even of occurence of two heads }}{\text { Total number of possible outcomes }}=\frac{1}{4} $

(ii) The probability of getting at least on head

$ =\frac{\text { Favourable outcomes }}{\text { Total no. of outcomes }}=\frac{3}{4} $

(iii) The probability of getting no head $P(TT)=\frac{1}{4}$

Ex. 3 A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. If one ball is drawn at random, find the probability that it is

(i) Black

(ii) Not red

(iii) Green

Sol. Number of red balls in the bag $=5$

Number of white balls in the bag $=8$

Number of green balls in the bag $=4$

Number of black balls in the bag $=7$

$\therefore$ Total number of balls in the bag $\quad=5+8+4+7=24$.

Drawing balls randomly are equally likely outcomes.

$\therefore \quad$ Total number of possible outcomes $=24$

Now,

(i) There are 7 black balls, hence the number of such favourable outcomes $=7$

$\therefore \quad \text{Probability of drawing a black ball} =\frac{\text{Number of favourable outcomes}}{\text {Total number of possible outcomes}}=\frac{7}{24}$

Ans.

(ii) There are 5 red balls, hence the number of such favourable outcomes $=5$.

$\therefore \quad$ Probability of drawing a red ball $=\frac{\text { Numbe of favourable outcomes }}{\text { Total number of possible outcoes }}=\frac{5}{24}$

Ans.

$\therefore \quad$ Probability of drawing not a red ball $=P($ Not Red ball $)=1-\frac{5}{24}=\frac{19}{24}$

Ans.

(iii) There are 4 green balls.

$\therefore \quad$ Number of such favourable outcomes $=4$

Probability of drawing a green ball $=\frac{\text { Number of favourable outcomes }}{\text { Total number of possibl outcomes }}=\frac{4}{24}=\frac{1}{6}$

Ans.

Ex. 4 A card is drawn from a well - shuffled deck of playing cards. Find the probability of drawing

(i) a face card

(ii) a red face card

Sol. Random drawing of cards ensures equally likely outcomes

(i) Number of face cards (King, Queen and jack of each suits) $=4 \times 3=12$

Total number of cards in deck $=52$

$\therefore \quad$ Total number of possible outcomes $=52$

$P($ drawing a face card $)=\frac{12}{52}=\frac{3}{13}$

(ii) Number of red face cards $=2 \times 3=6$

Number of favourable outcomes of drawing red face card $=6$

$P($ drawing of red face red $)=\frac{6}{52}=\frac{3}{26} \quad$Ans.

$$EXERCISE$$

OBJECTIVE DPP - 20.1

~~ 1. 3 Coins are tossed simultaneously. The probability of getting at least 2 heads is

(A) $\frac{3}{10}$

(B) $\frac{3}{4}$

(D) $\frac{1}{2}$

~~ 2. Two cards are drawn successively with replacement from a pack of 52 cards. The probability of drowsing two aces is

(A) $\frac{1}{169}$

(B) $\frac{1}{221}$

(D) $\frac{4}{663}$

~~ 3. In a single throw of two dice, the probability of getting more than 7 is

(A) $\frac{7}{36}$

(B) $\frac{7}{12}$

(D) $\frac{5}{36}$

~~ 4. Two cards are drawn at random from a pack of 52 cards. The probability that both are the cards of space is

(A) $\frac{1}{26}$

(B) $\frac{1}{4}$

(D) Nome of these

~~ 5. Two dice are thrown together. The probability that sum of the two numbers will be a multiple of 4 is

(A) $\frac{1}{9}$

(B) $\frac{1}{3}$

(D) $\frac{5}{9}$

~~ 6. If the odds in favour of an event be $3: 5$ then the probability of non-happening of the event is

(A) $\frac{3}{5}$

(B) $\frac{5}{3}$

(D) $\frac{5}{8}$

~~ 7. In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.

(A) 0.8

(B) 0.6

(D) 0.2

~~ 8. If the three coins are simultaneously tossed again compute the probability of 2 heads coming up.

(A) $\frac{3}{8}$

(B) $\frac{1}{4}$

(D) $\frac{3}{4}$

~~ 9. A coin is tossed successively three times. The probability of getting one head or two heads is :

(A) $2 / 3$

(B) $3 / 4$

(D) $1 / 9$

~~ 10. One card is drawn from a pack of 52 cards. What is the probability that the drawn card is either red or king:

(A) $15 / 26$

(B) $1 / 2$

(D) $17 / 32$

SUBJECTIVE DPP - 20.2

~~ 1. Two dice are thrown together. Find the probability of getting a total of 9 .

~~ 2. A coin and a dice are tossed simultaneously find the sample space.

~~ 3. A dice is thrown repeatedly until a six comes up. What is the sample space for this experiment.

~~ 4. On a simultaneous toss of three coins, find the probability of getting

(i) at least 2 heads

(ii) at most 2 heads

(iii) exactly 2 heads

~~ 5. Two dice are thrown simultaneously. Find the probability of getting

(i) an even number s the sum

(ii) the sum as a prime number

(iii) a doubled of even number

~~ 6. Three dice are thrown together. Find the probability of getting a total of a least 6 .

~~ 7. Find the probability that a leap year selected at random will contain 53 Tuesday.

~~ 8. A coin is tossed 80 times with the following outcomes :

(i) head : 35

(ii) tail : 45

Find the probability of each event.

~~ 9. Two coins are tossed simultaneously 150 times and we get the following outcomes.

(a) No tail $=45$

(b) One tail $=55$

Find the probability of each event.

~~ 10. In a cricket match a batsman hits a boundary 10 times out of 36 balls be play. Find the probability that he did not hit the boundary.

~~ 11. In a cricket match a batsman hits a boundary 3 times in 3 over he play. Find the probability that the did not hit the boundary.

~~ 12. A bag which contains 7 blue marbles, 4 black marbles and 9 white marbles. A marbles drawn at random from the bag then what is the probability that the drawn marble is

(i) blue

(ii) white or black

~~ 13. The odds in favour of an event are $3: 5$ find the probability of occurrence of this event.

~~ 14. Cards marked with the numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box, find the probability that the number on card is

(i) An even number

(ii) A number less than 14

(iii) A number which is a prefect square.

(iv) A prime number less than 20

~~ 15. An urn contains 6 oranges, 7 apples \And 11 mango. A fruit is drawn at random, what is the probability of drawing.

(i) An orange

(ii) Not apple

(iii) An apple or a mango

~~ 16. A card is drawn at random from a well shuffled desk of playing cards. Find the probability that the card drawn is

(i) A card of spade or an ace

(ii) A red king

(iii) Neither a king nor a queen

(iv) Either a king or a queen

~~ 17. A box contains 19 balls bearing numbers $1,2,3 \ldots . .19$. A ball is drawn at random from the box. Find the probability that the number on the balls is

(i) A prime number

(ii) Divisible by 3 or 5

(iii) Neither divisible by 5 nor by 10

(iv) An even number

~~ 18. There are 30 cards of same size in a bag containing numbers 1 to 30 . One card is taken out from the bag at random. Find the probability that the number on the selected card is not divisible by ~~ **3. **

$$EXERCISE$$

(Objective DPP # 20.1)

Qus.	1	2	3	4	5	6	7	8	9	10
Ans.	D	A	C	C	C	D	A	A	B	C

(Subjective DPP # 20.2)

~~ 1. $\frac{1}{9}$

~~ 2. $(H, 1)(H, 2)(H, 3)(H, 4)(H, 5)(T, 1)(T, 2)(T, 3)(T, 4)(T, 5)(T, 6)$

~~ 3. ${6,(1,6)(2,6)(3,6)(4,6)(5,6)(1,1,6)(1,2,6) \ldots}$.

~~ 4. $\Big(\frac{1}{2}, \frac{7}{8}, \frac{3}{8}\Big)$

~~ 5. $\Big(\frac{1}{2}, \frac{5}{12}, \frac{1}{12}\Big)$

~~ 6. $\Big(\frac{103}{108}\Big)$

~~ 7. $\Big(\frac{2}{7}\Big)$

~~ 8. (i) $\frac{7}{16}$ (ii) $\frac{9}{16}$

~~ 9. $\begin{matrix} \text { (a) } \frac{3}{10} & \text { (b) } \frac{11}{30} & \text { (c) } \frac{1}{3}\end{matrix} $

~~ 10. $\frac{13}{18}$

~~ 11. $\frac{3}{8}$

~~ 12. $\begin{matrix} \text { (i) } \frac{7}{20} & \text { (ii) } \frac{13}{20}\end{matrix} $

~~ 13. $\frac{3}{8}$

~~ 14. (i) $\frac{1}{2}$

(ii) $\frac{3}{25}$

(iii) $\frac{2}{25} \frac{9}{100}$

(iv) $\frac{2}{25}$

~~ 15. (i) $\frac{1}{4}$

(ii) $\frac{17}{24}$ (iii) $\frac{3}{4}$

~~ 16. (i) $\frac{4}{13}$

(ii) $\frac{1}{26}$ (iii) $\frac{11}{13}$ (iv) $\frac{2}{13}$

~~ 17. $\frac{8}{19}, \frac{8}{19}, \frac{16}{19}, \frac{9}{19}$ ~~ 18. $\frac{2}{3}$

$\ggg$
PROOF IN MATHEMATICS $\lll$

ML - 21

STATEMENT

is a sentence which is neither an order nor a question nor an exclamatory sentence.

A sentence or statement can be

(a) a true statement (b) a false statement (c) an ambiguous statement

Examples for true statement

(i) $\quad 1+3=4,7+3=10$.

(ii) The number of days in a week is seven.

(iii) When $2 x=10$, then $x=5$.

(iv) There are three sides in a triangle.

(v) New Delhi the capital of India.

Examples for false statement

(i) $2+7=5$ is a false statement

$9 \times 2=15$ is a false statement

(iii) $\quad 1 m=1000 cm$ is a false statement

(iv) Patna is the capital of West Bengal is a false statement

(v) Sunday comes after Monday is a false statement

Examples for ambiguous statement

$\begin{matrix} \text { (i) The 7th of Mach falls on Monday. } & \text { (ii) The sum of any two angles of a triangle is } 110^{\circ} \text {. }\end{matrix} $

(iii) Today is Friday.

Mathematically valid statement

Mathematically, a statement is valid or acceptable only if it is either always true or always false.

Deduction :

Deductive reasoning : To find the truth value of an unambiguous statement we use the deductive reasoning. This is the main logical tool.

AXIOM CONJECTURE AND THEOREM

(a) Axiom : Axiom is a statement which is accepted as a true statement. An axiom does not require a proof.

Example :

(i) $\quad a=b, b=c \Rightarrow \quad a=c$

(ii) $\quad a>b, b>c \Rightarrow \quad a>c$

(iii) $\quad a=b \Rightarrow \quad \frac{1}{2} a=\frac{1}{2} b$

(b) Conjecture : It is a statement whose truth ness or falseness has not been established mathematically.

(c) Theorem : A theorem is a mathematical statement whose truth has been established logically.

Proof of a theorem

The main parts of a proof are as under.

(i) The Hypothesis (i.e. what is given)

(ii) The conclusion (i.e. what is to be proved)

(iii) Consists of successive mathematical statements derived logically from the previous statement or axiom or hypothesis.

Ex. 1 State whether the following statements are always true, always false or ambiguous, Justify your answer.

(i) There are 13 months in a year.

(ii) Diwali falls on Friday.

(iii) The temperature in Magadi is $26^{\circ} C$.

(iv) Dogs can fly.

(v) February has only 28 days.

Sol. (i) This statement is false because there are 12 months in a year.

(ii) This statement is always ambiguous because Diwali can fall on any day.

(iii) This statement is always ambiguous because it is not fixed.

(iv) This statement is always false.

(v) This is a false statement because February has 29 days in a leap year.

Ex. 2 State whether the following statements are true or false. Give reasons for your answers.

(i) The sum of the interior angles of a quadrilateral is $350^{\circ}$.

(ii) For any real number $x, x^{2} \geq 0$.

(iii) A rhombus is a parallelogram.

(iv) The sum of two even numbers is even.

(v) The sum of two odd numbers is odd.

Sol. (i) This statement is false because the sum of the interior angles of a quadrilateral is $360^{\circ}$.

(ii) This statement in always true. For example $(-2)^{2}=4$, then we can say $x^{2} \geq 0$ for any real number $x$.

(iii) This statement is always true.

(iv) This statement is always true. For example, $2+2=4$ and $6+4=10$.

(v) This statement is always false. For example, $3+5=7$ and $3+9=10$.

Ex. 3 State whether the following statements are true or false :

(i) Opposite angles of a cyclic quadrilateral are supplementary.

(ii) Every odd number greater than 1 is prime.

(iii) Exterior angle of a cyclic quadrilateral is equal to the opposite angle.

(iv) For any real number $x, 5 x+x=6 x$.

(v) For every real number $x, x^{3} \geq x$.

(vi) An exterior angle is greater than each interior opposite angle.

Sol. (i) This statement is true.

(ii) This statement is false ; for example, 9 is not a prime number

(iii) This statement is true.

(iv) This statement is true.

(v) This statement is false, for example $\Big(\frac{1}{2}\Big)^{3}=\frac{1}{8}$ and $\frac{1}{8}$ is not greater than $\frac{1}{2}$.

(vi) This statement is true.

Ex. 4 Restate the following statements with appropriate condition so that they become true statements.

(i) Square of a real number is always greater than the number.

(ii) In a parallelogram the diagonals are equal.

(iii) There are four angles is a triangle.

Sol. (i) Square of a real number is always greater than the number when the magnitude of the number is greater than one.

(ii) In a rectangle, the diagonals are equal.

(iii) There are three and only three angles in a triangle.

Ex. 5 Restate the following statements with appropriate conditions, so that they become true statements.

(i) All prime numbers are odd.(ii) Two times a real numbers is always even.

(iii) For any $x, 3 x+1>4$. (iv) For any $x, x^{3} \geq 0$.

(v) In an equilateral triangle the medians are also an angle bisector.

Ex. 6 The sum of the angles of a triangle is $180^{\circ}$

Sol. Statement : The sum of the angles of a triangles is $180^{\circ}$

Given : $A \triangle ABC$

To prove : $\angle 1+\angle 2+\angle 3=180^{\circ}$

Construction : Through A, draw a line DE parallel to BC.

Proof :

S.No.	Statement	Reason
~~
1.	DE \|\| BC and AC is the traversal $\therefore \angle 1 = \angle 4$	Alternative interior $\angle$ s
~~
2.	Again DE \|\| BC and AC is the traversal $\therefore \angle 2 = \angle 2$ Adding (1) and (2)	Alternative interior $\angle$ s
~~
3.	$\angle 1+ \angle 2 = \angle 4 + \angle 5$	Adding the corresponding side of (1) \And (2)
~~
4.	$\angle 1+ \angle 2 + \angle 3 = \angle 4 + \angle 5 + \angle 3$	Adding $\angle$ 3 on the sides
~~
5.	But $\angle 4 + \angle 5 + \angle 3 = \angle DAE = 180 \degree$	$\angle$ DAE is a straight line angle
~~
6.	$\angle 1+ \angle 2 + \angle 3 = 180 \degree$	The sum of the angles of a triangle is 180 $\degree$

Ex. 7 For each natural number, $n(n+1)$ is multiple of 2 .

Sol. We have to prove that the product $(n+1)$ is divisible by 2 .

Now we have two cases. Either is even or odd. Let us examine each case. Suppose $n$ is even. Then we can write $n=2 m$, for same natural number $m$. And, then

$n(n+1)=2 m(2 m+1)$ which is clearly divisible by 2 .

Next, suppose $n$ is odd. Then $n+1$ is even and we can write $n+1=2 r$, for some natural number 2 .

We have $n(n+1)=(2 r-12 r=2 r(2 r-1)$ which is clearly divisible by 2 .

So, we can say that the natural number $n(n+1)$ is divisible by 2 .

$$EXERCISE$$

SUBJECTIVE DPP # 22

~~ 1. Write down the truth value of each of the following statements.

(i) India is a democratic country.

(ii) Each prime number has exactly two factor.

(iii) $\sqrt{2}$ is an irrational number.

(iv) Jaipur is in U.P.

~~ 2. Write down the negation of the following

(i) Hindi is the mother tongue of India.

(ii) India is progressing rapidly.

(iii) $(a+b)^{3}=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}$

(iv) 2 is the real part of $2+4 i$.

(v) 4 is multiple of ~~ 20. (vi) 2nd October is the birthday of Mahatma Gandhi.

(vii) Republic day of India held o 26th January.

(viii) The roots of the equation $x^{4}+4 x^{3}+6 x^{2}+4 x+1=0$ are equal.

(ix) New-York is in England.

~~ 3. State whether the following statements are true or false. Give reason for your answers

(i) $1 m=100 cm$

(ii) The isosceles triangles have to sides equal.

(iii) The sum of two odd number is even.

(iv) Three and three makes six.

(v) February has 30 days.

Prove that all $n \in N$ (Q. No. 4 to 6 )

~~ 4. $\quad 1^{2}+2^{2}+3^{2}+4^{2}+\ldots . .+n^{2}=\frac{n(n+1(2 n+1)}{6}$

~~ 5. $\quad 2^{2 n}-1$ is divisible by 8 .

~~ 6. $\quad 1+3+5+\ldots . .+(2 n-1)=n^{2}$.

~~ 7. Whish of the following sentences are statements :

(i) It is hot day.

(ii) Qutubminar is in Lucknow.

(iii) Don’t talk, please

(iv) Hurrah! India has won the match.

(v) Rasika is a sincere girl.

(vi) Will it rain today?

(vii) $2+3=5$.

(viii) $\quad 5+7=10$.

(ix) $x+2=11$.

(x) Every prime number has only one factor.

~~ 8. A number can be divided into three equal parts if the sum of its digits is divisible by 3 . Based on the above statement can 9875340 be divided into three equal groups.

~~ 9. Look at the following pattern :

$11^{1}=11=$

$11^{2}=121=$

$11^{3}=1331=$

$11^{4}=14641=$

$11^{5}=161051=$

$11^{6}=1771561=$

Is 19487171 power of ~~ 11. [Hint : Sum of digits at odd places - Sum of digits at even places $=0$ ]

~~ 10. $a^{2}+b^{2}$ is a prime for all whole numbers $a, b$.

$$\text{ANSWER KEY}$$

(Subjective DPP # 22)

~~ 1. (i) $T$

(ii) $T$

(iii) $T$

(iv) F

~~ 2. (i) Hindi is not the mother tongue of India

(ii) India is not progressing rapidly.

(iii) $(a+b)^{3} \neq a^{3}+3 a^{2} b+3 a b^{2}+b^{3}$.

(iv) 2 is not the real part of $2+4 i$.

(v) 4 is not multiple of 20 .

(vi) 2nd October is not the birthday of Mahatma Gandhi.

(vii) Republic day of India held not on 26th January.

(viii) The roots of the equation $x^{4}+4 x^{3}+6 x^{2}+4 x+1=0$ are not equal.

(ix) New-York is not in England.

~~ 3. (i) $T$

(ii) $T$

(iii) $T$

(iv) $F$

(v) F

~~ 7. (i), (ii). (v), (vii), (ix)

~~ 8. Number 9875340 can be divided into three equal groups.

~~ 9. $Y$ Yes $11^{7}=19487171=[1+4+7+7]-[9+8+1+1]=0$

~~ 10. For $a=3, b=4, a^{2}+b^{2}$, is not a prime.

$\ggg$
MATHEMATICAL MODELLING
$\lll$

ML-22

Definition : mathematical model is a mathematical relation that describes some real life situation. e.g. To find the area of an equilateral triangle we can use

Area $=\frac{\sqrt{3}}{4}(\text { side })^{2}$

This formula is an example of mathematical model.

Ex. 1 A car travelled 416 kilometres on 52 litres of petrol. I have to go by same car to a place which is $96 km$ away. How much petrol do I need?

Following steps include to solve the problem.

Sol. Formulation : Farther we travel, the more petrol we require, that is, the amount of petrol we need varies directly with the distance we travel.

Petrol needed for travelling $416 km=52$ litres

Petrol needed for travelling $96 km=$ ?

Mathematical Description :

Let $x=$ distance traveled

$y=$ need of Petrol and $y$ varies directly with $x$

So, $y=Kx$, where $K$ is a constant.

I can travel 416 kilometres with 52 litres of petrol.

So, $y=52, x=416$

$\therefore \quad K=\frac{y}{x}=\frac{52}{416}=\frac{1}{8}$

$\therefore \quad y=\frac{1}{8} x \quad \ldots \text{(i)}$

Equation (1) describes the relationship between the petrol needed and distance travelled.

Step 2 : We want to find the petrol we need to travel 96 kilometres. So we have to find the value of $y$ when $x=96$. Putting $x=96$ in (1), we have $=\frac{96}{8}=12$

Step 3 . Interpretation:

Since $y=12$, we need 12 litres of petrol to travel 96 kilometres.

Step ~~ 4. Validation of the result :

This result is valid only if all the conditions remain same i.e. mileage of car track on which car is running, gradient of the track (road), etc.

Ex. 2 Suppose Rakesh has invested Rs. 20,000 at 12% simple interest per year. With the return from the investment, he wants to buy a colour. T.V. that cost Rs. 25,~~ 000. For what period should he invest Rs. 20,000 so that he has enough money to buy a colour T.V.?

Sol. Step ~~ 1. Formulation of the problem :

Here, we know the principal and the rte of interest in the amount Rakesh needs in addition to Rs. 20,000 to buy the colour T.V. We have to find the number of years.

Mathematical Description :

The formula for simple interest is

$ \text { S.I }=\frac{\text { Pnr. }}{100} $

Where $P=$ Principal

$n=$ Number of years

$r %=$ Rate of interest

S.I. $=$ Interest earned

Here, the principal $=$ Rs. 20,000

The money required by Rakesh for buying a colour T.V. $=$ Rs. 25000

So, the interest to be earned $=$ Rs. $(25,000-20,000)=$ Rs. 5,000

The number of years for which Rs. 20,000 is deposited $=n$

The Interest of Rs. 20,000 for $n$ years at the rate of $12 %=$ S.I.

Then, $\quad$ S.I. $=\frac{20,000 \times n \times 12}{100}$

So, $\quad$ S.I. $=2400 n \quad \ldots (2)$

Give the relationship between the number of years and interest, if Rs. 20,000 is invested an annual interest rate of $12 %$. We have to find the period in which the interest earned is Rs. 5,000 . Putting S.I. $=5,000$ in (1), we have

$ 5,000=2400 n $

Step ~~ 2. Solution of the problem :

Solving equations (2), we get

$ n=\frac{5000}{2400}=\frac{50}{24}=2 \frac{1}{12} $

Step 3 : Interpretation : Since $n=2 \frac{1}{12}$ and one twelfth of a year is one month Rakesh can buy a colour T.V. after 2 years and one month.

Step ~~ 4. : Validation of result

We have to assume that the interest rate remains the same for the period for which we calculate the interest.

Otherwise, the formula S.I. $=\frac{pnr}{100}$ will into be valid. We have also assumed that the price of the colour T.V. machine does not increase by the time $2 \frac{1}{12}$ year.

Ex. 3 A motor boat goes upstream on a river and covers the distance between two town on the river bank in \And hours. It covers this distance downstream in five hours. If the speed of the stream is $4 km / h$, find the speed of the boat in still water.

Sol. Step 1 : Formulation : We know the speed of the river and the time taken to cover the distance between two places. We have to find the speed of the boat in still water.

Mathematical Description : Let us write $x$ for the speed of the boat, $t$ for the time taken and $y$ for the distance travelled. Then $y=tx \quad \ldots (1)$

Let $d$ be the distance between the two places. While going upstream.

The actual speed of boat $=$ speed of the boat $\boldsymbol{\text { speed }}$ of the river

$\therefore$ The boat is travelling against the flow of the river.

So, the speed of the boat in upstream $=(x-4) km / h$.

It takes 8 hours to cover the distance between the towns upstream. So from (1), we have

$ d=8(x+4) \quad \ldots (2) $

When going downstream,

The speed of the boat in downstream $=(x+4) km / h$

The boat takes five hours to cover the same distance downstream, so

$ d=5(x+4) \quad \ldots (3) $

From (2) and (3), we have

$ 5(x+4)=8(x-4) $

Step ~~ 2. Finding the solution.

Solving for a in equation (4), we get $x=\frac{52}{3}$

Step ~~ 3. Interpretation.

Since $x=\frac{52}{3}$, therefore the speed of the motorboat in still water is $\frac{52}{3} km / h$.

We have assumed that

~~ 1. The speed of the river and the boat remains constant all the time.

~~ 2. The effect of the friction between the boat and water and the friction due to air is negligible.

Step ~~ 4. Validation of result :

The speed of the motor boat is $\frac{52}{3} km / h$ and the distance between two towns,

$ \begin{aligned} & y=8(x-4) \\ \Rightarrow \quad & y=8\Big(\frac{52}{3}-4\Big) \\ \Rightarrow \quad & y=8\Big(\frac{52-12}{3}\Big) \\ \Rightarrow \quad & y=\frac{8 \times 40}{3}=\frac{320}{3} \\ \Rightarrow \quad & y=106.66 km . \end{aligned} $

Hence the distance between two towns $=106.66 km$.

Ex. 4 Four hundred entrance tickets were sold for a school fair. The cost of the ticket for adults was Rs. 20 and that for students was Rs. ~~ 10. The total collection from of the sale of entrance tickets was Rs. 6000 . How many adults visited the fair?

Sol. Step ~~ 1. Formulation of the problem

We know that

The total number of tickets sold $=400$

The cost of a ticket for adults $=$ Rs. 20

The cost of the ticket for student $=$ Rs. 10

and the total proceedings were $=$ Rs. 6000

Mathematical formulation :

Let the number of adults who visited the fair be $x$

$\therefore \quad$ number of students visited $=(400-x)$

Total amount received from adults $=$ Rs. $20 x$

Total amount received from students $=$ Rs. $10(400-x)$

Total amount collected $=$ Rs. 6000

$\therefore \quad$ the model (relation) is

$ 20 x+10(400-x)=6000 $

Step ~~ 2. Finding the solutions

$ 20 x+4000-10 x=6000 $

$\Rightarrow 10 x+4000=6000$

$\Rightarrow 10 x=2000$

$\Rightarrow x=200$

Step ~~ 3. Interpretation of the solution

We assumed that the number of adults who visited the fair was $x \therefore 200$ adults visited the fair.

Step ~~ 4. Validation of the result

No. of adults who visited the fair $=200$

No. of students who visited the fair $=(400-200)=200$

$ \begin{aligned} \therefore \quad & \text { Total receipts } \\ & =\text { Rs }(20 \times 200)+\text { Rs. }(10 \times 200) \\ & =\text { Rs. } 4000+\text { Rs. } 2000 \\ & =\text { Rs. } 6000 \end{aligned} $

Thus, the total collection from 200 adults and 200 students is Rs. 6000.

Ex. 5 The price of sugar has gone up by $40 %$. By what percent should a family reduce the consumption of sugar so that the expenditure on sugar may remain the same?

Sol. Let us consider various steps of mathematical modeling and solve this problem.

Step ~~ 1. Formulation of the problem : Price of sugar goes up by $40 %$ i.e. if the family spends Rs. 100 on sugar, then with the increase in price, the family will have to spend Rs. 140. But the family decides not to increase the expenditure, instead it prefers to reduce the consumption of sugar.

Mathematical formulation :

Suppose the family consumed $\times kg$ of sugar for Rs. 100 before the price hike.

$\therefore \quad$ the increased prince of $x k$ of sugar is Rs. 140 .

Quantity of sugar that can be bought for Rs. 100 at the increased price $=\frac{x}{140} \times 100=\frac{10 x}{14} kg$

$\therefore \quad$ reduction in quantity of sugar

$=x-\frac{10 x}{14}=\frac{4 x}{14} kg$

Step ~~ 2. Finding the solution : Percent reduction in consumption of sugar $=\frac{\frac{4 x}{14}}{x} \times 100=\frac{400}{14}=\frac{200}{7}$

Step ~~ 3. Interpretation of the solution :

The family should reduce the consumption of sugar by $28 \frac{4}{7} %$

Step ~~ 4. Validation of result :

After increase in price of sugar, the amount of sugar bought by the family.

$=\Big(100-28 \frac{4}{7}\Big) kg=\Big(100-\frac{200}{7}\Big) kg$

$=\frac{700-200}{7} kg=\frac{500 \times 2}{7 \times 2}=\frac{1000}{14} kg$

The cost of $\frac{1000}{14} kg$ sugar at Rs. 140 for $100 kg$

$=$ Rs. $\frac{1000}{14} \times \frac{140}{100}$ Rs. 100

Hence the result

Ex. 6 Suppose company need a computer for some period of time. The company can either hire a computer for Rs. 2,000 per month or buy one for Rs. 25,000. If the company has to use the computer for a long period, the company will pay such a high rent, that buying a computer will be cheaper. On the other hand, if the company has to use the computer for say, just one month, then hiring a computer will be cheaper. For the number of months beyond which it will be cheaper to buy a computer.

Sol. Step ~~ 1. Formulation :

We know that the company can hire a computer for Rs. 2,000 per month or the company can buy the compute for Rs. 25,000 . The company has to use the computer for 0 just one month then the hiring the computer will be cheaper. Here we have find out number of months beyond which it will be cheaper to buy a computer.

Mathematical formulation :

Let the number of months beyond which it will be cheaper to buy a compute $=x$ months.

Rate of hiring computer $=$ Rs. 2,000 per month

The amount of hiring a computer for $x$ months $=$ Rs. $2,000 x$

The cost of the compute $=$ Rs. 25,000

The company will not have to pay more if cost of computer is less than the hiring charges for computer.

Step ~~ 2. Solution :

$\Rightarrow \frac{25,000}{2,000}<x$

$\Rightarrow \frac{25}{2}<x$

Step ~~ 3. Interpretation :

If $\frac{25}{2}<x$. The least value of $x$ is 13 month (more than $12 \frac{1}{2}$ ) It will be cheaper for the company to buy a computer if it has to hire a computer 13 months or more than 13 months.

we have assumed that :

(i) The rate of hiring a computer remains same throughout the period.

(ii) After 13 months the cost of computer may not increase

Ex. 7 We have given the timings of the gold medalists in the 400-metre race from the time the event was included in the Olympics, in the table below. Construct a mathematical model relating the years and timings. Use it to estimate the timing in the next Olympics.

Year	Timing (in seconds)
1964	52.01
1968	52.03
1972	51.08
1976	49.28
1978	48.88
1984	48.83
1988	48.65
1992	48.83
1996	48.25
2000	49.11
2004	49.41

Sol. Formulation : In the figure, the times of the gold medalist of 400 metres race are given of the Olympics (1964 - 2004). We take 1964 as zeroth years and write 1 for 1968, 2 for 1972 and 3 for 1976. We prepare a new table.

Year	Timing (in seconds)
0	52.01
1	52.03
2	51.08
3	49.28
4	48.88
5	48.83
6	48.65
7	48.83
8	48.25
9	49.11
10	49.41

The reduction in timings of gold medalist in 400 metres rave in Olympics given in the following table.

Year	Timings	Change in timings
0	52.01	0
1	52.03	+0.02
2	51.08	-0.95
3	49.28	+1.80
4	48.88	-0.40
5	48.3	-0.05
6	48.65	-0.18
7	48.83	+0.18
8	48.25	-0.58
9	49.11	+.086
10	49.41	+0.30

At the end of 4 years period from 1964 - 1968 the timing has increased by 0.02 second from 52.01 to 52.03 second.

At the end of second Olympic the reduction in timing is 0.95 second from 52.03 to 51.08 . From the table above we cannot find a definite relationship between the number of years and change in timing. But the reduction is fairly steady except in the first $7^{\text {th }}$ year, $9^{\text {th }}$ year and $10^{\text {th }}$ year.

The mean of the value is

$ \begin{aligned} & =\frac{0.02-0.95-1.80-0.40-0.05-0.18+0.18-0.58+0.86+0.30}{10} \\ & =\frac{-2.6}{10}=-0.26 \end{aligned} $

Le us assume that the timings in $400 m$ race of Olympic reduced at the rate of 0.26 per Olympic.

Mathematical description :

We have assumed that the timings reduces at the rate of 0.26 second per Olympic.

So, the reduction in timings in the first Olympic $=52.01-0.26$

Reduction in the second Olympic $=52.01-0.26-0.26=52.01-2 \times 0.26$

Reduction in the third Olympic $=52.01-0.26-0.26-0.26=52.01-3 \times 0.26$

So, the reduction in the $11^{\text {th }}$ Olympic $=52.01-11 \times 0.26$

Now, we have to estimate the timings in the next ( $11^{\text {th }}$ ) Olympic i.e., 2008

But the timings in the $n^{\text {th }}$ Olympic will be $=52.01-0.26 n \quad \ldots (1)$

Step ~~ 2. Solution : Substituting $n=11$, in (1), we get

$ 52.01-0.26 \times 11=52.01-2.86=49.15 $

Step ~~ 3. Interpretation : The timings for $400 m$ race in the next Olympic i.e., (2008) is estimated as $49.15 sec$.

Step ~~ 4. Validation : Let us check if formula (1) is in agreement with the reality. Let us find the values for the years we already know using formula (1) and compare it with known values by finding the difference.

$$EXERCISE$$

SUBJECTIVE DPP # 23

~~ 1. A sailor goes $8 kg$ downstream in 40 minutes and returns in 1 hour. Determine the speed of the sailor in still water and the speed of the current.

~~ 2. While covering a distance of $30 km$, Ajeet takes 2 hours more than Amit. If Ajeet doubles his speed,. he would take 1 hour less than Amit. Find their speed of walking.

~~ 3. Places A and B are $80 k m$ apart from each other on a highway. A car starts from A and other from B at the same speed. If they move in the same direction, they meet in 8 hours and if they more in opposite directions, they meet in 1 hour and 20 minutes. Find the speed of the cars.

~~ 4. A bag contains one rupee, 50 paise and 25 paise coin in the ratio $5: 6: 7$. if total amount is Rs. 390 , find the number of coins of each kind.

~~ 5. The ages of two person are in the ratio of $5: 7$. Sixteen years ago, the ratio was $3: 5$. Find their present ages.

~~ 6. Suppose Sudhir has invested Rs. 15,000 at $8 %$ simple interest per year with the return from the investment. He wants to buy a washing machine that costs Rs. 19,000 For what period should he invest Rs. 15,000 so that he has enough money to buy a washing machine?

~~ 7. A motorboat goes upstream on a river and covers the distance between two points on the riverbank in six hours. It covers this distance downstream in five hours. If the speed of the steam is $2 km / hr$. find the speed of the boat in still water.

~~ 8. Suppose you have a room of length $6 m$ and breadth $5 m$. You want to cover the floor of the room with square mosaic tiles of side $30 cm$. How many tiles will you need? Solve this by constructing a mathematical model.

~~ 9. A travelled 432 kilometers on 48 litres of petrol in my car. I have to go by my cat to a place which is $180 km$ away. how much petrol do I need ?

~~ 10. Suppose a car starts from Delhi at a speed of $70 km / h$ towards Chandigarh. At that instance, a motorcycle starts from Chandigarh towards Delhi at a speed of $55 km / h$. If the distance between Delhi and Chandigarh is $250 km$, after how much time will the car and motorcycle meet?

$$\text{ANSWER KEY}$$

(Subjective DPP # 23)

~~ 1. **Speed of sailor $=10 km / h$., Speed of current $=2 km / h$.

~~ 2. **Ajeet’s speed $=5 km / h$., Amit’s speed $= 7.5 km / h$.

~~ 3. ** $35 km / h ., 25 km / h$.

~~ 4. **No. of one rupee coins $=200$; No. of 50 paise coin $=240$; No. of 25 paise coin $=280$

~~ 5. **40 and 56 years.

~~ 6. **3 years, 4 months

~~ 7. ** $22 km / hr$.

~~ 8. ** $\quad 340$ tiles

~~ 9. **20 litres

~~ 10. **2 hours

knowledge-route Class 9 Maths

》 ) NUMBER SYSTEM $\ll«$

CLASSIFICATION OF NUMBERS

IDENTIFICATION PRIME NUMBER

REPRESENTATIO FO RATIONAL NUMBER OF A REAL NUMBER LINE

(b) Find Rational Numbers Between Two Integral Numbers :

RATIONAL NUMBER IN DECIMAL REPRESENTATION

PROPERTIES OF RATIONAL NUMBER

GEOMETRICAL REPRESENTATION OF REAL NUMBERS

> > > NUMBER SYSTEM < < <

SURDS

LAWS OF SURDS

OPERATION OF SURDS

RATIONALIZATION OF SURDS

EXPONENTS OF REAL NUMBER

RATIONAL EXPONENTS OR A REAL NUMBER

LAWS OF RATIONAL EXPONETNS

ILLUSTRATIONS :

$$EXERCISE$$

$$ANSWER KEY$$

> > > POLYNOMIALS < < <

POLYNOMIALS

ALGEBRAIC IDENTITY

ILLUSTRATION

FACTORS OF A POLYNOMIAL

(a) Factorisation by Making a Trinomial a Perfect Square :

(b) Factorisation by Using the Formula for the Difference of Two Squares :

(c) Factorisation by Using Formula of $a^{3}+b^{3}$ and $a^{3}-b^{3}$ :

$$EXERCISE$$

$\ggg$ POLYNOMIALS

$\lll$

ZEROS OR ROOTS OF A POLYNOMIAL

REMAINDER THEOREM

FACTORISATION OFA QUADRATIC POLYNOMIAL

$$EXERCISE$$

$$ANSWER KEY$$

$ \ggg \text{COORDINATE GEOMETRY} \lll $

CO-ORDINATE SYSTEM

CARTESIAN CO-ORDINATE SYSTEM

ILLUSTRATIONS :

DISTANCE BETWEEN TWO POINTS

$$EXERCISE$$

$$\text{ANSWER KEY}$$

LINEAR EQUATION IN $> > >$ TWO VARIABLES $< < <$

LINEAR EQUATIONS IN ONE VARIABLE

LINEAR EQUATION IN TWO VARIABLES

GRAPH OF A LINEAR EQUATION

DIFFERENCE FORMS OF A LINE

SOLUTION OF LINEAR EQUATION IN ONE VARIABLE

SOLUTION OF LINEAR EQUATIONS IN TWO VARIABLE

$EXERCISE$

$$\text{ANSWER KEY}$$

$> > >$ INTRODUCTION $< < <$

TO EUCLID’S GEOMETRY

INTRODUCTION

BASIC CONCEPTS IN GEOMETRY

SOME IMPORTANT DEFINITOINS

$$\text{EXERCISE}$$

$$\text{ANSWER KEY}$$

$> > > \text{LINES AND ANGLES} < < <$

IMPORTANT THEOREMS

ILLUSTRATIONS

$$\text{EXERCISE}$$

$$\text{ANSWER KEY}$$

$\quad \ggg$ TRIANGLES $\quad \lll$

TRIANGLE

CONGRUENT FIGURES

SOME INEQUALITY RELATIONS IN A TRIANGLE

$$EXERCISE$$

$$\text{ANSWER KEY}$$

$\ggg$ QUADRILATERIAL $\lll$

QUADRILATERL

ALTERNATE SOLUTION :

PARALLELOGRAM

$$EXERCISE$$

$\quad \ggg$ QUADRILATERAL $\lll$

MID-POINT THEOREM

ALTERNATIVE METHOD :

$$EXERCISE$$

$$\text{ANSWER KEY}$$

> > >
NUMBER SYSTEM < < <

$\ggg$
POLYNOMIALS

$\quad \ggg$
TRIANGLES
$\quad \lll$

$\ggg$
QUADRILATERIAL
$\lll$

$\quad \ggg$
QUADRILATERAL
$\lll$

$> > >$
CIRCLE
$< < <$

$> > >$
CONSTRCUTIONS
$< < <$

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HERON’S FORMULA
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$\ggg$
SURFACE AREA AND VOLUME $\lll«$

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STATISTICS
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PROBABILITY
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PROOF IN MATHEMATICS $\lll$

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MATHEMATICAL MODELLING
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