Chapter 1

REAL NUMBERS

(A) Main Concepts and Results

  • Euclid’s Division Lemma : Given two positive integers a and b, there exist unique integers q and r satisfying a=bq+r,0r<b.

  • Euclid’s Division Algorithm to obtain the HCF of two positive integers, say c and d, c>d.

Step 1 : Apply Euclid’s division lemma to c and d, to find whole numbers q and r, such that c=dq+r,0r<d.

Step 2 : If r=0,d is the HCF of c and d. If r0, apply the division lemma to d and r.

Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.

  • Fundamental Theorem of Arithmetic : Every composite number can be expressed as a product of primes, and this expression (factorisation) is unique, apart from the order in which the prime factors occur.

  • Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer.

  • 2,3,5 are irrational numbers.

  • The sum or difference of a rational and an irrational number is irrational.

  • The product or quotient of a non-zero rational number and an irrational number is irrational.

  • For any two positive integers a and bHCF(a,b)×LCM(a,b)=a×b.

  • Let x=pq,p and q are co-prime, be a rational number whose decimal expansion terminates. Then, the prime factorisation of q is of the form 2m.5n;m,n are non-negative integers.

  • Let x=pq be a rational number such that the prime factorisation of q is not of the form 2m.5n;m,n being non-negative integers. Then, x has a non-terminating repeating decimal expansion.

(B) Multiple Choice Questions

Choose the correct answer from the given four options:

Sample Question 1 : The decimal expansion of the rational number 3322.5 will terminate after

(A) one decimal place

(B) two decimal places

(C) three decimal places

(D) more than 3 decimal places

Solution : Answer (B)

Sample Question 2 : Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that a=bq+r, where r must satisfy

(A) 1<r<b

(B) 0<rb

(C) 0r<b

(D) 0<r<b

Solution : Answer (C)

EXERCISE 1.1

Choose the correct answer from the given four options in the following questions:

~~ 1. For some integer m, every even integer is of the form

(A) m

(B) m+1

(C) 2m

(D) 2m+1

~~ 2. For some integer q, every odd integer is of the form

(A) q

(B) q+1

(C) 2q

(D) 2q+1

~~ 3. n21 is divisible by 8 , if n is

(A) an integer

(B) a natural number

(C) an odd integer

(D) an even integer

~~ 4. If the HCF of 65 and 117 is expressible in the form 65m117, then the value of m is

(A) 4

(B) 2

(C) 1

(D) 3

~~ 5. The largest number which divides 70 and 125 , leaving remainders 5 and 8 , respectively, is

(A) 13

(B) 65

(C) 875

(D) 1750

~~ 6. If two positive integers a and b are written as a=x3y2 and b=xy3;x,y are prime numbers, then HCF(a,b) is

(A) xy

(B) xy2

(C) x3y3

(D) x2y2

~~ 7. If two positive integers p and q can be expressed as p=ab2 and q=a3b;a,b being prime numbers, then LCM(p,q) is

(A) ab

(B) a2b2

(C) a3b2

(D) a3b3

~~ 8. The product of a non-zero rational and an irrational number is

(A) always irrational

(B) always rational

(C) rational or irrational

(D) one

~~ 9. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

(A) 10

(B) 100

(C) 504

(D) 2520

~~ 10 The decimal expansion of the rational number 145871250 will terminate after:

(A) one decimal place

(B) two decimal places

(C) three decimal places

(D) four decimal places

(C) Short Answer Questions with Reasoning

Sample Question 1: The values of the remainder r, when a positive integer a is divided by 3 are 0 and 1 only. Justify your answer.

Solution : No.

According to Euclid’s division lemma,

a=3q+r, where 0r<3

and r is an integer. Therefore, the values of r can be 0,1 or 2 .

Sample Question 2: Can the number 6n,n being a natural number, end with the digit 5? Give reasons.

Solution : No, because 6n=(2×3)n=2n×3n, so the only primes in the factorisation of 6n are 2 and 3 , and not 5 .

Hence, it cannot end with the digit 5.

EXERCISE 1.2

~~ 1. Write whether every positive integer can be of the form 4q+2, where q is an integer. Justify your answer.

~~ 2. “The product of two consecutive positive integers is divisible by 2 “. Is this statement true or false? Give reasons.

~~ 3. “The product of three consecutive positive integers is divisible by 6 “. Is this statement true or false”? Justify your answer.

~~ 4. Write whether the square of any positive integer can be of the form 3m+2, where m is a natural number. Justify your answer.

~~ 5. A positive integer is of the form 3q+1,q being a natural number. Can you write its square in any form other than 3m+1, i.e., 3m or 3m+2 for some integer m ? Justify your answer.

~~ 6. The numbers 525 and 3000 are both divisible only by 3,5,15,25 and 75 . What is HCF (525,3000) ? Justify your answer.

~~ 7. Explain why 3×5×7+7 is a composite number.

~~ 8. Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons.

~~ 9. Without actually performing the long division, find if 98710500 will have terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer.

~~ 10 A rational number in its decimal expansion is 327.7081 . What can you say about the prime factors of q, when this number is expressed in the form pq ? Give reasons.

(D) Short Answer Questions

Sample Question 1: Using Euclid’s division algorithm, find which of the following pairs of numbers are co-prime:

(i) 231,396

(ii) 847,2160

Solution : Let us find the HCF of each pair of numbers.

(i) 396=231×1+165

231=165×1+66

165=66×2+33

66=33×2+0

Therefore, HCF=33. Hence, numbers are not co-prime.

(ii) 2160=847×2+466

847=466×1+381

466=381×1+85

381=85×4+41

85=41×2+3

41=3×13+2

3=2×1+1

2=1×2+0

Therefore, the HCF=1. Hence, the numbers are co-prime.

Sample Question 2: Show that the square of an odd positive integer is of the form 8m+1, for some whole number m.

Solution : Any positive odd integer is of the form 2q+1, where q is a whole number.

Therefore, (2q+1)2=4q2+4q+1=4q(q+1)+1,

q(q+1) is either 0 or even. So, it is 2m, where m is a whole number.

Therefore, (2q+1)2=4.2m+1=8m+1.

[From (1)]

Sample Question 3: Prove that 2+3 is irrational.

Solution : Let us suppose that 2+3 is rational. Let 2+3=a, where a is rational.

Therefore, 2=a3

Squaring on both sides, we get

2=a2+32a3

Therefore, 3=a2+12a, which is a contradiction as the right hand side is a rational number while 3 is irrational. Hence, 2+3 is irrational.

EXERCISE 1.3

~~ 1. Show that the square of any positive integer is either of the form 4q or 4q+1 for some integer q.

~~ 2. Show that cube of any positive integer is of the form 4m,4m+1 or 4m+3, for some integer m.

~~ 3. Show that the square of any positive integer cannot be of the form 5q+2 or 5q+3 for any integer q.

~~ 4. Show that the square of any positive integer cannot be of the form 6m+2 or 6m+5 for any integer m.

~~ 5. Show that the square of any odd integer is of the form 4q+1, for some integer q.

~~ 6. If n is an odd integer, then show that n21 is divisible by 8 .

~~ 7. Prove that if x and y are both odd positive integers, then x2+y2 is even but not divisible by 4.

~~ 8. Use Euclid’s division algorithm to find the HCF of 441, 567, 693.

~~ 9. Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively.

~~ 10 Prove that 3+5 is irrational.

~~ 11 Show that 12n cannot end with the digit 0 or 5 for any natural number n.

~~ 12 On a morning walk, three persons step off together and their steps measure 40cm, 42cm and 45cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?

~~ 13 Write the denominator of the rational number 2575000 in the form 2m×5n, where m,n are non-negative integers. Hence, write its decimal expansion, without actual division.

~~ 14 Prove that p+q is irrational, where p,q are primes.

(E) Long Answer Questions

Sample Question 1: Show that the square of an odd positive integer can be of the form 6q+1 or 6q+3 for some integer q.

Solution : We know that any positive integer can be of the form 6m,6m+1,6m+2, 6m+3,6m+4 or 6m+5, for some integer m.

Thus, an odd positive integer can be of the form 6m+1,6m+3, or 6m+5 Thus we have:

(6m+1)2=36m2+12m+1=6(6m2+2m)+1=6q+1,q is an integer (6m+3)2=36m2+36m+9=6(6m2+6m+1)+3=6q+3,q is an integer (6m+5)2=36m2+60m+25=6(6m2+10m+4)+1=6q+1,q is an integer. 

Thus, the square of an odd positive integer can be of the form 6q+1 or 6q+3.

EXERCISE 1.4

~~ 1. Show that the cube of a positive integer of the form 6q+r,q is an integer and r=0,1,2,3,4,5 is also of the form 6m+r.

~~ 2. Prove that one and only one out of n,n+2 and n+4 is divisible by 3 , where n is any positive integer.

~~ 3. Prove that one of any three consecutive positive integers must be divisible by 3 .

~~ 4. For any positive integer n, prove that n3n is divisible by 6 .

~~ 5. Show that one and only one out of n,n+4,n+8,n+12 and n+16 is divisible by 5 , where n is any positive integer.

[Hint: Any positive integer can be written in the form 5q,5q+1,5q+2,5q+3, 5q+4].

Chapter 2

POLYNOMIALS

(A) Main Concepts and Results

  • Geometrical meaning of zeroes of a polynomial: The zeroes of a polynomial p(x) are precisely the x-coordinates of the points where the graph of y=p(x) intersects the x-axis.

  • Relation between the zeroes and coefficients of a polynomial: If α and β are the zeroes of a quadratic polynomial ax2+bx+c, then α+β=ba,αβ=ca.

  • If α,β and γ are the zeroes of a cubic polynomial ax3+bx2+cx+d, then α+β+γ=ba,αβ+βγ+γα=ca and αβγ=da.

  • The division algorithm states that given any polynomial p(x) and any non-zero polynomial g(x), there are polynomials q(x) and r(x) such that p(x)=g(x)q(x)+r(x), where r(x)=0 or degree r(x)< degree g(x).

(B) Multiple Choice Questions

Choose the correct answer from the given four options:

Sample Question 1: If one zero of the quadratic polynomial x2+3x+k is 2 , then the value of k is

(A) 10

(B) -10

(C) 5

(D) -5

Solution : Answer (B)

Sample Question 2: Given that two of the zeroes of the cubic polynomial ax3+bx2+cx+d are 0 , the third zero is

(A) ba

(B) ba

(C) ca

(D) da

Solution : Answer (A). [ Hint: Because if third zero is α, sum of the zeroes .=α+0+0=ba]

EXERCISE 2.1

Choose the correct answer from the given four options in the following questions:

~~ 1. If one of the zeroes of the quadratic polynomial (k1)x2+kx+1 is -3 , then the value of k is

(A) 43

(B) 43

(C) 23

(D) 23

~~ 2. A quadratic polynomial, whose zeroes are -3 and 4 , is

(A) x2x+12

(B) x2+x+12

(C) x22x26

(D) 2x2+2x24

~~ 3. If the zeroes of the quadratic polynomial x2+(a+1)x+b are 2 and -3 , then

(A) a=7,b=1

(B) a=5,b=1

(C) a=2,b=6

(D) a=0,b=6

~~ 4. The number of polynomials having zeroes as -2 and 5 is

(A) 1

(B) 2

(C) 3

(D) more than 3

~~ 5. Given that one of the zeroes of the cubic polynomial ax3+bx2+cx+d is zero, the product of the other two zeroes is

(A) ca

(B) ca

(C) 0

(D) ba

~~ 6. If one of the zeroes of the cubic polynomial x3+ax2+bx+c is -1 , then the product of the other two zeroes is

(A) ba+1

(B) ba1

(C) ab+1

(D) ab1

~~ 7. The zeroes of the quadratic polynomial x2+99x+127 are

(A) both positive

(B) both negative

(C) one positive and one negative

(D) both equal

~~ 8. The zeroes of the quadratic polynomial x2+kx+k,k0,

(A) cannot both be positive

(B) cannot both be negative

(C) are always unequal

(D) are always equal

~~ 9. If the zeroes of the quadratic polynomial ax2+bx+c,c0 are equal, then

(A) c and a have opposite signs

(B) c and b have opposite signs

(C) c and a have the same sign

(D) c and b have the same sign

~~ 10 If one of the zeroes of a quadratic polynomial of the form x2+ax+b is the negative of the other, then it

(A) has no linear term and the constant term is negative.

(B) has no linear term and the constant term is positive.

(C) can have a linear term but the constant term is negative.

(D) can have a linear term but the constant term is positive.

~~ 11 Which of the following is not the graph of a quadratic polynomial?

alt text

(C) Short Answer Questions with Reasoning

Sample Question 1: Can x1 be the remainder on division of a polynomial p(x) by 2x+3 ? Justify your answer.

Solution : No, since degree (x1)=1= degree (2x+3).

Sample Question 2: Is the following statement True or False? Justify your answer. If the zeroes of a quadratic polynomial ax2+bx+c are both negative, then a,b and c all have the same sign.

Solution : True, because ba= sum of the zeroes <0, so that ba>0. Also the product of the zeroes =ca>0.

EXERCISE 2.2

~~ 1. Answer the following and justify:

(i) Can x21 be the quotient on division of x6+2x3+x1 by a polynomial in x of degree 5 ?

(ii) What will the quotient and remainder be on division of ax2+bx+c by px3+qx2+rx+s,p0 ?

(iii) If on division of a polynomial p(x) by a polynomial g(x), the quotient is zero, what is the relation between the degrees of p(x) and g(x) ?

(iv) If on division of a non-zero polynomial p(x) by a polynomial g(x), the remainder is zero, what is the relation between the degrees of p(x) and g(x) ?

(v) Can the quadratic polynomial x2+kx+k have equal zeroes for some odd integer k>1 ?

~~ 2. Are the following statements ‘True’ or ‘False’? Justify your answers.

(i) If the zeroes of a quadratic polynomial ax2+bx+c are both positive, then a,b and c all have the same sign.

(ii) If the graph of a polynomial intersects the x-axis at only one point, it cannot be a quadratic polynomial.

(iii) If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.

(iv) If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.

(v) If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign.

(vi) If all three zeroes of a cubic polynomial x3+ax2bx+c are positive, then at least one of a,b and c is non-negative.

(vii) The only value of k for which the quadratic polynomial kx2+x+k has equal zeros is 12

(D) Short Answer Questions

Sample Question 1: Find the zeroes of the polynomial x2+16x2, and verify the relation between the coefficients and the zeroes of the polynomial.

Solution : x2+16x2=16(6x2+x12)=16[6x2+9x8x12]

=16[3x(2x+3)4(2x+3)]=16(3x4)(2x+3)

Hence, 43 and 32 are the zeroes of the given polynomial.

The given polynomial is x2+16x2.

The sum of zeroes =43+32=16= Coefficient of x Coefficient of x2 and

the product of zeroes =43×32=2= Constant term  Coefficient of x2

EXERCISE 2.3

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:

~~ 1. 4x23x1

~~ 2. 3x2+4x4

~~ 3. 5t2+12t+7

~~ 4. 2x2+72x+34

~~ 5. 2s2(1+22)s+2

~~ 6. y2+325y5

~~ 7. t32t215t

~~ 8. 4x2+52x3

~~ 9. v2+43v15

~~ 10 7y2113y23

(E) Long Answer Questions

Sample Question 1: Find a quadratic polynomial, the sum and product of whose zeroes are 2 and 32, respectively. Also find its zeroes.

Solution : A quadratic polynomial, the sum and product of whose zeroes are

2 and 32 is x22x32x22x32=12[2x222x3]=12[2x2+2x32x3]=12[2x(2x+1)3(2x+1)]=12[2x+1][2x3]

Hence, the zeroes are 12 and 32.

Sample Question 2: If the remainder on division of x3+2x2+kx+3 by x3 is 21 , find the quotient and the value of k. Hence, find the zeroes of the cubic polynomial x3+2x2+kx18.

Solution : Let p(x)=x3+2x2+kx+3

Then, p(3)=33+2×32+3k+3=21

i.e., 3k=27

i.e., k=9

Hence, the given polynomial will become x3+2x29x+3.

Now,

x3)x3+2x29x+3(x2+5x+6.x33x25x29x+35x215x6x+36x1821

So, x3+2x29x+3=(x2+5x+6)(x3)+21

i.e., x3+2x29x18=(x3)(x2+5x+6)

=(x3)(x+2)(x+3)

So, the zeroes of x3+2x2+kx18 are 3,2,3.

EXERCISE 2.4

~~ 1. For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also find the zeroes of these polynomials by factorisation.

(i) 83,43

(ii) 218,516

(iii) 23,9

(iv) 325,12

~~ 2. Given that the zeroes of the cubic polynomial x36x2+3x+10 are of the form a, a+b,a+2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.

~~ 3. Given that 2 is a zero of the cubic polynomial 6x3+2x210x42, find its other two zeroes.

~~ 4. Find k so that x2+2x+k is a factor of 2x4+x314x2+5x+6. Also find all the zeroes of the two polynomials.

~~ 5. Given that x5 is a factor of the cubic polynomial x335x2+13x35, find all the zeroes of the polynomial.

~~ 6. For which values of a and b, are the zeroes of q(x)=x3+2x2+a also the zeroes of the polynomial p(x)=x5x44x3+3x2+3x+b ? Which zeroes of p(x) are not the zeroes of q(x) ?

Chapter 3

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

(A) Main Concepts and Results

  • Two linear equations in the same two variables are said to form a pair of linear equations in two variables.

  • The most general form of a pair of linear equations is

a1x+b1y+c1=0

a2x+b2y+c2=0,

where a1,a2,b1,b2,c1,c2 are real numbers, such that a12+b120,a22+b220.

  • A pair of linear equations is consistent if it has a solution - either a unique or infinitely many.

In case of infinitely many solutions, the pair of linear equations is also said to be dependent. Thus, in this case, the pair of linear equations is dependent and consistent.

  • A pair of linear equations is inconsistent, if it has no solution.

  • Let a pair of linear equations in two variables be a1x+b1y+c1=0 and a2x+b2y+c2=0.

(I) If a1a2b1b2, then

(i) the pair of linear equations is consistent,

(ii) the graph will be a pair of lines intersecting at a unique point, which is the solution of the pair of equations.

(II) If a1a2=b1b2c1c2, then

(i) the pair of linear equations is inconsistent,

(ii) the graph will be a pair of parallel lines and so the pair of equations will have no solution.

(III) If a1a2=b1b2=c1c2, then

(i) the pair of linear equations is dependent, and consistent,

(ii) the graph will be a pair of coincident lines. Each point on the lines will be a solution, and so the pair of equations will have infinitely many solutions.

  • A pair of linear equations can be solved algebraically by any of the following methods:

(i) Substitution Method

(ii) Elimination Method

(iii) Cross- multiplication Method

  • The pair of linear equations can also be solved geometrically/graphically.

(B) Multiple Choice Questions

Choose the correct answer from the given four options:

Sample Question 1 : The pair of equations 5x15y=8 and 3x9y=245 has

(A) one solution

(B) two solutions

(C) infinitely many solutions

(D) no solution

Solution : Answer (C)

Sample Question 2 : The sum of the digits of a two-digit number is 9 . If 27 is added to it, the digits of the number get reversed. The number is

(A) 25

(B) 72

(C) 63

(D) 36

Solution : Answer (D)

EXERCISE 3.1

Choose the correct answer from the given four options:

~~ 1. Graphically, the pair of equations

6x3y+10=02xy+9=0

represents two lines which are

(A) intersecting at exactly one point.

(B) intersecting at exactly two points.

(C) coincident.

(D) parallel.

~~ 2. The pair of equations x+2y+5=0 and 3x6y+1=0 have

(A) a unique solution

(B) exactly two solutions

(C) infinitely many solutions

(D) no solution

~~ 3. If a pair of linear equations is consistent, then the lines will be

(A) parallel

(B) always coincident

(C) intersecting or coincident

(D) always intersecting

~~ 4. The pair of equations y=0 and y=7 has

(A) one solution

(B) two solutions

(C) infinitely many solutions

(D) no solution

~~ 5. The pair of equations x=a and y=b graphically represents lines which are

(A) parallel

(B) intersecting at (b,a)

(C) coincident

(D) intersecting at (a,b)

~~ 6. For what value of k, do the equations 3xy+8=0 and 6xky=16 represent coincident lines?

(A) 12

(B) 12

(C) 2

(D) -2

~~ 7. If the lines given by 3x+2ky=2 and 2x+5y+1=0 are parallel, then the value of k is

(A) 54

(B) 25

(C) 154

(D) 32

~~ 8. The value of c for which the pair of equations cxy=2 and 6x2y=3 will have infinitely many solutions is

(A) 3

(B) -3

(C) -12

(D) no value

~~ 9. One equation of a pair of dependent linear equations is 5x+7y=2. The second equation can be

(A) 10x+14y+4=0

(B) 10x14y+4=0

(C) 10x+14y+4=0

(D) 10x14y=4

~~ 10 A pair of linear equations which has a unique solution x=2,y=3 is

(A) x+y=1

(B) 2x+5y=11 2x3y=5 4x+10y=22

(C) 2xy=1

(D) x4y14=0 3x+2y=0 5xy13=0

~~ 11 If x=a,y=b is the solution of the equations xy=2 and x+y=4, then the values of a and b are, respectively

(A) 3 and 5

(B) 5 and 3

(C) 3 and 1

(D) -1 and -3

~~ 12 Aruna has only Re 1 and Rs 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs 75, then the number of Re1 and Rs 2 coins are, respectively

(A) 35 and 15

(B) 35 and 20

(C) 15 and 35

(D) 25 and 25

~~ 13 The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, in years, of the son and the father are, respectively

(A) 4 and 24

(B) 5 and 30

(C) 6 and 36

(D) 3 and 24

(C) Short Answer Questions with Reasoning

Sample Question 1: Is it true to say that the pair of equations

x+2y+2=0 and 12x14y1=0

has a unique solution? Justify your answer.

Solution : Yes.

Here, a1a2=112=2,b1b2=214=8

As a1a2b1b2, the pair of equations has a unique solution.

Sample Question 2: Do the equations 4x+3y1=5 and 12x+9y=15 represent a pair of coincident lines? Justify your answer.

Solution : No.

We may rewrite the equations as

4x+3y=6

12x+9y=15

Here, a1a2=13,b1b2=13 and c1c2=25

As a1a2=b1b2c1c2, the given equations do not represent a pair of coincident lines.

Sample Question 3: Is the pair of equations x+2y3=0 and 6y+3x9=0 consistent? Justify your answer.

Solution : Yes.

Rearranging the terms in the equations, we get

x+2y3=03x+6y9=0

Here, a1a2=13,b1b2=13,c1c2=13. As a1a2=b1b2=c1c2, the pair of equations is consistent.

EXERCISE 3.2

~~ 1. Do the following pair of linear equations have no solution? Justify your answer.

(i) 2x+4y=3 12y+6x=6

(ii) x=2y y=2x

(iii) 3x+y3=0 2x+23y=2

~~ 2. Do the following equations represent a pair of coincident lines? Justify your answer.

(i) 3x+17y=3 7x+3y=7

(ii) -2 x-3 y=1 6 y+4 x=-2

(iii) x2+y+25=0

4x+8y+516=0

~~ 3. Are the following pair of linear equations consistent? Justify your answer.

(i) 3x4y=12

4y+3x=12

(ii) 35xy=12

15x3y=16

(iii) 2ax+by=a

4ax+2by2a=0;a,b0

(iv) x+3y=11

2(2x+6y)=22

~~ 4. For the pair of equations

λx+3y=72x+6y=14

to have infinitely many solutions, the value of λ should be 1 . Is the statement true? Give reasons.

~~ 5. For all real values of c, the pair of equations

x2y=8

5x10y=c

have a unique solution. Justify whether it is true or false.

~~ 6. The line represented by x=7 is parallel to the x-axis. Justify whether the statement is true or not.

(D) Short Answer Questions

Sample Question 1 : For which values of p and q, will the following pair of linear equations have infinitely many solutions?

4x+5y=2(2p+7q)x+(p+8q)y=2qp+1

Solution :

Here, a1a2=42p+7q

b1b2=5p+8qc1c2=22qp+1

For a pair of linear equations to have infinitely many solutions

a1a2=b1b2=c1c2

So, 42p+7q=5p+8q=22qp+1

So, 42p+7q=5p+8q and 42p+7q=22qp+1

i.e., 4p+32q=10p+35q and 8q4p+4=4p+14q

i.e., 6p+3q=0 and 8p+6q=4

i.e., q=2p (1) and 4p+3q=2

Substituting the value of q obtained from Equation(1) in Equation(2), we get

4p6p=2

orp=1

Substituting the value of p in Equation (1), we get

q=2

So, for p=1,q=2, the given pair of linear equations will have infinitely many solutions.

Sample Question 2: Solve the following pair of linear equations:

21x+47y=11047x+21y=162

Solution : We have

21x+47y=110(1)

47x+21y=162(2)

Multiplying Equation (1) by 47 and Equation (2) by 21, we get

987x+2209y=5170(3)

987x+441y=3402(4)

Subtracting Equation (4) from Equation (3), we get

1768y=1768ory=1

Substituting the value of y in Equation (1), we get

21x+47=110

or 21x=63

or x=3

So, x=3,y=1

Alternative Solution : We have

21x+47y=110(1)

47x+21y=162(2)

Adding Equations (1) and (2), we have

68x+68y=272

x+y=4(5)

Subtracting Equation (1) from Equation (2), we have

26 x-26 y=52 $

x-y=2 (6)

On adding and subtracting Equations (5) and (6), we get

x=3,y=1

Sample Question 3 : Draw the graphs of the pair of linear equations xy+2=0 and 4xy4=0. Calculate the area of the triangle formed by the lines so drawn and the x-axis.

Solution :

For drawing the graphs of the given equations, we find two solutions of each of the equations, which are given in Table 3.1

Table 3.1

x 0 -2
y=x+2 2 0

x 0 1
y=4x4 -4 0

Plot the points A(0,2),B(2,0),P(0,4) and Q(1,0) on the graph paper, and join the points to form the lines AB and PQ as shown in Fig 3.1

Fig. 3.1

We observe that there is a point R(2,4) common to both the lines AB and PQ. The triangle formed by these lines and the x-axis is BQR.

The vertices of this triangle are B (2,0),Q(1,0) and R(2,4).

We know that ;

Area of triangle =12 Base × Altitude

Here, Base =BQ=BO+OQ=2+1=3 units.

Altitude =RM= Ordinate of R=4 units.

So, area of ΔBQR=12×3×4=6 sq. units.

EXERCISE 3.3

~~ 1. For which value(s) of λ, do the pair of linear equations

λx+y=λ2 and x+λy=1 have 

(i) no solution?

(ii) infinitely many solutions?

(iii) a unique solution?

~~ 2. For which value(s) of k will the pair of equations

kx+3y=k312x+ky=k

have no solution?

~~ 3. For which values of a and b, will the following pair of linear equations have infinitely many solutions?

x+2y=1(ab)x+(a+b)y=a+b2

~~ 4. Find the value(s) of p in (i) to (iv) and p and q in (v) for the following pair of equations:

(i) 3xy5=0 and 6x2yp=0, if the lines represented by these equations are parallel.

(ii) x+py=1 and pxy=1, if the pair of equations has no solution.

(iii) 3x+5y=7 and 2px3y=1,

if the lines represented by these equations are intersecting at a unique point.

(iv) 2x+3y5=0 and px6y8=0, if the pair of equations has a unique solution.

(v) 2x+3y=7 and 2px+py=28qy, if the pair of equations have infinitely many solutions.

~~ 5. Two straight paths are represented by the equations x3y=2 and 2x+6y=5. Check whether the paths cross each other or not.

~~ 6. Write a pair of linear equations which has the unique solution x=1,y=3. How many such pairs can you write?

~~ 7. If 2x+y=23 and 4xy=19, find the values of 5y2x and yx2.

~~ 8. Find the values of x and y in the following rectangle [see Fig. 3.2].

Fig. 3.2

~~ 9. Solve the following pairs of equations:

(i) x+y=3.3

0.63x2y=1,3x2y0

(ii) x3+y4=4

5x6y8=4

(iii) 4x+6y=15

6x8y=14,y0

(iv) 12x1y=1

1x+12y=8,x,y0

(v) 43x+67y=24

67x+43y=24

(vi) xa+yb=a+b

xa2+yb2=2,a,b0

(vii) 2xyx+y=32

xy2xy=310,x+y0,2xy0

~~ 10 Find the solution of the pair of equations x10+y51=0 and x8+y6=15.

Hence, find λ, if y=λx+5.

~~ 11 By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them.

(i) 3x+y+4=0

(ii) x2y=6 6x2y+4=0 3x6y=0

(iii) x+y=3

3x+3y=9

~~ 12 Draw the graph of the pair of equations 2x+y=4 and 2xy=4. Write the vertices of the triangle formed by these lines and the y-axis. Also find the area of this triangle.

~~ 13 Write an equation of a line passing through the point representing solution of the pair of linear equations x+y=2 and 2xy=1. How many such lines can we find?

~~ 14 If x+1 is a factor of 2x3+ax2+2bx+1, then find the values of a and b given that 2a3b=4.

~~ 15 The angles of a triangle are x,y and 40. The difference between the two angles x and y is 30. Find x and y.

~~ 16 Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than twice her age. How old are they now?

~~ 17 The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.

~~ 18 Two numbers are in the ratio 5:6. If 8 is subtracted from each of the numbers, the ratio becomes 4:5. Find the numbers.

~~ 19 There are some students in the two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from A to B. But if 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in the two halls.

~~ 20 A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid Rs 22 for a book kept for six days, while Anand paid Rs 16 for the book kept for four days. Find the fixed charges and the charge for each extra day.

~~ 21 In a competitive examination, one mark is awarded for each correct answer while 12 mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?

~~ 22 The angles of a cyclic quadrilateral ABCD are

A=(6x+10),B=(5x)C=(x+y),D=(3y10)

Find x and y, and hence the values of the four angles.

(E) Long Answer Questions

Sample Question 1: Draw the graphs of the lines x=2 and y=3. Write the vertices of the figure formed by these lines, the x-axis and the y-axis. Also, find the area of the figure.

Solution :

We know that the graph of x=2 is a line parallel to y-axis at a distance of 2 units to the left of it.

So, the line l is the graph of x=2 [see Fig. 3.3]

Fig. 3.3

The graph of y=3 is a line parallel to the x-axis at a distance of 3 units above it.

So, the line m is the graph of y=3.

The figure enclosed by the lines x=2,y=3, the x-axis and the y-axis is OABC, which is a rectangle. (Why?)

A is a point on the y-axis at a distance of 3 units above the x-axis. So, the coordinates of A are (0,3);

C is a point on the x-axis at a distance of 2 units to the left of y-axis. So, the coordinates of C are (2,0)

B is the solution of the pair of equations x=2 and y=3. So, the coordinates of B are (2,3)

So, the vertices of the rectangle OABC are O(0,0),A(0,3),B(2,3),C(2,0)

The length and breadth of this rectangle are 2 units and 3 units, respectively.

As the area of a rectangle = length × breadth,

the area of rectangle OABC=2×3=6 sq. units.

Sample Question 2: Determine, algebraically, the vertices of the triangle formed by the lines

5xy=5,x+2y=1 and 6x+y=17

Solution :

The vertex of a triangle is the common solution of the two equations forming its two sides. So, solving the given equations pairwise will give the vertices of the triangle.

From the given equations, we will have the following three pairs of equations:

5xy=5 and x+2y=1x+2y=1 and 6x+y=175xy=5 and 6x+y=17

Solving the pair of equations

5xy=5x+2y=1

we get, x=1,y=0

So, one vertex of the triangle is (1,0)

Solving the second pair of equations

x+2y=16x+y=17

we get x=3,y=1

So, another vertex of the triangle is (3,1)

Solving the third pair of equations

5xy=56x+y=17

we get x=2,y=5.

So, the third vertex of the triangle is (2,5). So, the three vertices of the triangle are (1,0),(3,1) and (2,5).

Sample Question 3 : Jamila sold a table and a chair for Rs 1050, thereby making a profit of 10 on the table and 25 on the chair. If she had taken a profit of 25 on the table and 10 on the chair she would have got Rs 1065. Find the cost price of each.

Solution : Let the cost price of the table be Rs x and the cost price of the chair be Rs y.

The selling price of the table, when it is sold at a profit of 10

=Rsx+10100x=Rs110100x

The selling price of the chair when it is sold at a profit of 25

=Rsy+25100y=Rs125100y

So, 110100x+125100y=1050

When the table is sold at a profit of 25, its selling price =Rs(x+25100x)=Rs125100x

When the chair is sold at a profit of 10, its selling price =Rs(y+10100y)=Rs110100y

So, 125100x+110100y=1065

From Equations (1) and (2), we get

110x+125y=105000 and 125x+110y=106500

On adding and subtracting these equations, we get

235x+235y=211500 and 15x15y=1500 i.e., x+y=900(3) and xy=100(4)

Solving Equations (3) and (4), we get

x=500,y=400

So, the cost price of the table is Rs 500 and the cost price of the chair is Rs 400 .

Sample Question 4: It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half the pool can be filled.How long would it take for each pipe to fill the pool separately?

Solution :

Let the time taken by the pipe of larger diameter to fill the pool be x hours and that taken by the pipe of smaller diameter pipe alone be y hours.

In x hours, the pipe of larger diameter fills the pool.

So, in 1 hour the pipe of larger diameter fills 1x part of the pool, and so, in 4 hours, the pipe of larger diameter fills 4x parts of the pool.

Similarly, in 9 hours, the pipe of smaller diameter fills 9y parts of the pool.

According to the question,

4x+9y=121

Also, using both the pipes, the pool is filled in 12 hours.

So, 12x+12y=12

Let 1x=u and 1y=v. Then Equations (1) and (2) become

4u+9v=123

12u+12v=14

Multiplying Equation (3) by 3 and subtracting Equation (4) from it, we get

15v=12 or v=130

Substituting the value of v in Equation (4), we get u=120

So, u=120,v=130

So, 1x=120,1y=130

or, x=20,y=30.

So, the pipe of larger diameter alone can fill the pool in 20 hours and the pipe of smaller diameter alone can fill the pool in 30 hours.

EXERCISE 3.4

~~ 1. Graphically, solve the following pair of equations:

2x+y=62xy+2=0

Find the ratio of the areas of the two triangles formed by the lines representing these equations with the x-axis and the lines with the y-axis.

~~ 2. Determine, graphically, the vertices of the triangle formed by the lines

y=x,3y=x,x+y=8

~~ 3. Draw the graphs of the equations x=3,x=5 and 2xy4=0. Also find the area of the quadrilateral formed by the lines and the x-axis.

~~ 4. The cost of 4 pens and 4 pencil boxes is Rs 100 . Three times the cost of a pen is Rs 15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box.

~~ 5. Determine, algebraically, the vertices of the triangle formed by the lines

3xy=32x3y=2x+2y=8

~~ 6. Ankita travels 14km to her home partly by rickshaw and partly by bus. She takes half an hour if she travels 2km by rickshaw, and the remaining distance by bus.

On the other hand, if she travels 4km by rickshaw and the remaining distance by bus, she takes 9 minutes longer. Find the speed of the rickshaw and of the bus.

~~ 7. A person, rowing at the rate of 5km/h in still water, takes thrice as much time in going 40km upstream as in going 40km downstream. Find the speed of the stream.

~~ 8. A motor boat can travel 30km upstream and 28km downstream in 7 hours. It can travel 21km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.

~~ 9. A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3 . Find the number.

~~ 10 A railway half ticket costs half the full fare, but the reservation charges are the same on a half ticket as on a full ticket. One reserved first class ticket from the station A to B costs Rs 2530. Also, one reserved first class ticket and one reserved first class half ticket from A to B costs Rs 3810. Find the full first class fare from station A to B, and also the reservation charges for a ticket.

~~ 11 A shopkeeper sells a saree at 8 profit and a sweater at 10 discount, thereby, getting a sum Rs 1008 . If she had sold the saree at 10 profit and the sweater at 8 discount, she would have got Rs 1028 . Find the cost price of the saree and the list price (price before discount) of the sweater.

~~ 12 Susan invested certain amount of money in two schemes A and B, which offer interest at the rate of 8 per annum and 9 per annum, respectively. She received Rs 1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received Rs 20 more as annual interest. How much money did she invest in each scheme?

~~ 13 Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot at the rate of Rs 2 for 3 bananas and the second lot at the rate of Re 1 per banana, and got a total of Rs 400. If he had sold the first lot at the rate of Re 1 per banana, and the second lot at the rate of Rs 4 for 5 bananas, his total collection would have been Rs 460. Find the total number of bananas he had.

Chapter 4

QUADRATIC EQUATIONS

(A) Main Concepts and Results

  • Quadratic equation : A quadratic equation in the variable x is of the form ax2+bx+c=0, where a,b,c are real numbers and a0.

  • Roots of a quadratic equation : A real number α is said to be a root of the quadratic equation ax2+bx+c=0, if aα2+bα+c=0.

  • The roots of the quadratic equation ax2+bx+c=0 are the same as the zeroes of the quadratic polynomial ax2+bx+c.

  • Finding the roots of a quadratic equation by the method of factorisation : If we can factorise the quadratic polynomial ax2+bx+c, then the roots of the quadratic equation ax2+bx+c=0 can be found by equating to zero the linear factors of ax2+bx+c.

  • Finding the roots of a quadratic equation by the method of completing the square : By adding and subtracting a suitable constant, we club the x2 and x terms in the quadratic equation so that they become a complete square, and solve for x.

  • Quadratic Formula: If b24ac0, then the real roots of the quadratic equation ax2+bx+c=0 are given by b2a±b24ac2a.

  • The expression b24ac is called the discriminant of the quadratic equation.

  • Existence of roots of a quadratic equation: A quadratic equation ax2+bx+c=0 has

(i) two distinct real roots if b24ac>0

(ii) two equal real roots if b24ac=0

(iii) no real roots if b24ac<0.

(B) Multiple Choice Questions

Choose the correct answer from the given four options:

Sample Question 1 : Which one of the following is not a quadratic equation?

(A) (x+2)2=2(x+3)

(B) x2+3x=(1)(13x)2

(C) (x+2)(x1)=x22x3

(D) x3x2+2x+1=(x+1)3

Solution : Answer (C)

Sample Question 2 : Which constant should be added and subtracted to solve the quadratic equation 4x23x5=0 by the method of completing the square?

(A) 916

(B) 316

(C) 34

(D) 34

Solution : Answer (B)

EXERCISE 4.1

Choose the correct answer from the given four options in the following questions:

~~ 1. Which of the following is a quadratic equation?

(A) x2+2x+1=(4x)2+3

(B) 2x2=(5x)(2x25)

(C) (k+1)x2+32x=7, where k=1

(D) x3x2=(x1)3

~~ 2. Which of the following is not a quadratic equation?

(A) 2(x1)2=4x22x+1

(B) 2xx2=x2+5

(C) (2x+3)2+x2=3x25x

(D) (x2+2x)2=x4+3+4x3

~~ 3. Which of the following equations has 2 as a root?

(A) x24x+5=0

(B) x2+3x12=0

(C) 2x27x+6=0

(D) 3x26x2=0

~~ 4. If 12 is a root of the equation x2+kx54=0, then the value of k is

(A) 2

(B) -2

(C) 14

(D) 12

~~ 5. Which of the following equations has the sum of its roots as 3 ?

(A) 2x23x+6=0

(B) x2+3x3=0

(C) 2x232x+1=0

(D) 3x23x+3=0

~~ 6. Values of k for which the quadratic equation 2x2kx+k=0 has equal roots is

(A) 0 only

(B) 4

(C) 8 only

(D) 0,8

~~ 7. Which constant must be added and subtracted to solve the quadratic equation 9x2+34x2=0 by the method of completing the square?

(A) 18

(B) 164

(C) 14

(D) 964

~~ 8. The quadratic equation 2x25x+1=0 has

(A) two distinct real roots

(B) two equal real roots

(C) no real roots

(D) more than 2 real roots

~~ 9. Which of the following equations has two distinct real roots?

(A) 2x232x+94=0

(B) x2+x5=0

(C) x2+3x+22=0

(D) 5x23x+1=0

~~ 10 Which of the following equations has no real roots?

(A) x24x+32=0

(B) x2+4x32=0

(C) x24x32=0

(D) 3x2+43x+4=0

~~ 11 (x2+1)2x2=0 has

(A) four real roots

(B) two real roots

(C) no real roots

(D) one real root.

(C) Short Answer Questions with Reasoning

Sample Question 1 : Does (x1)2+2(x+1)=0 have a real root? Justify your answer.

Solution : No, since the equation is simplified to x2+3=0 whose discriminant is -12 .

Sample Question 2 : Is the following statement ‘True’ or ‘False’?Justify your answer. If in a quadratic equation the coefficient of x is zero, then the quadratic equation has no real roots.

Solution : False, since the discriminant in this case is 4ac which can still be nonnegative if a and c are of opposite signs or if one of a or c is zero.

EXERCISE 4.2

~~ 1. State whether the following quadratic equations have two distinct real roots. Justify your answer.

(i) x23x+4=0

(ii) 2x2+x1=0

(iii) 2x26x+92=0

(iv) 3x24x+1=0

(v) (x+4)28x=0

(vi) (x2)22(x+1)=0

(vii) 2x232x+12=0

(viii) x(1x)2=0

(ix) (x1)(x+2)+2=0

(x) (x+1)(x2)+x=0

~~ 2. Write whether the following statements are true or false. Justify your answers.

(i) Every quadratic equation has exactly one root.

(ii) Every quadratic equation has at least one real root.

(iii) Every quadratic equation has at least two roots.

(iv) Every quadratic equations has at most two roots.

(v) If the coefficient of x2 and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.

(vi) If the coefficient of x2 and the constant term have the same sign and if the coefficient of x term is zero, then the quadratic equation has no real roots.

~~ 3. A quadratic equation with integral coefficient has integral roots. Justify your answer.

~~ 4. Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.

~~ 5. Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rationals? Why?

~~ 6. Is 0.2 a root of the equation x20.4=0 ? Justify.

~~ 7. If b=0,c<0, is it true that the roots of x2+bx+c=0 are numerically equal and opposite in sign? Justify.

(D) Short Answer Questions

Sample Question 1 : Find the roots of the quadratic equation 2x25x2=0 using the quadratic formula.

Solution : b24ac=54×2×(2)=21

Therefore, the roots are 5±214, i.e., 5+214 and 5214

Sample Question 2 : Find the roots of 6x22x2=0 by the factorisation of the corresponding quadratic polynomial.

Solution : 6x22x2=6x232x+22x2

=3x(2x2)+2(2x2)=(3x+2)(2x2)

Now, 6x22x2=0 gives (3x+2)(2x2)=0, i.e., 3x+2=0 or 2x2=0

So, the roots are 23 and 22.

EXERCISE 4.3

~~ 1. Find the roots of the quadratic equations by using the quadratic formula in each of the following:

(i) 2x23x5=0

(ii) 5x2+13x+8=0

(iii) 3x2+5x+12=0

(iv) x2+7x10=0

(v) x2+22x6=0

(vi) x235x+10=0

(vii) 12x211x+1=0

~~ 2. Find the roots of the following quadratic equations by the factorisation method:

(i) 2x2+53x2=0

(ii) 25x2x35=0

(iii) 32x25x2=0

(iv) 3x2+55x10=0

(v) 21x22x+121=0

(E) Long Answer Questions

Sample Question 1: Check whether the equation 6x27x+2=0 has real roots, and if it has, find them by the method of completing the squares.

Solution : The discriminant =b24ac=494×6×2=1>0

So, the given equation has two distinct real roots.

Now, 6x27x+2=0

i.e., 36x242x+12=0

i.e., 6x722+12494=0

i.e., 6x722122=0 or (6x72)2=(12)2

The roots are given by 6x72=±12

i.e., 6x=4,3

i.e., x=23,12.

Sample Question 2 : Had Ajita scored 10 more marks in her mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test?

Solution : Let her actual marks be x

Therefore, 9(x+10)=x2

i.e., x29x90=0

i.e., x215x+6x90=0

i.e., x(x15)+6(x15)=0

i.e., (x+6)(x15)=0

Therefore, x=6 or x=15

Since x is the marks obtained, x6. Therefore, x=15.

So, Ajita got 15 marks in her mathematics test.

Sample Question 3 : A train travels at a certain average speed for a distance of 63km and then travels a distance of 72km at an average speed of 6km/h more than its original speed. If it takes 3 hours to complete the total journey, what is its original average speed?

Solution : Let its original average speed be x km/h. Therefore,

63x+72x+6=3

i.e., 7x+8x+6=39=13

i.e., 7(x+6)+8xx(x+6)=13

 i.e., 21(x+6)+24x=x(x+6) i.e., 21x+126+24x=x2+6x i.e., x239x126=0 i.e., (x+3)(x42)=0 i.e., x=3 or x=42

Since x is the average speed of the train, x cannot be negative.

Therefore, x=42.

So, the original average speed of the train is 42km/h.

EXERCISE 4.4

~~ 1. Find whether the following equations have real roots. If real roots exist, find them.

(i) 8x2+2x3=0

(ii) 2x2+3x+2=0

(iii) 5x22x10=0

(iv) 12x3+1x5=1,x32,5

(v) x2+55x70=0

~~ 2. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

~~ 3. A natural number, when increased by 12 , equals 160 times its reciprocal. Find the number.

~~ 4. A train, travelling at a uniform speed for 360km, would have taken 48 minutes less to travel the same distance if its speed were 5km/h more. Find the original speed of the train.

~~ 5. If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?

~~ 6. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.

~~ 7. In the centre of a rectangular lawn of dimensions 50m×40m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184m2 [see Fig. 4.1]. Find the length and breadth of the pond.

Fig. 4.1

~~ 8. At t minutes past 2pm, the time needed by the minutes hand of a clock to show 3pm was found to be 3 minutes less than t24 minutes. Find t.

Chapter 5

ARITHMETIC PROGRESSIONS

(A) Main Concepts and Results

  • An arithmetic progression (AP) is a list of numbers in which each term is obtained by adding a fixed number d to the preceding term, except the first term a. The fixed number d is called its common difference.

The general form of an AP is a,a+d,a+2d,a+3d,

  • In the list of numbers a1,a2,a3, if the differences a2a1,a3a2,a4a3, give the same value, i.e., if ak+1ak is the same for different values of k, then the given list of numbers is an AP.

  • The nth  term an (or the general term) of an AP is an=a+(n1)d, where a is the first term and d is the common difference. Note that a1=a.

  • The sum Sn of the first n terms of an AP is given by

Sn=n2[2a+(n1)d]

If l is the last term of an AP of n terms, then the sum of all the terms can also be given by

Sn=n2[a+l]

Sometimes Sn is also denoted by S.

  • If Sn is the sum of the first n terms of an AP, then its nth  term an is given by

an=SnSn1

(B) Multiple Choice Questions

Choose the correct answer from the given four options:

Sample Question 1 : The 10th  term of the AP: 5,8,11,14, is

(A) 32

(B) 35

(C) 38

(D) 185

Solution : Answer (A)

Sample Question 2 : In an AP if a=7.2,d=3.6,an=7.2, then n is

(A) 1

(B) 3

(C) 4

(D) 5

Solution : Answer (D)

EXERCISE 5.1

Choose the correct answer from the given four options:

~~ 1. In an AP, if d=4,n=7,an=4, then a is

(A) 6

(B) 7

(C) 20

(D) 28

~~ 2. In an AP, if a=3.5,d=0,n=101, then an will be

(A) 0

(B) 3.5

(C) 103.5

(D) 104.5

~~ 3. The list of numbers 10,6,2,2, is

(A) an AP with d=16

(B) an AP with d=4

(C) an AP with d=4

(D) not an AP

~~ 4. The 11th  term of the AP: 5,52,0,52, is

(A) -20

(B) 20

(C) -30

(D) 30

~~ 5. The first four terms of an AP, whose first term is -2 and the common difference is -2 , are

(A) 2,0,2,4

(B) 2,4,8,16

(C) 2,4,6,8

(D) 2,4,8,16

~~ 6. The 21st  term of the AP whose first two terms are -3 and 4 is

(A) 17

(B) 137

(C) 143

(D) -143

~~ 7. If the 2nd  term of an AP is 13 and the 5th  term is 25 , what is its 7th  term?

(A) 30

(B) 33

(C) 37

(D) 38

~~ 8. Which term of the AP: 21,42,63,84, is 210 ?

(A) 9th 

(B) 10th 

(C) 11th 

(D) 12th 

~~ 9. If the common difference of an AP is 5 , then what is a18a13 ?

(A) 5

(B) 20

(C) 25

(D) 30

~~ 10 What is the common difference of an AP in which a18a14=32 ?

(A) 8

(B) -8

(C) -4

(D) 4

~~ 11 Two APs have the same common difference. The first term of one of these is -1 and that of the other is -8 . Then the difference between their 4th  terms is

(A) -1

(B) -8

(C) 7

(D) -9

~~ 12 If 7 times the 7th  term of an AP is equal to 11 times its 11th  term, then its 18 th term will be

(A) 7

(B) 11

(C) 18

(D) 0

~~ 13 The 4th  term from the end of the AP: 11,8,5,,49 is

(A) 37

(B) 40

(C) 43

(D) 58

~~ 14 The famous mathematician associated with finding the sum of the first 100 natural numbers is

(A) Pythagoras

(B) Newton

(C) Gauss

(D) Euclid

~~ 15 If the first term of an AP is -5 and the common difference is 2 , then the sum of the first 6 terms is

(A) 0

(B) 5

(C) 6

(D) 15

~~ 16 The sum of first 16 terms of the AP: 10,6,2, is

(A) -320

(B) 320

(C) -352

(D) -400

~~ 17 In an AP if a=1,an=20 and Sn=399, then n is

(A) 19

(B) 21

(C) 38

(D) 42

~~ 18 The sum of first five multiples of 3 is

(A) 45

(B) 55

(C) 65

(D) 75

(C) Short Answer Questions with Reasoning

Sample Question 1: In the AP: 10,5,0,5, the common difference d is equal to 5 . Justify whether the above statement is true or false.

Solution :

a2a1=510=5a3a2=05=5a4a3=50=5

Although the given list of numbers forms an AP, it is with d=5 and not with d=5 So, the given statement is false.

Sample Question 2 : Divya deposited Rs 1000 at compound interest at the rate of 10 per annum. The amounts at the end of first year, second year, third year, …, form an AP. Justify your answer.

Solution : Amount at the end of the 1 st year = Rs 1100

Amount at the end of the 2nd year = Rs 1210

Amount at the end of 3rd year = Rs 1331 and so on.

So, the amount (in Rs) at the end of 1st year, 2nd year, 3rd year, … are

1100,1210,1331,

Here, a2a1=110

a3a2=121

As, a2a1a3a2, it does not form an AP.

Sample Question 3: The nth  term of an AP cannot be n2+1. Justify your answer.

Solution :

Here, an=n2+1

So, a1=12+1=2

a2=22+1=5

a3=32+1=10

List of numbers becomes 2, 5, 10, ..

Here, 52105, so it does not form an AP.

Alternative Solution 1:

We know that in an AP, d=anan1

Here, an=n2+1

So, anan1=(n2+1)(n1)2+1

=2n1

As anan1 depends upon n,d is not a fixed number.

So, an=n2+1 cannot be the nth  term of an AP.

Alternative Solution 2:

We know that in an AP

an=a+(n1)d. We observe that an is a linear polynomial in n.

Here, an=n2+1 is not a linear polynomial in n. So, it cannot be the nth  term of an AP.

EXERCISE 5.2

~~ 1. Which of the following form an AP? Justify your answer.

(i) 1,1,1,1,

(ii) 0,2,0,2,

(iii) 1,1,2,2,3,3,

(iv) 11,22,33,

(v) 12,13,14,

(vi) 2,22,23,24,

(vii) 3,12,27,48,

~~ 2. Justify whether it is true to say that 1,32,2,52, forms an AP as a2a1=a3a2.

~~ 3. For the AP: 3,7,11,, can we find directly a30a20 without actually finding a30 and a20 ? Give reasons for your answer.

~~ 4. Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7 . The difference between their 10th  terms is the same as the difference between their 21st  terms, which is the same as the difference between any two corresponding terms. Why?

~~ 5. Is 0 a term of the AP: 31,28,25, ? Justify your answer.

~~ 6. The taxi fare after each km, when the fare is Rs 15 for the first km and Rs 8 for each additional km, does not form an AP as the total fare (in Rs) after each km is

15,8,8,8,

Is the statement true? Give reasons.

~~ 7. In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.

(i) The fee charged from a student every month by a school for the whole session, when the monthly fee is Rs 400 .

(ii) The fee charged every month by a school from Classes I to XII, when the monthly fee for Class I is Rs 250, and it increases by Rs 50 for the next higher class.

(iii) The amount of money in the account of Varun at the end of every year when Rs 1000 is deposited at simple interest of 10 per annum.

(iv) The number of bacteria in a certain food item after each second, when they double in every second.

~~ 8. Justify whether it is true to say that the following are the nth  terms of an AP.

(i) 2n3

(ii) 3n2+5

(iii) 1+n+n2

(D) Short Answer Questions

Sample Question 1 : If the numbers n2,4n1 and 5n+2 are in AP, find the value of n.

Solution :

As n2,4n1,5n+2 are in AP,

so (4n1)(n2)=(5n+2)(4n1)

i.e, 3n+1=n+3

i.e, n=1

Sample Question 2 : Find the value of the middle most term (s) of the AP : 11,7,3,,49.

Solution :

Here, a=11,d=7(11)=4,an=49

We have an=a+(n1)d

So, 49=11+(n1)×4

i.e., 60=(n1)×4

i.e., n=16

As n is an even number, there will be two middle terms which are 162 th and (162+1) th, i.e., the 8th  term and the 9th  term.

a8=a+7d=11+7×4=17a9=a+8d=11+8×4=21

So, the values of the two middle most terms are 17 and 21 , respectively.

Sample Question 3: The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, find the AP.

Solution : Let the three terms in AP be

ad,a,a+d

So, ad+a+a+d=33

or a=11

Also, (ad)(a+d)=a+29

i.e., a2d2=a+29

i.e., 121d2=11+29

i.e., d2=81

i.e., d=±9

So there will be two APs and they are : 2,11,20,

and 20,11,2,

EXERCISE 5.3

~~ 1. Match the APs given in column A with suitable common differences given in column B.

Column A Column B

(A1)2,2,6,10, (B1)23

(A2) a=18,n=10,an=0 (B2)5

(A3)a=0,a10=6 (B3)4

(A4)a2=13,a4=3 (B4)4

(B5)2

(B6)12

(B7)5

~~ 2. Verify that each of the following is an AP, and then write its next three terms.

(i) 0,14,12,34,

(ii) 5,143,133,4,

(iii) 3,23,33,

(iv) a+b,(a+1)+b,(a+1)+(b+1),

(v) a,2a+1,3a+2,4a+3,

~~ 3. Write the first three terms of the APs when a and d are as given below:

(i) a=12,d=16

(ii) a=5,d=3

(iii) a=2,d=12

~~ 4. Find a,b and c such that the following numbers are in AP: a,7,b,23,c.

~~ 5. Determine the AP whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20 .

~~ 6. The 26th ,11th  and the last term of an AP are 0,3 and 15, respectively. Find the common difference and the number of terms.

~~ 7. The sum of the 5th  and the 7th  terms of an AP is 52 and the 10th  term is 46 . Find the AP.

~~ 8. Find the 20th  term of the AP whose 7th  term is 24 less than the 11th  term, first term being 12 .

~~ 9. If the 9th  term of an AP is zero, prove that its 29th  term is twice its 19th  term.

~~ 10 Find whether 55 is a term of the AP: 7,10,13,— or not. If yes, find which term it is.

~~ 11 Determine k so that k2+4k+8,2k2+3k+6,3k2+4k+4 are three consecutive terms of an AP.

~~ 12 Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623 .

~~ 13 The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.

~~ 14 If the nth terms of the two APs: 9,7,5, and 24,21,18, are the same, find the value of n. Also find that term.

~~ 15 If sum of the 3rd  and the 8th  terms of an AP is 7 and the sum of the 7th  and the 14th  terms is -3 , find the 10th  term.

~~ 16 Find the 12th  term from the end of the AP: 2,4,6,,100.

~~ 17 Which term of the AP: 53,48,43, is the first negative term?

~~ 18 How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3 ?

~~ 19 Find the sum of the two middle most terms of the AP: 43,1,23,,413.

~~ 20 The first term of an AP is -5 and the last term is 45 . If the sum of the terms of the AP is 120 , then find the number of terms and the common difference.

~~ 21 Find the sum:

(i) 1+(2)+(5)+(8)++(236)

(ii) 41n+42n+43n+ upto n terms

(iii) aba+b+3a2ba+b+5a3ba+b+ to 11 terms.

~~ 22 Which term of the AP: 2,7,12, will be -77 ? Find the sum of this AP upto the term -77 .

~~ 23 If an=34n, show that a1,a2,a3, form an AP. Also find S20.

~~ 24 In an AP, if Sn=n(4n+1), find the AP.

~~ 25 In an AP, if Sn=3n2+5n and ak=164, find the value of k.

~~ 26 If Sn denotes the sum of first n terms of an AP, prove that

S12=3(S8S4)

~~ 27 Find the sum of first 17 terms of an AP whose 4th  and 9th  terms are -15 and -30 respectively.

~~ 28 If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

~~ 29 Find the sum of all the 11 terms of an AP whose middle most term is 30 .

~~ 30 Find the sum of last ten terms of the AP: 8, 10, 12,—, 126.

~~ 31 Find the sum of first seven numbers which are multiples of 2 as well as of 9 .

[Hint: Take the LCM of 2 and 9]

~~ 32 How many terms of the AP: 15,13,11, are needed to make the sum -55 ? Explain the reason for double answer.

~~ 33 The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is -30 and the common difference is 8 . Find n.

~~ 34 Kanika was given her pocket money on Jan 1st ,2008. She puts Re 1 on Day 1 , Rs 2 on Day 2, Rs 3 on Day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money, and found that at the end of the month she still had Rs 100 with her. How much was her pocket money for the month?

~~ 35 Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000 ?

(E) Long Answer Questions

Sample Question 1: The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last terms to the product of the two middle terms is 7:15. Find the numbers.

Solution : Let the four consecutive numbers in AP be

a3d,ad,a+d,a+3d

So, a3d+ad+a+d+a+3d=32

or, 4 a = 32

or, a = 8

Also, (a3d)(a+3d)(ad)(a+d)=715

or, a29d2a2d2=715

or, 15a2135d2=7a27d2

or, 8a2128d2=0

or, d2=8×8×8128=4

or, d=±2

So, when a=8,d=2, the numbers are 2,6,10,14.

Sample Question 2: Solve the equation :

1+4+7+10++x=287

Solution :

Here, 1,4,7,10,,x form an AP with a=1,d=3,an=x

We have, an=a+(n1)d

So, x=1+(n1)×3=3n2

Also, S=n2(a+l)

So, 287=n2(1+x)

=n2(1+3n2)

or, 574=n(3n1)

or, 3n2n574=0

Therefore, n=1±1+68886

=1±836=846,826=14,413

As n cannot be negative, so n=14

Therefore, x=3n2=3×142=40.

Alternative Solution :

Here, 1,4,7,10,x form an AP with a=1,d=3,S=287

We have, S=n22a+(n1)d

So, 287=n22+(n1)×3

or, 574=n(3n1)

or, 3n2n574=0

Now proceed as above.

EXERCISE 5.4

~~ 1. The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.

~~ 2. Find the

(i) sum of those integers between 1 and 500 which are multiples of 2 as well as of 5 .

(ii) sum of those integers from 1 to 500 which are multiples of 2 as well as of 5 .

(iii) sum of those integers from 1 to 500 which are multiples of 2 or 5.

[Hint (iii) : These numbers will be : multiples of 2+ multiples of 5 - multiples of 2 as well as of 5 ]

~~ 3. The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by 1 . Find the 15th  term.

~~ 4. An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429 . Find the AP.

~~ 5. Find the sum of the integers between 100 and 200 that are

(i) divisible by 9

(ii) not divisible by 9

[Hint (ii) : These numbers will be : Total numbers - Total numbers divisible by 9]

~~ 6. The ratio of the 11th  term to the 18th  term of an AP is 2:3. Find the ratio of the 5th  term to the 21st  term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.

~~ 7. Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to

(a+c)(b+c2a)2(ba)

~~ 8. Solve the equation

4+(1)+2++x=437

~~ 9. Jaspal Singh repays his total loan of Rs 118000 by paying every month starting with the first instalment of Rs 1000 . If he increases the instalment by Rs 100 every month, what amount will be paid by him in the 30th  instalment? What amount of loan does he still have to pay after the 30th  instalment?

~~ 10 The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?

Chapter 6

TRIANGLES

(A) Main Concepts and Results

Congruence and similarity, Conditions for similarity of two polygons, Similarity of Triangles, Similarity and correspondence of vertices, Criteria for similarity of triangles; (i) AAA or AA (ii) SSS (iii) SAS

  • If a line is drawn parallel to one side of a triangle to intersect the other two sides, then these two sides are divided in the same ratio (Basic Proportionality Theorem) and its converse.

  • Ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

  • Perpendicular drawn from the vertex of the right angle of a right triangle to its hypotenuse divides the triangle into two triangles which are similar to the whole triangle and to each other.

  • In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides (Pythagoras Theorem) and its converse.

(B) Multiple Choice Questions

Choose the correct answer from the given four options:

Sample Question 1: If in Fig 6.1, O is the point of intersection of two chords AB and CD such that OB=OD, then triangles OAC and ODB are

Fig. 6.1

(A) equilateral but not similar

(B) isosceles but not similar

(C) equilateral and similar

(D) isosceles and similar

Solution : Answer (D)

Sample Question 2: D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD=2cm,BD=3cm,BC=7.5cm and DE|BC. Then, length of DE (in cm ) is

(A) 2.5

(B) 3

(C) 5

(D) 6

Solution : Answer (B)

EXERCISE 6.1

Choose the correct answer from the given four options:

~~ 1. In Fig. 6.2, BAC=90 and ADBC. Then,

Fig. 6.2

(A) BDCD=BC2

(B) ABAC=BC2

(C) BDCD=AD2

(D) ABAC=AD2

~~ 2. The lengths of the diagonals of a rhombus are 16cm and 12cm. Then, the length of the side of the rhombus is

(A) 9cm

(B) 10cm

(C) 8cm

(D) 20cm

~~ 3. If ABCEDF and ABC is not similar to DEF, then which of the following is not true?

(A) BCEF=ACFD

(B) ABEF=ACDE

(C) BCDE=ABEF

(D) BCDE=AB. FD

~~ 4. If in two triangles ABC and PQR,ABQR=BCPR=CAPQ, then

(A) PQRCAB

(B) PQRABC

(C) CBAΔPQR

(D) BCAPQR

~~ 5. In Fig.6.3, two line segments AC and BD intersect each other at the point P such that PA=6cm,PB=3cm,PC=2.5cm,PD=5cm,APB=50 and CDP=30. Then, PBA is equal to

alt text

Fig. 6.3

(A) 50

(B) 30

(C) 60

(D) 100

~~ 6. If in two triangles DEF and PQR,D=Q and R=E, then which of the following is not true?

(A) EFPR=DFPQ

(B) DEPQ=EFRP

(C) DEQR=DFPQ

(D) EFRP=DEQR

~~ 7. In triangles ABC and DEF,B=E,F=C and AB=3DE. Then, the two triangles are

(A) congruent but not similar

(B) similar but not congruent

(C) neither congruent nor similar

(D) congruent as well as similar

~~ 8. It is given that ABCΔPQR, with BCQR=13. Then, ar(PRQ)ar(BCA) is equal to

(A) 9

(B) 3

(C) 13

(D) 19

~~ 9. It is given that ABCDFE,A=30,C=50,AB=5cm,AC=8cm and DF=7.5cm. Then, the following is true:

(A) DE=12cm,F=50

(B) DE=12cm,F=100

(C) EF=12cm,D=100

(D) EF=12cm,D=30

~~ 10 If in triangles ABC and DEF,ABDE=BCFD, then they will be similar, when

(A) B=E

(B) A=D

(C) B=D

(D) A=F

~~ 11 If ABCΔQRP,ar(ABC)ar(PQR)=94,AB=18cm and BC=15cm, then PR is equal to

(A) 10cm

(B) 12cm

(C) 203cm

(D) 8cm

~~ 12 If S is a point on side PQ of a PQR such that PS=QS=RS, then

(A) PRQR=RS2

(B) QS2+RS2=QR2

(C) PR2+QR2=PQ2

(D) PS2+RS2=PR2

(C) Short Answer Questions with Reasoning

Sample Question 1: In ABC,AB=24cm,BC=10cm and AC=26cm. Is this triangle a right triangle? Give reasons for your answer.

Solution : Here AB2=576,BC2=100 and AC2=676. So, AC2=AB2+BC2

Hence, the given triangle is a right triangle.

Sample Question 2: P and Q are the points on the sides DE and DF of a triangle DEF such that DP=5cm,DE=15cm,DQ=6cm and QF=18cm. Is PQ|EF ? Give reasons for your answer.

Solution : Here, DPPE=5155=12 and DQQF=618=13

As DPPEDQQF, therefore PQ is not parallel to EF.

Sample Question 3: It is given that ΔFEDΔSTU. Is it true to say that DEST=EFTU ? Why?

Solution : No, because the correct correspondence is FS,ET,DU.

With this correspondence, EFST=DETU.

EXERCISE 6.2

~~ 1. Is the triangle with sides 25cm,5cm and 24cm a right triangle? Give reasons for your answer.

~~ 2. It is given that DEFRPQ. Is it true to say that D=R and F=P ? Why?

~~ 3. A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ=12.5cm,PA=5cm,BR=6cm and PB=4cm. Is AB|QR ? Give reasons for your answer.

~~ 4. In Fig 6.4, BD and CE intersect each other at the point P. Is ΔPBCPDE ? Why?

alt text

~~ 5. In triangles PQR and MST,P=55,Q=25,M=100 and S=25. Is QPRTSM ? Why?

~~ 6. Is the following statement true? Why?

“Two quadrilaterals are similar, if their corresponding angles are equal”.

~~ 7. Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?

~~ 8. If in two right triangles, one of the acute angles of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles will be similar? Why?

~~ 9. The ratio of the corresponding altitudes of two similar triangles is 35. Is it correct to say that ratio of their areas is 65? Why?

~~ 10 D is a point on side QR of PQR such that PDQR. Will it be correct to say that PQDΔRPD ? Why?

~~ 11 In Fig. 6.5, if D=C, then is it true that ADEACB ? Why?

~~ 12 Is it true to say that if in two triangles, an angle of one triangle is equal to an angle of another triangle and two sides of one triangle are proportional to the two sides of the other triangle, then the triangles are similar? Give reasons for your answer.

(D) Short Answer Questions

Sample Question 1: Legs (sides other than the hypotenuse) of a right triangle are of lengths 16cm and 8cm. Find the length of the side of the largest square that can be inscribed in the triangle.

Solution : Let ABC be a right triangle right angled at B with AB=16cm and BC= 8cm. Then, the largest square BRSP which can be inscribed in this triangle will be as shown in Fig.6.6.

Let PB=xcm. So., AP=(16x)cm. In APS and ABC,A=A and APS=ABC(. Each .90)

Fig. 6.6

So, APSABC (AA similarity)

Therefore, APAB=PSBC

or 16x16=x8

or 1288x=16x

or x=12824=163

Thus, the side of the required square is of length 163cm.

Sample Question 2: Hypotenuse of a right triangle is 25cm and out of the remaining two sides, one is longer than the other by 5cm. Find the lengths of the other two sides.

Solution : Let one side be xcm. Then the other side will be (x+5)cm.

Therefore, from Pythagoras Theorem

x2+(x+5)2=(25)2

x2+x2+10x+25=625

x2+5x300=0

x(x+20)15(x+20)=0

(x15)(x+20)=0

So x=15,x=20

Rejecting x=20, we have length of one side =15cm and that of the other side =(15+5)cm=20cm

Sample Question 3: In Fig 6.7,

D=E and ADDB=AEEC. Prove that BAC is an isosceles triangle.

Solution : ADDB=AEEC (Given)

Fig. 6.7

Therefore, DE || BC (Converse of Basic Proportionality Theorem)

So, D=B and E=C (Corresponding angles)

But D=E (Given)

Therefore, B=C [ From (1)]

So, AB=AC (Sides opposite to equal angles)

i.e., BAC is an isosceles triangle.

EXERCISE 6.3

~~ 1. In a ΔPQR,PR2PQ2=QR2 and M is a point on side PR such that QMPR. Prove that QM2=PM×MR.

~~ 2. Find the value of x for which DE|AB in Fig. 6.8.

Fig. 6.8

~~ 3. In Fig. 6.9, if 1=2 and ΔNSQMTR, then prove that ΔPTSΔPRQ.

Fig. 6.9

~~ 4. Diagonals of a trapezium PQRS intersect each other at the point O,PQ|RS and PQ=3RS. Find the ratio of the areas of triangles POQ and ROS.

~~ 5. In Fig. 6.10, if AB|DC and AC and PQ intersect each other at the point O, prove that OA.CQ=OC.AP.

Fig. 6.10

~~ 6. Find the altitude of an equilateral triangle of side 8cm.

~~ 7. If ABCΔDEF,AB=4cm,DE=6cm,EF=9cm and FD=12cm, find the perimeter of ABC.

~~ 8. In Fig. 6.11, if DE|BC, find the ratio of ar (ADE) and ar (DECB).

Fig. 6.11

~~ 9. ABCD is a trapezium in which AB|DC and P and Q are points on AD and BC, respectively such that PQ|DC. If PD=18cm,BQ=35cm and QC=15cm, find AD.

~~ 10 Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of the smaller triangle is 48cm2, find the area of the larger triangle.

~~ 11 In a triangle PQR,N is a point on PR such that QNPR. If PN. NR=QN2, prove that PQR=90.

~~ 12 Areas of two similar triangles are 36cm2 and 100cm2. If the length of a side of the larger triangle is 20cm, find the length of the corresponding side of the smaller triangle.

~~ 13 In Fig. 6.12, if ACB=CDA,AC=8cm and AD=3cm, find BD.

~~ 14 A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.

~~ 15 Foot of a 10m long ladder leaning against a vertical wall is 6m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.

(E) Long Answer Questions

Sample Question 1: In Fig 6.13, OB is the perpendicular bisector of the line segment DE, FA OB and F E intersects OB at the point C. Prove that 1OA+1OB=2OC.

Solution : In ΔAOF and ΔBOD.

O=O( Same angle ) and A=B(. each .90)

Therefore, ΔAOFΔBOD (AA similarity)

So, OAOB=FADB

Also, in FAC and EBC,A=B(. Each .90)

and FCA=ECB (Vertically opposite angles).

Therefore, ΔFACΔEBC (AA similarity).

So, FAEB=ACBC

But EB=DB(B is mid-point of DE)

So, FADB=ACBC

Therefore, from (1) and (2), we have:

ACBC=OAOB

i.e., OCOAOBOC=OAOB

or OBOCOAOB=OAOBOAOC

or OBOC+OAOC=2OAOB

or (OB+OA)OC=2OAOB

or 1OA+1OB=2OC [ Dividing both the sides by OA. OB . OC]

Sample Question 2: Prove that if in a triangle square on one side is equal to the sum of the squares on the other two sides, then the angle opposite the first side is a right angle.

Solution : See proof of Theorem 6.9 of Mathematics Textbook for Class X.

Sample Question 3: An aeroplane leaves an Airport and flies due North at 300km/h. At the same time, another aeroplane leaves the same Airport and flies due West at 400km/h. How far apart the two aeroplanes would be after 112 hours?

Solution : Distance travelled by first aeroplane in 112 hours =300×32km=450km and that by second aeroplane =400×32km=600km

Position of the two aeroplanes after 112 hours would be A and B as shown in Fig. 6.14.

Fig. 6.14

That is, OA=450km and OB=600km.

From ΔAOB, we have

AB2=OA2+OB2 or AB2=(450)2+(600)2=(150)2×32+(150)2×42=1502(32+42)=1502×52

 or AB=150×5=750

Thus, the two aeroplanes will be 750km apart after 112 hours.

Sample Question 4: In Fig. 6.15, if ΔABCΔDEF and their sides are of lengths (in cm ) as marked along them, then find the lengths of the sides of each triangle.

alt text

Solution : ABCΔDEF (Given)

Therefore, ABDE=BCEF=CAFD

So, 2x118=2x+23x+9=3x6x

Now, taking 2x118=3x6x, we have

2x118=12

or

4x2=18

or

x=5

Therefore, AB=2×51=9,BC=2×5+2=12,

CA=3×5=15,DE=18,EF=3×5+9=24 and FD=6×5=30

Hence, AB=9cm,BC=12cm,CA=15cm,

DE=18cm,EF=24cm and FD=30cm

EXERCISE 6.4

~~ 1. In Fig. 6.16, if A=C,AB=6cm,BP=15cm, AP=12cm and CP=4cm, then find the lengths of PD and CD.

~~ 2. It is given that ABCEDF such that AB=5cm, AC=7cm,DF=15cm and DE=12cm. Find the lengths of the remaining sides of the triangles.

~~ 3. Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.

~~ 4. In Fig 6.17, if PQRS is a parallelogram and AB|PS, then prove that OC|SR.

Fig. 6.17

~~ 5. A 5m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4m high. If the foot of the ladder is moved 1.6m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.

~~ 6. For going to a city B from city A, there is a route via city C such that ACCB, AC=2xkm and CB=2(x+7)km. It is proposed to construct a 26km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.

~~ 7. A flag pole 18m high casts a shadow 9.6m long. Find the distance of the top of the pole from the far end of the shadow.

~~ 8. A street light bulb is fixed on a pole 6m above the level of the street. If a woman of height 1.5m casts a shadow of 3m, find how far she is away from the base of the pole.

~~ 9. In Fig. 6.18, ABC is a triangle right angled at B and BDAC. If AD=4cm, and CD=5cm, find BD and AB.

Fig. 6.18

~~ 10 In Fig. 6.19, PQR is a right triangle right angled at Q and QSPR. If PQ=6cm and PS=4cm, find QS,RS and QR.

~~ 11 In ΔPQR,PDQR such that D lies on QR. If PQ=a,PR=b,QD=c and DR=d, prove that (a+b)(ab)=(c+d)(cd).

~~ 12 In a quadrilateral ABCD,A+D=90. Prove that AC2+BD2=AD2+BC2

[Hint: Produce AB and DC to meet at E.]

~~ 13 In fig. 6.20, l|m and line segments AB,CD and EF are concurrent at point P.

Prove that AEBF=ACBD=CEFD.

Fig. 6.20

~~ 14 In Fig. 6.21, PA, QB, RC and SD are all perpendiculars to a line l,AB=6cm, BC=9cm,CD=12cm and SP=36cm. Find PQ,QR and RS.

Fig. 6.21

~~ 15 O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB|DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO=QO.

~~ 16 In Fig. 6.22, line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and AEF=AFE. Prove that BDCD=BFCE.

[Hint: Take point G on AB such that CG|DF.]

Fig. 6.22

~~ 17 Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.

~~ 18 Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.

Chapter 7

COORDINATE GEOMETRY

(A) Main Concepts and Results

Distance Formula , Section Formula, Area of a Triangle.

  • The distance between two points P(x1,y1) and Q(x2,y2) is

(x2x1)2+(y2y1)2

  • The distance of a point P(x,y) from the origin is x2+y2

  • The coordinates of the point P which divides the line segment joining the points A (x1,y1) and B(x2,y2) internally in the ratio m1:m2 are (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)

  • The coordinates of the mid-point of the line segment joining the points P(x1,y1)

and Q(x2,y2) are (x1+x22,y1+y22)

  • The area of a triangle with vertices A(x1,y1),B(x2,y2) and C(x3,y3) is

12[x1(y2y3)+x2(y3y1)+x3(y1y2)]

which is non-zero unless the points A,B and C are collinear.

(B) Multiple Choice Questions

Choose the correct answer from the given four options:

Sample Question 1: If the distance between the points (2,2) and (1,x) is 5 , one of the values of x is

(A) -2

(B) 2

(C) -1

(D) 1

Solution : Answer (B)

Sample Question 2: The mid-point of the line segment joining the points A(2,8) and B(6,4) is

(A) (4,6)

(B) (2,6)

(C) (4,2)

(D) (4,2)

Solution : Answer (C)

Sample Question 3: The points A (9,0), B (9,6),C(9,6) and D (9,0) are the vertices of a

(A) square

(B) rectangle

(C) rhombus

(D) trapezium

Solution : Answer (B)

EXERCISE 7.1

Choose the correct answer from the given four options:

~~ 1. The distance of the point P(2,3) from the x-axis is

(A) 2

(B) 3

(C) 1

(D) 5

~~ 2. The distance between the points A(0,6) and B(0,2) is

(A) 6

(B) 8

(C) 4

(D) 2

~~ 3. The distance of the point P(6,8) from the origin is

(A) 8

(B) 27

(C) 10

(D) 6

~~ 4. The distance between the points (0,5) and (5,0) is

(A) 5

(B) 52

(C) 25

(D) 10

~~ 5. AOBC is a rectangle whose three vertices are vertices A(0,3),O(0,0) and B(5,0). The length of its diagonal is

(A) 5

(B) 3

(C) 34

(D) 4

~~ 6. The perimeter of a triangle with vertices (0,4),(0,0) and (3,0) is

(A) 5

(B) 12

(C) 11

(D) 7+5

~~ 7. The area of a triangle with vertices A(3,0),B(7,0) and C(8,4) is

(A) 14

(B) 28

(C) 8

(D) 6

~~ 8. The points (4,0),(4,0),(0,3) are the vertices of a

(A) right triangle

(C) equilateral triangle

(B) isosceles triangle

(D) scalene triangle

~~ 9. The point which divides the line segment joining the points (7,6) and (3,4) in ratio 1:2 internally lies in the

(A) I quadrant

(B) II quadrant

(C) III quadrant

(D) IV quadrant

~~ 10 The point which lies on the perpendicular bisector of the line segment joining the points A(2,5) and B(2,5) is

(A) (0,0)

(B) (0,2)

(C) (2,0)

(D) (2,0)

~~ 11 The fourth vertex D of a parallelogram ABCD whose three vertices are A(2,3),B(6,7) and C(8,3) is

(A) (0,1)

(B) (0,1)

(C) (1,0)

(D) (1,0)

~~ 12 If the point P(2,1) lies on the line segment joining points A(4,2) and B(8,4), then

(A) AP=13AB

(B) AP=PB

(C) PB=13AB

(D) AP=12AB

~~ 13 If Pa3,4 is the mid-point of the line segment joining the points Q(6,5) and R(2,3), then the value of a is

(A) -4

(B) -12

(C) 12

(D) -6

~~ 14 The perpendicular bisector of the line segment joining the points A(1,5) and B (4,6) cuts the y-axis at

(A) (0,13)

(B) (0,13)

(C) (0,12)

(D) (13,0)

~~ 15 The coordinates of the point which is equidistant from the three vertices of the AOB as shown in the Fig. 7.1 is

Fig. 7.1

(A) (x,y)

(B) (y,x)

(C) x2,y2

(D) y2,x2

~~ 16 A circle drawn with origin as the centre passes through (132,0). The point which does not lie in the interior of the circle is

(A) 34,1

(B) 2,73

(C) 5,12

(D) (6,52)

~~ 17 A line intersects the y-axis and x-axis at the points P and Q, respectively. If (2,5) is the mid-point of PQ, then the coordinates of P and Q are, respectively

(A) (0,5) and (2,0)

(B) (0,10) and (4,0)

(C) (0,4) and (10,0)

(D) (0,10) and (4,0)

~~ 18 The area of a triangle with vertices (a,b+c),(b,c+a) and (c,a+b) is

(A) (a+b+c)2

(B) 0

(C) a+b+c

(D) abc

~~ 19 If the distance between the points (4,p) and (1,0) is 5 , then the value of p is

(A) 4 only

(B) ±4

(C) -4 only

(D) 0

~~ 20 If the points A(1,2),O(0,0) and C(a,b) are collinear, then

(A) a=b

(B) a=2b

(C) 2a=b

(D) a=b

(C) Short Answer Questions with Reasoning

State whether the following statements are true or false. Justify your answer.

Sample Question 1: The points A (1,0), B (3,1),C(2,2) and D (2,1) are the vertices of a parallelogram.

Solution : True. The coordinates of the mid-points of both the diagonals AC and BD are 12,1, i.e., the diagonals bisect each other.

Sample Question 2 : The points (4,5),(7,6) and (6,3) are collinear.

Solution : False. Since the area of the triangle formed by the points is 4 sq. units, the points are not collinear.

Sample Question 3 : Point P (0,7) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points A(1,0) and B(7,6).

Solution : True. P(0,7) lies on the y-axis. It is at a distance of 50 units from both the points (1,0) and (7,6).

EXERCISE 7.2

State whether the following statements are true or false. Justify your answer.

~~ 1. ABC with vertices A(2,0),B(2,0) and C(0,2) is similar to DEF with vertices D(4,0)E(4,0) and F(0,4).

~~ 2. Point P(4,2) lies on the line segment joining the points A(4,6) and B(4,6).

~~ 3. The points (0,5),(0,9) and (3,6) are collinear.

~~ 4. Point P(0,2) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points A(1,1) and B(3,3).

~~ 5. Points A(3,1),B(12,2) and C(0,2) cannot be the vertices of a triangle.

~~ 6. Points A(4,3),B(6,4),C(5,6) and D(3,5) are the vertices of a parallelogram.

~~ 7. A circle has its centre at the origin and a point P(5,0) lies on it. The point Q(6,8) lies outside the circle.

~~ 8. The point A(2,7) lies on the perpendicular bisector of line segment joining the points P(6,5) and Q(0,4).

~~ 9. Point P(5,3) is one of the two points of trisection of the line segment joining the points A(7,2) and B(1,5).

~~ 10 Points A(6,10),B(4,6) and C(3,8) are collinear such that AB=29AC.

~~ 11 The point P(2,4) lies on a circle of radius 6 and centre C(3,5).

~~ 12 The points A(1,2),B(4,3),C(2,5) and D(3,0) in that order form a rectangle.

(D) Short Answer Questions

Sample Question 1 : If the mid-point of the line segment joining the points A(3,4) and B(k,6) is P(x,y) and x+y10=0, find the value of k.

Solution : Mid-point of the line segment joining A (3,4) and B(k,6)=3+k2,4+62

=3+k2,5

Then, 3+k2,5=(x,y)

Therefore, 3+k2=x and 5=y.

Since x+y10=0, we have

3+k2+510=0 i.e., 3+k=10

Therefore, k=7.

Sample Question 2 : Find the area of the triangle ABC with A (1,4) and the mid-points of sides through A being (2,1) and (0,1).

Solution : Let the coordinates of B and C be (a,b) and (x,y), respectively.

Then, (1+a2,4+b2)=(2,1)

Therefore, 1+a=4,4+b=2

a=3b=2

Also, (1+x2,4+y2)=(0,1)

Therefore, 1+x=0,4+y=2

i.e., x=1 i.e., y=2

The coordinates of the vertices of ABC are A(1,4),B(3,2) and C(1,2).

Area of ΔABC=12[1(22)+3(2+4)1(42)]

=12[18+6]=12 sq. units. 

Sample Question 3 : Name the type of triangle PQR formed by the points P(2,2),

Q(2,2) and R(6,6).

Solution : Using distance formula

PQ=(2+2)2+(2+2)2=(22)2+(22)2=16=4PR=(2+6)2+(26)2=2+6+212+2+6212=16=4RQ=(2+6)2+(26)2=2+6212+2+6+212=16=4

Since PQ=PR=RQ=4, points P,Q,R form an equilateral triangle.

Sample Question 4 : ABCD is a parallelogram with vertices A (x1,y1),B(x2,y2) and C(x3,y3). Find the coordinates of the fourth vertex D in terms of x1,x2,x3,y1,y2 and y3.

Solution : Let the coordinates of D be (x,y). We know that diagonals of a parallelogram bisect each other.

Fig. 7.2

Therefore, mid-point of AC= mid-point of BDx1+x32,y1+y32=x2+x2,y2+y2

i.e., x1+x3=x2+x and y1+y3=y2+y

i.e., x1+x3x2=x and y1+y3y2=y

Thus, the coordinates of D are (x1+x3x2,y1+y3y2)

EXERCISE 7.3

~~ 1. Name the type of triangle formed by the points A(5,6),B(4,2) and C(7,5).

~~ 2. Find the points on the x-axis which are at a distance of 25 from the point (7,4). How many such points are there?

~~ 3. What type of a quadrilateral do the points A(2,2),B(7,3),C(11,1) and D(6,6) taken in that order, form?

~~ 4. Find the value of a, if the distance between the points A(3,14) and B(a,5) is 9 units.

~~ 5. Find a point which is equidistant from the points A(5,4) and B(1,6) ? How many such points are there?

~~ 6. Find the coordinates of the point Q on the x-axis which lies on the perpendicular bisector of the line segment joining the points A(5,2) and B(4,2). Name the type of triangle formed by the points Q, A and B.

~~ 7. Find the value of m if the points (5,1),(2,3) and (8,2m) are collinear.

~~ 8. If the point A(2,4) is equidistant from P(3,8) and Q(10,y), find the values of y. Also find distance PQ.

~~ 9. Find the area of the triangle whose vertices are (8,4),(6,6) and (3,9).

~~ 10 In what ratio does the x-axis divide the line segment joining the points (4,6) and (1,7) ? Find the coordinates of the point of division.

~~ 11 Find the ratio in which the point P(34,512) divides the line segment joining the points A12,32 and B(2,5).

~~ 12 If P(9a2,b) divides line segment joining A(3a+1,3) and B(8a,5) in the ratio 3:1, find the values of a and b.

~~ 13 If (a,b) is the mid-point of the line segment joining the points A(10,6) and B(k,4) and a2b=18, find the value of k and the distance AB.

~~ 14 The centre of a circle is (2a,a7). Find the values of a if the circle passes through the point (11,9) and has diameter 102 units.

~~ 15 The line segment joining the points A(3,2) and B(5,1) is divided at the point P in the ratio 1:2 and it lies on the line 3x18y+k=0. Find the value of k.

~~ 16 If D(12,52),E(7,3) and F(72,72) are the midpoints of sides of ABC, find the area of the ABC.

~~ 17 The points A(2,9),B(a,5) and C(5,5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of ABC.

~~ 18 Find the coordinates of the point R on the line segment joining the points P(1,3) and Q(2,5) such that PR=35PQ.

~~ 19 Find the values of k if the points A(k+1,2k),B(3k,2k+3) and C(5k1,5k) are collinear.

~~ 20 Find the ratio in which the line 2x+3y5=0 divides the line segment joining the points (8,9) and (2,1). Also find the coordinates of the point of division.

(E) Long Answer Questions

Sample Question 1: The mid-points D, E, F of the sides of a triangle ABC are (3,4), (8,9) and (6,7). Find the coordinates of the vertices of the triangle.

Solution : Since D and F are the mid-points of AB and AC, respectively, by mid-point theorem, we can prove that DFEB is a parallelogram. Let the coordinates of B be (x,y).

Fig. 7.3

Refer to Sample Question 4 of Section (D) to get

x=3+86=5y=4+97=6

Therefore, B(5,6) is one of the vertices of the triangle.

Similarly DFCE and DAFE are also parallelograms, and the coordinates of A are (3+68,4+79)=(1,2). Coordinates of C are (8+63,9+74)=(11,12). Thus, the coordinates of the vertices of the triangle are A(1,2),B(5,6) and C(11,12).

EXERCISE 7.4

~~ 1. If (4,3) and (4,3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

~~ 2. A(6,1),B(8,2) and C(9,4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ADE.

~~ 3. The points A(x1,y1),B(x2,y2) and C(x3y3) are the vertices of ΔABC.

(i) The median from A meets BC at D. Find the coordinates of the point D.

(ii) Find the coordinates of the point P on AD such that AP:PD=2:1

(iii) Find the coordinates of points Q and R on medians BE and CF, respectively such that BQ:QE=2:1 and CR:RF=2:1

(iv) What are the coordinates of the centroid of the triangle ABC ?

~~ 4. If the points A(1,2),B(2,3)C(a,2) and D(4,3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.

~~ 5. Students of a school are standing in rows and columns in their playground for a drill practice. A, B, C and D are the positions of four students as shown in figure 7.4. Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students A, B, C and D? If so, what should be his position?

Fig. 7.4

~~ 6. Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distances covered are in straight lines).

If the house is situated at (2,4), bank at (5,8), school at (13,14) and office at (13,26) and coordinates are in km.

Chapter 8

INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS

(A) Main Concepts and Results

  • Trigonometric Ratios of the angle A in a triangle ABC right angled at B are defined as:

Fig. 8.1

sine of A=sinA= side opposite to A hypotenuse =BCAC

cosine of A=cosA= side adjacent to A hypotenuse =ABAC

tangent of A=tanA= side opposite to A side adjacent to angle A=BCAB

cosecant of A=cosecA=1sinA=ACBC

secant of A=secA=1cosA=ACAB

cotangent of A=cotA=1tanA=ABBC

tanA=sinAcosA,cotA=cosAsinA

  • The values of trigonometric ratios of an angle do not vary with the lengths of the sides of the triangle, if the angle remains the same.

  • If one trigonometric ratio of an angle is given, the other trigonometric ratios of the angle can be determined.

  • Trigonometric ratios of angles: 0,30,45,60 and 90.

A 0 30 45 60 90
sinA 0 12 12 32 1
cosA 1 32 12 12 0
tanA 0 13 1 3 Not defined
\cosecA Not defined 2 2 23 1
secA 1 23 2 2 Not defined
cotA Not defined 3 1 13 0
  • The value of sinA or cosA never exceeds 1 , whereas the value of secA or cosecA is always greater than or equal to 1 .

  • Trigonometric ratios of complementary angles:

sin(90A)=cosA,cos(90A)=sinA

tan(90A)=cotA,cot(90A)=tanA

sec(90A)=cosecA,cosec(90A)=secA

  • Trigonometric identities:

cos2A+sin2A=1

1+tan2A=sec2A

cot2A+1=cosec2A

  • The ’line of sight’ is the line from the eye of an observer to the point in the object viewed by the observer.

  • The ‘angle of elevation’ of an object viewed, is the angle formed by the line of sight with the horizontal when it is above the horizontal level.

  • The angle of depression of an object viewed, is the angle formed by the line of sight with the horizontal when it is below the horizontal level.

  • The height or length of an object or the distance between two distinct objects can be determined with the help of trigonometric ratios.

(B) Multiple Choice Questions

Choose the correct answer from the given four options:

Sample Question 1 : The value of (sin30+cos30)(sin60+cos60) is

(A) -1

(B) 0

(C) 1

(D) 2

Solution : Answer (B)

Sample Question 2 : The value of tan30cot60 is

(A) 12

(B) 13

(C) 3

(D) 1

Solution : Answer (D)

Sample Question 3 : The value of (sin45+cos45) is

(A) 12

(B) 2

(C) 32

(D) 1

Solution : Answer (B)

EXERCISE 8.1

Choose the correct answer from the given four options:

~~ 1. If cosA=45, then the value of tanA is

(A) 35

(B) 34

(C) 43

(D) 53

~~ 2. If sinA=12, then the value of cotA is

(A) 3

(B) 13

(C) 32

(D) 1

~~ 3. The value of the expression [cosec(75+θ)sec(15θ)tan(55+θ)+. .cot(35θ)] is

(A) -1

(B) 0

(C) 1

(D) 32

~~ 4. Given that sinθ=ab, then cosθ is equal to

(A) bb2a2

(B) ba

(C) b2a2b

(D) ab2a2

~~ 5. If cos(α+β)=0, then sin(αβ) can be reduced to

(A) cosβ

(B) cos2β

(C) sinα

(D) sin2α

~~ 6. The value of (tan1tan2tan3tan89) is

(A) 0

(B) 1

(C) 2

(D) 12

~~ 7. If cos9α=sinα and 9α<90, then the value of tan5α is

(A) 13

(B) 3

(C) 1

(D) 0

~~ 8. If ABC is right angled at C, then the value of cos(A+B) is

(A) 0

(B) 1

(C) 12

(D) 32

~~ 9. If sinA+sin2A=1, then the value of the expression (cos2A+cos4A) is

(A) 1

(B) 12

(C) 2

(D) 3

~~ 10 Given that sinα=12 and cosβ=12, then the value of (α+β) is

(A) 0

(B) 30

(C) 60

(D) 90

~~ 11 The value of the expression [sin222+sin268cos222+cos268+sin263+cos63sin27] is

(A) 3

(B) 2

(C) 1

(D) 0

~~ 12 If 4tanθ=3, then (4sinθcosθ4sinθ+cosθ) is equal to

(A) 23

(B) 13

(C) 12

(D) 34

~~ 13 If sinθcosθ=0, then the value of (sin4θ+cos4θ) is

(A) 1

(B) 34

(C) 12

(D) 14

~~ 14 sin(45+θ)cos(45θ) is equal to

(A) 2cosθ

(B) 0

(C) 2sinθ

(D) 1

~~ 15 A pole 6m high casts a shadow 23m long on the ground, then the Sun’s elevation is

(A) 60

(B) 45

(C) 30

(D) 90

(C) Short Answer Questions with Reasoning

Write ‘True’ or ‘False’ and justify your answer.

Sample Question 1 : The value of sinθ+cosθ is always greater than 1 .

Solution: False.

The value of (sinθ+cosθ) for θ=0 is 1 .

Sample Question 2 : The value of tanθ(θ<90) increases as θ increases.

Solution : True.

alt text Fig. 8.2

In Fig. 8.2, B is moved closer to C along BC. It is observed that

(i) θ increases (as θ1>θ,θ2>θ1, ) and

(ii) BC decreases (B1C<BC,B2C<B1C,)

Thus the perpendicular AC remains fixed and the base BC decreases. Hence tanθ increases as θ increases.

Sample Question 3: tanθ increases faster than sinθ as θ increases.

Solution : True

We know that sinθ increases as θ increases but cosθ decreases as θ increases.

We have tanθ=sinθcosθ

Now as θ increases, sinθ increases but cosθ decreases. Therefore, in case of tanθ, the numerator increases and the denominator decreases. But in case of sinθ which can be seen as sinθ1, only the numerator increases but the denominator remains fixed at 1 . Hence tanθ increases faster than sinθ as θ increases.

Sample Question 4 : The value of sinθ is a+1a, where ’ a ’ is a positive number.

Solution : False.

We know that a1a20 or a+1a2, but sinθ is not greater than 1 .

Alternatively, there exists the following three posibilities :

Case 1. If a<1, then a+1a>1

Case 2. If a=1, then a+1a>1

Case 3. If a>1, then a+1a>1

However, sinθ cannot be greater than 1 .

EXERCISE 8.2

Write ‘True’ or ‘False’ and justify your answer in each of the following:

~~ 1. tan47cot43=1

~~ 2. The value of the expression (cos223sin267) is positive.

~~ 3. The value of the expression (sin80cos80) is negative.

~~ 4. (1cos2θ)sec2θ=tanθ

~~ 5. If cosA+cos2A=1, then sin2A+sin4A=1.

~~ 6. (tanθ+2)(2tanθ+1)=5tanθ+sec2θ.

~~ 7. If the length of the shadow of a tower is increasing, then the angle of elevation of the sun is also increasing.

~~ 8. If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection.

~~ 9. The value of 2sinθ can be a+1a, where a is a positive number, and a1.

~~ 10 cosθ=a2+b22ab, where a and b are two distinct numbers such that ab>0.

~~ 11 The angle of elevation of the top of a tower is 30. If the height of the tower is doubled, then the angle of elevation of its top will also be doubled.

~~ 12 If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10, then the angle of elevation of its top remains unchanged.

(D) Short Answer Questions

Sample Question 1 : Prove that sin6θ+cos6θ+3sin2θcos2θ=1

Solution : We know that sin2θ+cos2θ=1

Therefore,

(sin2θ+cos2θ)3=1

or, (sin2θ)3+(cos2θ)3+3sin2θcos2θ(sin2θ+cos2θ)=1

or,

sin6θ+cos6θ+3sin2θcos2θ=1

Sample Question 2 : Prove that (sin4θcos4θ+1)cosec2θ=2

Solution :

L.H.S. =(sin4θcos4θ+1)cosec2θ

=[(sin2θcos2θ)(sin2θ+cos2θ)+1]cosec2θ=(sin2θcos2θ+1)cosec2θ

[Because sin2θ+cos2θ=1 ]

=2sin2θcosec2θ[ Because 1cos2θ=sin2θ]=2= RHS 

Sample Question 3 : Given that α+β=90, show that

cosαcosecβcosαsinβ=sinα

Solution :

cosαcosecβcosαsinβ=cosαcosec(90α)cosαsin(90α)

[Given α+β=90 ]

=cosαsecαcosαcosα=1cos2α=sinα

Sample Question 4 : If sinθ+cosθ=3, then prove that tanθ+cotθ=1

Solution :

sinθ+cosθ=3

 or (sinθ+cosθ)2=3

 or sin2θ+cos2θ+2sinθcosθ=3

2sinθcosθ=2

 or sinθcosθ=1=sin2θ+cos2θ

 or 1=sin2θ+cos2θsinθcosθ

Therefore, tanθ+cotθ=1

EXERCISE 8.3

Prove the following (from Q. 1 to Q.7):

~~ 1. sinθ1+cosθ+1+cosθsinθ=2cosecθ

~~ 2. tanA1+secAtanA1secA=2cosecA

~~ 3. If tanA=34, then sinAcosA=1225

~~ 4. (sinα+cosα)(tanα+cotα)=secα+cosecα

~~ 5. (3+1)(3cot30)=tan3602sin60

~~ 6. 1+cot2α1+cosecα=cosecα

~~ 7. tanθ+tan(90θ)=secθsec(90θ)

~~ 8. Find the angle of elevation of the sun when the shadow of a pole h metres high is 3h metres long.

~~ 9. If 3tanθ=1, then find the value of sin2θcos2θ.

~~ 10 A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60 with the wall, find the height of the wall.

~~ 11 Simplify (1+tan2θ)(1sinθ)(1+sinθ)

~~ 12 If 2sin2θcos2θ=2, then find the value of θ.

~~ 13 Show that cos2(45+θ)+cos2(45θ)tan(60+θ)tan(30θ)=1

~~ 14 An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.

~~ 15 Show that tan4θ+tan2θ=sec4θsec2θ.

(E) Long Answer Questions

Sample Question 1: A spherical balloon of radius r subtends an angle θ at the eye of an observer. If the angle of elevation of its centre is ϕ, find the height of the centre of the balloon.

Solution : In Fig. 8.3, O is the centre of balloon, whose radius OP=r and PAQ=θ. Also, OAB=ϕ.

Fig. 8.3

Let the height of the centre of the balloon be h. Thus OB=h.

Now, from OAP,sinθ2=rd, where OA=d

Also from OAB,sinϕ=hd.

From (1) and (2), we get sinϕsinθ2=hrrd=hr

or h=rsinϕcosecθ2.

Sample Question 2 : From a balloon vertically above a straight road, the angles of depression of two cars at an instant are found to be 45 and 60. If the cars are 100m apart, find the height of the balloon.

Solution : Let the height of the balloon at P be h meters (see Fig. 8.4). Let A and B be the two cars. Thus AB=100m. From PAQ,AQ=PQ=h

Now from PBQ,PQBQ=tan60=3 or hh100=3

or h=3(h100)

Therefore, h=100331=50(3+3)

i.e., the height of the balloon is 50(3+3)m.

Sample Question 3: The angle of elevation of a cloud from a point h metres above the surface of a lake is θ and the angle of depression of its reflection in the lake is ϕ.

Prove that the height of the cloud above the lake is h(tanϕ+tanθtanϕtanθ).

Solution : Let P be the cloud and Q be its reflection in the lake (see Fig. 8.5). Let A be the point of observation such that AB=h.

Fig. 8.5

Let the height of the cloud above the lake be x. Let AL=d.

Now from PAL, xhd=tanθ

From ΔQAL,x+hd=tanϕ

From (1) and (2), we get

x+hxh=tanϕtanθ or 2x2h=tanϕ+tanθtanϕtanθ

Therefore, x=h(tanϕ+tanθtanϕtanθ).

EXERCISE 8.4

~~ 1. If cosecθ+cotθ=p, then prove that cosθ=p21p2+1.

~~ 2. Prove that sec2θ+cosec2θ=tanθ+cotθ

~~ 3. The angle of elevation of the top of a tower from certain point is 30. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15. Find the height of the tower.

~~ 4. If 1+sin2θ=3sinθcosθ, then prove that tanθ=1 or 12.

~~ 5. Given that sinθ+2cosθ=1, then prove that 2sinθcosθ=2.

~~ 6. The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is st.

~~ 7. The shadow of a tower standing on a level plane is found to be 50m longer when Sun’s elevation is 30 than when it is 60. Find the height of the tower.

~~ 8. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are α and β, respectively. Prove that the height of the tower is (htanαtanβtanα).

~~ 9. If tanθ+secθ=l, then prove that secθ=l2+12l.

~~ 10 If sinθ+cosθ=p and secθ+cosecθ=q, then prove that q(p21)=2p.

~~ 11 If asinθ+bcosθ=c, then prove that acosθbsinθ=a2+b2c2.

~~ 12 Prove that 1+secθtanθ1+secθ+tanθ=1sinθcosθ

~~ 13 The angle of elevation of the top of a tower 30m high from the foot of another tower in the same plane is 60 and the angle of elevation of the top of the second tower from the foot of the first tower is 30. Find the distance between the two towers and also the height of the other tower.

~~ 14 From the top of a tower hm high, the angles of depression of two objects, which are in line with the foot of the tower are α and β(β>α). Find the distance between the two objects.

~~ 15 A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal.

Show that pq=cosβcosαsinαsinβ.

~~ 16 The angle of elevation of the top of a vertical tower from a point on the ground is 60. From another point 10m vertically above the first, its angle of elevation is 45. Find the height of the tower.

~~ 17 A window of a house is h metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be α and β, respectively. Prove that the height of the other house is h(1+tanαcotβ) metres.

~~ 18 The lower window of a house is at a height of 2m above the ground and its upper window is 4m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60 and 30, respectively. Find the height of the balloon above the ground.

Chapter 9

CIRCLES

(A) Main Concepts and Results

  • The meaning of a tangent and its point of contact on a circle.
  • Tangent is perpendicular to the radius through the point of contact.
  • Only two tangents can be drawn to a circle from an external point.
  • Lengths of tangents from an external point to a circle are equal.

(B) Multiple Choice Questions

Choose the correct answer from the given four options:

Sample Question 1 : If angle between two radii of a circle is 130, the angle between the tangents at the ends of the radii is :

(A) 90

(B) 50

(C) 70

(D) 40

Solution : Answer (B)

Sample Question 2 : In Fig. 9.1, the pair of tangents AP and AQ drawn from an external point A to a circle with centre O are perpendicular to each other and length of each tangent is 5cm. Then the radius of the circle is

Fig. 9.1

(A) 10cm

(B) 7.5cm

(C) 5cm

(D) 2.5cm

Solution : Answer (C)

Sample Question 3: In Fig. 9.2, PQ is a chord of a circle and PT is the tangent at P such that QPT=60. Then PRQ is equal to

Fig. 9.2

(A) 135

(B) 150

(C) 120

(D) 110

Solution : Answer (C)

[Hint: OPQ=OQP=30, i.e., POQ

=120. Also, .PRQ=12reflexPOQ]

EXERCISE 9.1

Choose the correct answer from the given four options:

~~ 1. If radii of two concentric circles are 4cm and 5cm, then the length of each chord of one circle which is tangent to the other circle is

(A) 3cm

(B) 6cm

(C) 9cm

(D) 1cm

~~ 2. In Fig. 9.3, if AOB=125, then COD is equal to

(A) 62.5

(B) 45

(C) 35

(D) 55

Fig. 9.3

~~ 3. In Fig. 9.4, AB is a chord of the circle and AOC is its diameter such that ACB=50. If AT is the tangent to the circle at the point A, then BAT is equal to

(A) 65

(B) 60

(C) 50

(D) 40

Fig. 9.4

~~ 4. From a point P which is at a distance of 13cm from the centre O of a circle of radius 5cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is

(A) 60cm2

(B) 65cm2

(C) 30cm2

(D) 32.5cm2

~~ 5. At one end A of a diameter AB of a circle of radius 5cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8cm from A is

(A) 4cm

(B) 5cm

(C) 6cm

(D) 8cm

~~ 6. In Fig. 9.5, AT is a tangent to the circle with centre O such that OT=4cm and OTA=30. Then AT is equal to

(A) 4cm

(B) 2cm

(C) 23cm

(D) 43cm

Fig. 9.5

~~ 7. In Fig. 9.6, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50 with PQ, then POQ is equal to

(A) 100

(B) 80

(C) 90

(D) 75

~~ 8. In Fig. 9.7, if PA and PB are tangents to the circle with centre O such that APB=50, then OAB is equal to

(A) 25

(B) 30

(C) 40

(D) 50

Fig. 9.7

~~ 9. If two tangents inclined at an angle 60 are drawn to a circle of radius 3cm, then length of each tangent is equal to

(A) 323cm

(B) 6cm

(C) 3cm

(D) 33cm

~~ 10 In Fig. 9.8, if PQR is the tangent to a circle at Q whose centre is O,AB is a chord parallel to PR and BQR=70, then AQB is equal to

Fig. 9.8

(A) 20

(B) 40

(C) 35

(D) 45

(C) Short Answer Questions with Reasoning

Write ‘True’ or ‘False’ and give reasons for your answer.

Sample Question 1 : In Fig. 9.9, BOA is a diameter of a circle and the tangent at a point P meets BA extended at T. If PBO=30, then PTA is equal to 30.

Fig. 9.9

Solution : True. As BPA=90,PAB=OPA=60. Also, OPPT. Therefore, APT=30 and PTA=6030=30.

Sample Question 2 : In Fig. 9.10, PQL and PRM are tangents to the circle with centre O at the points Q and R, respectively and S is a point on the circle such that SQL=50 and SRM=60. Then QSR is equal to 40.

Solution : False. Here OSQ=OQS=9050=40 and RSO=SRO= 9060=30. Therefore, QSR=40+30=70.

EXERCISE 9.2

Write ‘True’ or ‘False’ and justify your answer in each of the following :

~~ 1. If a chord AB subtends an angle of 60 at the centre of a circle, then angle between the tangents at A and B is also 60.

~~ 2. The length of tangent from an external point on a circle is always greater than the radius of the circle.

~~ 3. The length of tangent from an external point P on a circle with centre O is always less than OP.

~~ 4. The angle between two tangents to a circle may be 0.

~~ 5. If angle between two tangents drawn from a point P to a circle of radius a and centre O is 90, then OP=a2.

~~ 6. If angle between two tangents drawn from a point P to a circle of radius a and centre O is 60, then OP=a3.

~~ 7. The tangent to the circumcircle of an isosceles triangle ABC at A, in which AB=AC, is parallel to BC.

~~ 8. If a number of circles touch a given line segment PQ at a point A, then their centres lie on the perpendicular bisector of PQ.

~~ 9. If a number of circles pass through the end points P and Q of a line segment PQ, then their centres lie on the perpendicular bisector of PQ.

~~ 10 AB is a diameter of a circle and AC is its chord such that BAC=30. If the tangent at C intersects AB extended at D, then BC=BD.

(D) Short Answer Questions

Sample Question 1 : If d1,d2(d2>d1) be the diameters of two concentric circles and c be the length of a chord of a circle which is tangent to the other circle, prove that d22=c2+d12.

Solution : Let AB be a chord of a circle which touches the other circle at C. Then OCB is right triangle (see Fig.9.11). By Pythagoras theorem OC2+CB2=OB2.

 i.e., 12d12+12c2=12d22

(As C bisects AB )

Therefore, d22=c2+d12.

Fig. 9.11

Sample Question 2: If a,b,c are the sides of a right triangle where c is the hypotenuse, prove that the radius r of the circle which touches the sides of the triangle is given by r=a+bc2.

Solution : Let the circle touches the sides BC, CA, AB of the right triangle ABC at D,E and F respectively, where BC=a,CA=b and AB=c (see Fig. 9.12). Then AE=AF and BD=BF. Also CE=CD=r.

Fig. 9.12

i.e., br=AF,ar=BF

or AB=c=AF+BF=br+ar

This gives r=a+bc2

EXERCISE 9.3

~~ 1. Out of the two concentric circles, the radius of the outer circle is 5cm and the chord AC of length 8cm is a tangent to the inner circle. Find the radius of the inner circle.

~~ 2. Two tangents PQ and PR are drawn from an external point to a circle with centre O. Prove that QORP is a cyclic quadrilateral.

~~ 3. If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that DBC=120, prove that BC+BD=BO, i.e., BO=2BC.

~~ 4. Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.

~~ 5. In Fig. 9.13, AB and CD are common tangents to two circles of unequal radii. Prove that AB=CD.

Fig. 9.13

~~ 6. In Question 5 above, if radii of the two circles are equal, prove that AB=CD.

~~ 7. In Fig. 9.14, common tangents AB and CD to two circles intersect at E. Prove that AB=CD.

Fig. 9.14

~~ 8. A chord PQ of a circle is parallel to the tangent drawn at a point R of the circle. Prove that R bisects the arc PRQ.

~~ 9. Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

~~ 10 Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.

(E) Long Answer Questions

Sample Question 1 : In Fig. 9.15, from an external point P, a tangent PT and a line segment PAB is drawn to a circle with centre O. ON is perpendicular on the chord AB. Prove that :

(i) PAPB=PN2AN2

(ii) PN2AN2=OP2OT2

(iii) PA.PB=PT2

Fig. 9.15

Solution :

(i) PAPB=(PNAN)(PN+BN)

=(PNAN)(PN+AN)(AsAN=BN)=PN2AN2

(ii) PN2AN2=(OP2ON2)AN2(AsONPN)

=OP2(ON2+AN2)=OP2OA2(AsONAN)=OP2OT2(AsOA=OT)

(iii)

From (i) and (ii)

PA.PB=OP2OT2=PT2

(. As .OTP=90)

Sample Question 2 : If a circle touches the side BC of a triangle ABC at P and extended sides AB and AC at Q and R, respectively, prove that AQ=12(BC+CA+AB)

Solution : See Fig. 9.16.

By Theorem 10.2 of the textbook,

Fig. 9.16

BQ=BPCP=CR, and AQ=AR2AQ=AQ+AR=(AB+BQ)+(AC+CR)=AB+BP+AC+CP=(BP+CP)+AC+AB=BC+CA+AB

Now,

i.e., AQ=12(BC+CA+AB).

EXERCISE 9.4

~~ 1. If a hexagon ABCDEF circumscribe a circle, prove that AB+CD+EF=BC+DE+FA.

2. Let s denote the semi-perimeter of a triangle ABC in which BC=a,CA=b, AB=c. If a circle touches the sides BC,CA,AB at D,E,F, respectively, prove that BD=sb.

~~ 3. From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA=10cm, find the the perimeter of the triangle PCD.

~~ 4. If AB is a chord of a circle with centre O,AOC is a diameter and AT is the tangent at A as shown in Fig. 9.17. Prove that

BAT=ACB

Fig. 9.17

~~ 5. Two circles with centres O and O of radii 3cm and 4cm, respectively intersect at two points P and Q such that OP and OP are tangents to the two circles. Find the length of the common chord PQ.

~~ 6. In a right triangle ABC in which B=90, a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC.

~~ 7. In Fig. 9.18, tangents PQ and PR are drawn to a circle such that RPQ= 30. A chord RS is drawn parallel to the tangent PQ. Find the RQS.

[Hint: Draw a line through Q and perpendicular to QP.]

Fig. 9.18

~~ 8. AB is a diameter and AC is a chord of a circle with centre O such that BAC=30. The tangent at C intersects extended AB at a point D. Prove that BC=BD.

~~ 9. Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

~~ 10 In Fig. 9.19, the common tangent, AB and CD to two circles with centres O and O intersect at E. Prove that the points O,E,O ’ are collinear.

Fig. 9.19

~~ 11 In Fig. 9.20. O is the centre of a circle of radius 5cm,T is a point such that OT=13cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find the length of AB.

Fig. 9.20

~~ 12 The tangent at a point C of a circle and a diameter AB when extended intersect at P. If PCA=110, find CBA [see Fig. 9.21].

Fig. 9.21

[Hint: Join C with centre O.]

~~ 13 If an isosceles triangle ABC, in which AB=AC=6cm, is inscribed in a circle of radius 9cm, find the area of the triangle.

~~ 14 A is a point at a distance 13cm from the centre O of a circle of radius 5cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ABC.

Chapter 10

CONSTRUCTIONS

(A) Main Concepts and Results

  • Division of a line segment internally in a given ratio.

  • Construction of a triangle similar to a given triangle as per given scale factor which may be less than 1 or greater than 1 .

  • Construction of the pair of tangents from an external point to a circle.

(B) Multiple Choice Questions

Choose the correct answer from the given four options:

Sample Question 1: To divide a line segment AB in the ratio p:q(p,q are positive integers), draw a ray AX so that BAX is an acute angle and then mark points on ray AX at equal distances such that the minimum number of these points is

(A) greater of p and q

(B) p+q

(C) p+q1

(D) pq

Solution : Answer (B)

Sample Question 2: To draw a pair of tangents to a circle which are inclined to each other at an angle of 35, it is required to draw tangents at the end points of those two radii of the circle, the angle between which is

(A) 105

(B) 70

(C) 140

(D) 145

Solution : Answer (D)

EXERCISE 10.1

Choose the correct answer from the given four options:

~~ 1. To divide a line segment AB in the ratio 5:7, first a ray AX is drawn so that BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is

(A) 8

(B) 10

(C) 11

(D) 12

~~ 2. To divide a line segment AB in the ratio 4:7, a ray AX is drawn first such that BAX is an acute angle and then points A1,A2,A3,. are located at equal distances on the ray AX and the point B is joined to

(A) A12

(B) A11

(C) A10

(D) A9

~~ 3. To divide a line segment AB in the ratio 5:6, draw a ray AX such that BAX is an acute angle, then draw a ray BY parallel to AX and the points A1,A2,A3, and B1,B2,B3, are located at equal distances on ray AX and BY, respectively. Then the points joined are

(A) A5 and B6

(B) A6 and B5

(C) A4 and B5

(D) A5 and B4

~~ 4. To construct a triangle similar to a given ABC with its sides 37 of the corresponding sides of ABC, first draw a ray BX such that CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B1,B2,B3, on BX at equal distances and next step is to join

(A) B10 to C

(B) B3 to C

(C) B7 to C

(D) B4 to C

~~ 5. To construct a triangle similar to a given ABC with its sides 85 of the corresponding sides of ABC draw a ray BX such that CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is

(A) 5

(B) 8

(C) 13

(D) 3

~~ 6. To draw a pair of tangents to a circle which are inclined to each other at an angle of 60, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be

(A) 135

(B) 90

(C) 60

(D) 120

(C) Short Answer Questions with Reasoning

Write True or False and give reasons for your answer.

Sample Questions 1 : By geometrical construction, it is possible to divide a line segment in the ratio 2+3:23.

Solution : False. As 2+3:23 can be simplified as 7+43:1 and 7+43 is not a positive integer, while 1 is.

EXERCISE 10.2

Write True or False and give reasons for your answer in each of the following:

~~ 1. By geometrical construction, it is possible to divide a line segment in the ratio

3:13

~~ 2. To construct a triangle similar to a given ABC with its sides 73 of the corresponding sides of ABC, draw a ray BX making acute angle with BC and X lies on the opposite side of A with respect to BC. The points B1,B2,, B7 are located at equal distances on BX,B3 is joined to C and then a line segment B6C is drawn parallel to B3C where C lies on BC produced. Finally, line segment AC is drawn parallel to AC.

~~ 3. A pair of tangents can be constructed from a point P to a circle of radius 3.5cm situated at a distance of 3cm from the centre.

~~ 4. A pair of tangents can be constructed to a circle inclined at an angle of 170.

(D) Short Answer Questions

Sample Question 1 : Draw an equilateral triangle ABC of each side 4cm. Construct a triangle similar to it and of scale factor 35. Is the new triangle also an equilateral?

Solution : Follow the similar steps as given in Mathematics Textbook for Class X. Yes, the new triangle is also equilateral.

EXERCISE 10.3

~~ 1. Draw a line segment of length 7cm. Find a point P on it which divides it in the ratio 3:5.

~~ 2. Draw a right triangle ABC in which BC=12cm,AB=5cm and B=90. Construct a triangle similar to it and of scale factor 23. Is the new triangle also a right triangle?

~~ 3. Draw a triangle ABC in which BC=6cm,CA=5cm and AB=4cm. Construct a triangle similar to it and of scale factor 53.

~~ 4. Construct a tangent to a circle of radius 4cm from a point which is at a distance of 6cm from its centre.

(E) Long Answer Questions

Sample Questions 1 : Given a rhombus ABCD in which AB=4cm and ABC=60, divide it into two triangles say, ABC and ADC. Construct the triangle ABC similar to ABC with scale factor 23. Draw a line segment CD parallel to CD where D lies on AD. Is ABCD a rhombus? Give reasons.

Solution : First draw the rhombus ABCD in which AB=4cm and ABC=60 as given in Fig. 10.1 and join AC. Construct the triangle ABC similar to ABC with scale factor 23 as instructed in the Mathematics Textbook for Class X (See Fig. 10.1).

Finally draw the line segment

Now

ABAB=23=ACAC

Also ACAC=CDCD=ADAD=23

Therefore, AB=BC=CD=AD=23AB.

i.e., ABCD is a rhombus.

EXERCISE 10.4

~~ 1. Two line segments AB and AC include an angle of 60 where AB=5cm and AC=7cm. Locate points P and Q on AB and AC, respectively such that AP=34AB and AQ=14AC. Join P and Q and measure the length PQ.

~~ 2. Draw a parallelogram ABCD in which BC=5cm,AB=3cm and ABC=60, divide it into triangles BCD and ABD by the diagonal BD.

Construct the triangle BDC similar to BDC with scale factor 43. Draw the line segment DA parallel to DA where A lies on extended side BA. Is ABC2D a parallelogram?

~~ 3. Draw two concentric circles of radii 3cm and 5cm. Taking a point on outer circle construct the pair of tangents to the other. Measure the length of a tangent and verify it by actual calculation.

~~ 4. Draw an isosceles triangle ABC in which AB=AC=6cm and BC=5cm. Construct a triangle PQR similar to ABC in which PQ=8cm. Also justify the construction.

~~ 5. Draw a triangle ABC in which AB=5cm,BC=6cm and ABC=60. Construct a triangle similar to ABC with scale factor 57. Justify the construction.

~~ 6. Draw a circle of radius 4cm. Construct a pair of tangents to it, the angle between which is 60. Also justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.

~~ 7. Draw a triangle ABC in which AB=4cm,BC=6cm and AC=9cm. Construct a triangle similar to ABC with scale factor 32. Justify the construction. Are the two triangles congruent? Note that all the three angles and two sides of the two triangles are equal.

Chapter 11

(A) Main Concepts and Results

Perimeters and areas of simple closed figures. Circumference and area of a circle. Area of a circular path (i.e., ring). Sector of a circle and its central angle - Major and Minor sectors. Segment of a circle - Major and Minor segments.

  • Circumference of a circle =2πr and area of a circle =πr2, where r is the radius of the circle.

  • Area of the circular path formed by two concentric circles of radii r1 and r2(r1>r2)=πr12πr22=π(r12r22).

  • Area of the sector of a circle of radius r with central angle θ=θ360×πr2, where θ is measured in degrees.

  • Length of the arc of the sector of a circle of radius r with central angle θ=θ360×2πr, where θ is measured in degrees.

  • Area of the minor segment APB of the circle in Fig. 11.1= area of sector OAPB - area of OAB.

  • Area of the major sector of a circle of radius r =πr2 area of the corresponding minor sector.

Fig. 11.1

  • Area of the major segment of a circle of radius r=πr2 area of the corresponding minor segment.

Note: Unless stated otherwise, the value of π is to be taken as 227.

(B) Multiple Choice Questions

Choose the correct answer from the given four options:

Sample Question 1 : If the area of a circle is 154cm2, then its perimeter is

(A) 11cm

(B) 22cm

(C) 44cm

(D) 55cm

Solution : Answer (C)

Sample Question 2: If θ is the angle (in degrees) of a sector of a circle of radius r, then area of the sector is

(A) πr2θ360

(B) πr2θ180

(C) 2πrθ360

(D) 2πrθ180

Solution : Answer (A)

EXERCISE 11.1

Choose the correct answer from the given four options:

~~ 1. If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then

(A) R1+R2=R

(B) R12+R22=R2

(C) R1+R2<R

(D) R12+R22<R2

~~ 2. If the sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a circle of radius R, then

(A) R1+R2=R

(B) R1+R2>R

(C) R1+R2<R

(D) Nothing definite can be said about the relation among R1,R2 and R.

~~ 3. If the circumference of a circle and the perimeter of a square are equal, then

(A) Area of the circle = Area of the square

(B) Area of the circle > Area of the square

(C) Area of the circle < Area of the square

(D) Nothing definite can be said about the relation between the areas of the circle and square.

~~ 4. Area of the largest triangle that can be inscribed in a semi-circle of radius r units is

(A) r2 sq. units

(B) 12r2 sq. units

(C) 2r2 sq. units

(D) 2r2 sq. units

~~ 5. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is

(A) 22:7

(B) 14:11

(C) 7:22

(D) 11:14

~~ 6. It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16m and 12m in a locality. The radius of the new park would be

(A) 10m

(B) 15m

(C) 20m

(D) 24m

~~ 7. The area of the circle that can be inscribed in a square of side 6cm is

(A) 36πcm2

(B) 18πcm2

(C) 12πcm2

(D) 9πcm2

~~ 8. The area of the square that can be inscribed in a circle of radius 8cm is

(A) 256cm2

(B) 128cm2

(C) 642cm2

(D) 64cm2

~~ 9. The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36cm and 20cm is

(A) 56cm

(B) 42cm

(C) 28cm

(D) 16cm

~~ 10 The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24cm and 7cm is

(A) 31cm

(B) 25cm

(C) 62cm

(D) 50cm

(C) Short Answer Questions with Reasoning

Sample Question 1 : Is the following statement true? Give reasons for your answer.

Area of a segment of a circle = area of the corresponding sector - area of the corresponding triangle.

Solution : Statement is not true. It is true only for a minor segment. In the case of a major segment, area of the triangle will have to be added to the corresponding area of the sector.

Sample Question 2 : In Fig. 11.2, a circle is inscribed in a square of side 5cm and another circle is circumscribing the square. Is it true to say that area of the outer circle is two times the area of the inner circle? Give reasons for your answer.

Fig. 11.2

Solution : It is true, because diameter of the inner circle =5cm and that of outer circle = diagonal of the square =52cm.

So, A1=π(522)2 and A2=π(52)2, giving A1A2=2

EXERCISE 11.2

~~ 1. Is the area of the circle inscribed in a square of side acm,πa2cm2 ? Give reasons for your answer.

~~ 2. Will it be true to say that the perimeter of a square circumscribing a circle of radius acm is 8acm ? Give reasons for your answer.

~~ 3. In Fig 11.3, a square is inscribed in a circle of diameter d and another square is circumscribing the circle. Is the area of the outer square four times the area of the inner square? Give reasons for your answer.

Fig.11.3

~~ 4. Is it true to say that area of a segment of a circle is less than the area of its corresponding sector? Why?

~~ 5. Is it true that the distance travelled by a circular wheel of diameter dcm in one revolution is 2πdcm ? Why?

~~ 6. In covering a distance s metres, a circular wheel of radius r metres makes s2πr revolutions. Is this statement true? Why?

~~ 7. The numerical value of the area of a circle is greater than the numerical value of its circumference. Is this statement true? Why?

~~ 8. If the length of an arc of a circle of radius r is equal to that of an arc of a circle of radius 2r, then the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle. Is this statement false? Why?

~~ 9. The areas of two sectors of two different circles with equal corresponding arc lengths are equal. Is this statement true? Why?

~~ 10 The areas of two sectors of two different circles are equal. Is it necessary that their corresponding arc lengths are equal? Why?

~~ 11 Is the area of the largest circle that can be drawn inside a rectangle of length acm and breadth bcm(a>b) is πb2cm2 ? Why?

~~ 12 Circumferences of two circles are equal. Is it necessary that their areas be equal? Why?

~~ 13 Areas of two circles are equal. Is it necessary that their circumferences are equal? Why?

~~ 14 Is it true to say that area of a square inscribed in a circle of diameter pcm is p2cm2 ? Why?

(D) Short Answer Questions

Sample Question 1: Find the diameter of the circle whose area is equal to the sum of the areas of the two circles of diameters 20cm and 48cm.

Solution : Here, radius r1 of first circle =202cm=10cm and radius r2 of the second circle =482cm=24cm

Therefore, sum of their areas =πr12+πr22=π(10)2+π(24)2=π×676

Let the radius of the new circle be rcm. Its area =πr2

Therefore, from (1) and (2),

πr2=π×676

or r2=676

i.e., r=26

Thus, radius of the new circle =26cm

Hence, diameter of the new circle =2×26cm=52cm

Sample Question 2 : Find the area of a sector of circle of radius 21cm and central angle 120.

Solution : Area of the sector =θ360×πr2

=120360×227×(21)2cm2=22×21cm2=462cm2

Sample Question 3 : In Fig 11.4, a circle of radius 7.5cm is inscribed in a square.

Find the area of the shaded region (Use π=3.14 )

Fig.11.4

Solution : Area of the circle =πr2

=3.14×(7.5)2cm2=176.625cm2

Clearly, side of the square = diameter of the circle =15cm

So, area of the square =152cm2=225cm2

Therefore, area of the shaded region

=225cm2176.625cm2=48.375cm2

Sample Question 4 : Area of a sector of a circle of radius 36cm is 54πcm2. Find the length of the corresponding arc of the sector.

Solution : Let the central angle (in degrees) be θ.

So,

π×(36)2θ360=54πθ=54×36036×36=15

or

Now, length of the arc =θ360×2πr

=15360×2π×36cm=3πcm

EXERCISE 11.3

~~ 1. Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 15cm and 18cm.

~~ 2. In Fig. 11.5, a square of diagonal 8cm is inscribed in a circle. Find the area of the shaded region.

Fig.11.5

~~ 3. Find the area of a sector of a circle of radius 28cm and central angle 45.

~~ 4. The wheel of a motor cycle is of radius 35cm. How many revolutions per minute must the wheel make so as to keep a speed of 66km/h ?

~~ 5. A cow is tied with a rope of length 14m at the corner of a rectangular field of dimensions 20m×16m. Find the area of the field in which the cow can graze.

~~ 6. Find the area of the flower bed (with semi-circular ends) shown in Fig. 11.6.

Fig. 11.6

~~ 7. In Fig. 11.7, AB is a diameter of the circle, AC=6cm and BC=8cm. Find the area of the shaded region (Use π=3.14 ).

Fig. 11.7

~~ 8. Find the area of the shaded field shown in Fig. 11.8.

~~ 9. Find the area of the shaded region in Fig. 11.9.

Fig. 11.9

~~ 10 Find the area of the minor segment of a circle of radius 14cm, when the angle of the corresponding sector is 60.

~~ 11 Find the area of the shaded region in Fig. 11.10, where arcs drawn with centres A,B,C and D intersect in pairs at mid-points P,Q,R and S of the sides AB,BC, CD and DA, respectively of a square ABCD (Use π=3.14 ).

Fig. 11.10

~~ 12 In Fig. 11.11, arcs are drawn by taking vertices A,B and C of an equilateral triangle of side 10cm. to intersect the sides BC,CA and AB at their respective mid-points D,E and F. Find the area of the shaded region (Use π=3.14 ).

Fig. 11.11

~~ 13 In Fig. 11.12, arcs have been drawn with radii 14cm each and with centres P,Q and R. Find the area of the shaded region.

Fig. 11.12

~~ 14 A circular park is surrounded by a road 21m wide. If the radius of the park is 105m, find the area of the road.

~~ 15 In Fig. 11.13, arcs have been drawn of radius 21cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.

Fig. 11.13

~~ 16 A piece of wire 20cm long is bent into the form of an arc of a circle subtending an angle of 60 at its centre. Find the radius of the circle.

(E) Long Answer Questions

Sample Question 1: A chord of a circle of radius 20cm subtends an angle of 90 at the centre. Find the area of the corresponding major segment of the circle. (Use π=3.14 ).

Solution : Let AB be the chord of a circle of radius 10cm, with O as the centre of the circle (see Fig. 11.14).

Fig. 11.14

Here, AOB=90 and we have to find the area of the major segment (which is shaded). As AOB=90, therefore angle of the major sector =36090=270

So, area of the major sector =270360×π×(10)2cm2

=34×3.14×100cm2=75×3.14cm2=235.5cm2

Now, to find the area of OAB, draw OMAB.

So, AM=12AB and AOM=12×90=45.

Now, AMOA=sin45=12

So, AM=10×12cm.

Therefore, AB=102cm and OM=OAcos45=10×12cm=52cm So, area of ΔOAB=12 base × height

=12102×52cm2=50cm2

Therefore, the area of the required major segment

=235.5cm2+50cm2=285.5cm2

Another method for the area of OAB

 As, AOB=90

Therefore, area of ΔOAB=12OA×OB

=1210×10cm2=50cm2

Sample Question 2 : With the vertices A, B and C of a triangle ABC as centres, arcs are drawn with radii 5cm each as shown in Fig. 11.15. If AB=14cm,BC=48cm and CA=50cm, then find the area of the shaded region. (Use π=3.14 ).

Fig. 11.15

Solution : Area of the sector with angle A

=A360×πr2=A360×π×(5)2cm2

Area of the sector with angle B

=B360×πr2=B360×π×(5)2cm2

and the area of the sector with angle C=C360×π×(5)2cm2

Therefore, sum of the areas (in cm2 ) of the three sectors

=A360×π×(5)2+B360×π×(5)2+C360×π×(5)2=A+B+C360×25π=180360×25πcm2( Because A+B+C=180)=25×π2cm2=25×1.57cm2=39.25cm2

Now, to find area of ABC, we find

s=a+b+c2=48+50+142cm=56cm

By Heron’s Formula,

ar(ABC)=s(sa)(sb)(sc)=56×8×6×42cm2=336cm2

So, area of the shaded region = area of the ABC area of the three sectors

=(33639.25)cm2=296.75cm2

Alternate Method for ar (ABC)

Here, AB2+BC2=(14)2+(48)2=2500=(50)2=(CA)2

So, B=90 (By converse of Pythagoras Theorem)

Therefore, ar(ABC)=12AB×BC=12×14×48cm2=336cm2

Sample Question 3 : A calf is tied with a rope of length 6m at the corner of a square grassy lawn of side 20m. If the length of the rope is increased by 5.5m, find the increase in area of the grassy lawn in which the calf can graze.

Solution : Let the calf be tied at the corner A of the square lawn (see Fig. 11.16)

Fig. 11.16

Then, the increase in area = Difference of the two sectors of central angle 90 each and radii 11.5m(6m+5.5m) and 6m, which is the shaded region in the figure.

So, required increase in area

=[90360×π×11.5290360π×62]m2=π4×(11.5+6)(11.56)m2=227×4×17.5×5.5m2=75.625m2.

EXERCISE 11.4

~~ 1. The area of a circular playground is 22176m2. Find the cost of fencing this ground at the rate of Rs 50 per metre.

~~ 2. The diameters of front and rear wheels of a tractor are 80cm and 2m respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes 1400 revolutions.

~~ 3. Sides of a triangular field are 15m,16m and 17m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.

~~ 4. Find the area of the segment of a circle of radius 12cm whose corresponding sector has a central angle of 60 (Use .π=3.14).

~~ 5. A circular pond is 17.5m is of diameter. It is surrounded by a 2m wide path. Find the cost of constructing the path at the rate of Rs25 per m2

~~ 6. In Fig. 11.17, ABCD is a trapezium with AB|DC,AB=18cm,DC=32cm and distance between AB and DC=14cm. If arcs of equal radii 7cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure.

Fig. 11.17

~~ 7. Three circles each of radius 3.5cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

~~ 8. Find the area of the sector of a circle of radius 5cm, if the corresponding arc length is 3.5cm.

~~ 9. Four circular cardboard pieces of radii 7cm are placed on a paper in such a way that each piece touches other two pieces. Find the area of the portion enclosed between these pieces.

~~ 10 On a square cardboard sheet of area 784cm2, four congruent circular plates of maximum size are placed such that each circular plate touches the other two plates and each side of the square sheet is tangent to two circular plates. Find the area of the square sheet not covered by the circular plates.

~~ 11 Floor of a room is of dimensions 5m×4m and it is covered with circular tiles of diameters 50cm each as shown in Fig. 11.18. Find the area of floor that remains uncovered with tiles. (Use π=3.14 )

Fig. 11.18

~~ 12 All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is 1256cm2. (Use π=3.14 ).

~~ 13 An archery target has three regions formed by three concentric circles as shown in Fig. 11.19. If the diameters of the concentric circles are in the ratio 1:2:3, then find the ratio of the areas of three regions.

Fig. 11.19

~~ 14 The length of the minute hand of a clock is 5cm. Find the area swept by the minute hand during the time period 6:05am and 6:40am.

~~ 15 Area of a sector of central angle 200 of a circle is 770cm2. Find the length of the corresponding arc of this sector.

~~ 16 The central angles of two sectors of circles of radii 7cm and 21cm are respectively 120 and 40. Find the areas of the two sectors as well as the lengths of the corresponding arcs. What do you observe?

~~ 17 Find the area of the shaded region given in Fig. 11.20.

Fig. 11.20

~~ 18 Find the number of revolutions made by a circular wheel of area 1.54m2 in rolling a distance of 176m.

~~ 19 Find the difference of the areas of two segments of a circle formed by a chord of length 5cm subtending an angle of 90 at the centre.

~~ 20 Find the difference of the areas of a sector of angle 120 and its corresponding major sector of a circle of radius 21cm.

Chapter 11

SURFACE AREAS AND VOLUMES

(A) Main Concepts and Results

  • The surface area of an object formed by combining any two of the basic solids, namely, cuboid, cone, cylinder, sphere and hemisphere.

  • The volume of an object formed by combining any two of the basic solids namely, cuboid, cone, cylinder, sphere and hemisphere.

  • The formulae involving the frustum of a cone are:

(i) Volume of the frustum of the cone =13πh[r12+r22+r1r2]

(ii) Curved surface area of the frustum of the cone =π(r1+r2)l,

(iii) Total surface area of the frustum of the solid cone

=πl(r1+r2)+πr12+πr22, where l=h2+(r1r2)2,

h= vertical height of the frustum, l= slant height of the frustum and r1 and r2 are radii of the two bases (ends) of the frustum.

  • Solid hemisphere: If r is the radius of a hemisphere, then curved surface area =2πr2 total surface area =3πr2, and volume =23πr3

  • Volume of a spherical shell =43π(r13r23), where r1 and r2 are respectively its external and internal radii.

Throughout this Chapter, take π=227, if not stated otherwise.

(B) Multiple Choice Questions :

Choose the correct answer from the given four options:

Sample Question 1: A funnel (see Fig.12.1) is the combination of

Fig. 12.1

(A) a cone and a cylinder

(B) frustum of a cone and a cylinder

(C) a hemisphere and a cylinder

(D) a hemisphere and a cone

Solution : Answer (B)

Sample Question 2 : If a marble of radius 2.1cm is put into a cylindrical cup full of water of radius 5cm and height 6cm, then how much water flows out of the cylindrical cup?

(A) 38.8cm3

(B) 55.4cm3

(C) 19.4cm3

(D) 471.4cm3

Solution : Answer (A)

Sample Question 3 : A cubical ice cream brick of edge 22cm is to be distributed among some children by filling ice cream cones of radius 2cm and height 7cm upto its brim. How many children will get the ice cream cones?

(A) 163

(B) 263

(C) 363

(D) 463

Solution : Answer (C)

Sample Question 4 : The radii of the ends of a frustum of a cone of height hcm are r1cm and r2cm. The volume in cm3 of the frustum of the cone is

(A) 13πh[r12+r22+r1r2]

(B) 13πh[r12+r22r1r2]

(C) 13πh[r12r22+r1r2]

(D) 13πh[r12r22r1r2]

Solution : Answer (A)

Sample Question 5 : The volume of the largest right circular cone that can be cut out from a cube of edge 4.2cm is

(A) 9.7cm3

(B) 77.6cm3

(C) 58.2cm3

(D) 19.4cm3

Solution : Answer (D)

EXERCISE 12.1

Choose the correct answer from the given four options:

~~ 1. A cylindrical pencil sharpened at one edge is the combination of

(A) a cone and a cylinder

(C) a hemisphere and a cylinder

(B) frustum of a cone and a cylinder

(D) two cylinders.

~~ 2. A surahi is the combination of

(A) a sphere and a cylinder

(B) a hemisphere and a cylinder

(C) two hemispheres

(D) a cylinder and a cone.

~~ 3. A plumbline (sahul) is the combination of (see Fig. 12.2)

Fig. 12.2

(A) a cone and a cylinder

(B) a hemisphere and a cone

(C) frustum of a cone and a cylinder

(D) sphere and cylinder

~~ 4. The shape of a glass (tumbler) (see Fig. 12.3) is usually in the form of

(A) a cone

(B) frustum of a cone

(C) a cylinder

(D) a sphere

Fig. 12.3

~~ 5. The shape of a gilli, in the gilli-danda game (see Fig. 12.4), is a combination of

(A) two cylinders

(B) a cone and a cylinder

(C) two cones and a cylinder

(D) two cylinders and a cone

Fig. 12.4

~~ 6. A shuttle cock used for playing badminton has the shape of the combination of

(A) a cylinder and a sphere

(B) a cylinder and a hemisphere

(C) a sphere and a cone

(D) frustum of a cone and a hemisphere

~~ 7. A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is called

(A) a frustum of a cone

(B) cone

(C) cylinder

(D) sphere

~~ 8. A hollow cube of internal edge 22cm is filled with spherical marbles of diameter 0.5cm and it is assumed that 18 space of the cube remains unfilled. Then the number of marbles that the cube can accomodate is

(A) 142296

(B) 142396

(C) 142496

(D) 142596

~~ 9. A metallic spherical shell of internal and external diameters 4cm and 8cm, respectively is melted and recast into the form a cone of base diameter 8cm. The height of the cone is

(A) 12cm

(B) 14cm

(C) 15cm

(D) 18cm

~~ 10 A solid piece of iron in the form of a cuboid of dimensions 49cm×33cm×24cm, is moulded to form a solid sphere. The radius of the sphere is

(A) 21cm

(B) 23cm

(C) 25cm

(D) 19cm

~~ 11 A mason constructs a wall of dimensions 270cm×300cm×350cm with the bricks each of size 22.5cm×11.25cm×8.75cm and it is assumed that 18 space is covered by the mortar. Then the number of bricks used to construct the wall is

(A) 11100

(B) 11200

(C) 11000

(D) 11300

~~ 12 Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2cm and height 16cm. The diameter of each sphere is

(A) 4cm

(B) 3cm

(C) 2cm

(D) 6cm

~~ 13 The radii of the top and bottom of a bucket of slant height 45cm are 28cm and 7cm, respectively. The curved surface area of the bucket is

(A) 4950cm2

(B) 4951cm2

(C) 4952cm2

(D) 4953cm2

~~ 14 A medicine-capsule is in the shape of a cylinder of diameter 0.5cm with two hemispheres stuck to each of its ends. The length of entire capsule is 2cm. The capacity of the capsule is

(A) 0.36cm3

(B) 0.35cm3

(C) 0.34cm3

(D) 0.33cm3

~~ 15 If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is

(A) 4πr2

(B) 6πr2

(C) 3πr2

(D) 8πr2

~~ 16 A right circular cylinder of radius rcm and height hcm(h>2r) just encloses a sphere of diameter

(A) rcm

(B) 2rcm

(C) hcm

(D) 2hcm

~~ 17 During conversion of a solid from one shape to another, the volume of the new shape will

(A) increase

(B) decrease

(C) remain unaltered

(D) be doubled

~~ 18 The diameters of the two circular ends of the bucket are 44cm and 24cm. The height of the bucket is 35cm. The capacity of the bucket is

(A) 32.7 litres

(B) 33.7 litres

(C) 34.7 litres

(D) 31.7 litres

~~ 19 In a right circular cone, the cross-section made by a plane parallel to the base is a

(A) circle

(B) frustum of a cone

(C) sphere

(D) hemisphere

~~ 20 Volumes of two spheres are in the ratio 64:27. The ratio of their surface areas is

(A) 3:4

(B) 4:3

(C) 9:16

(D) 16:9

(C) Short Answer Questions with Reasoning

Write ‘True’ or ‘False’ and justify your answer.

Sample Question 1 : If a solid cone of base radius r and height h is placed over a solid cylinder having same base radius and height as that of the cone, then the curved surface area of the shape is πrh2+r2+2πrh.

Solution : True. Since the curved surface area taken together is same as the sum of curved surface areas measured separately.

Sample Question 2 : A spherical steel ball is melted to make eight new identical balls.

Then, the radius of each new ball be 18 th the radius of the original ball.

Solution : False. Let r be the radius of the original steel ball and r1 be the radius of the new ball formed after melting.

Therefore, 43πr3=8×43πr13. This implies r1=r2.

Sample Question 3 : Two identical solid cubes of side a are joined end to end. Then the total surface area of the resulting cuboid is 12a2.

Solution : False. The total surface area of a cube having side a is 6a2. If two identical faces of side a are joined together, then the total surface area of the cuboid so formed is 10a2.

Sample Question 4 : Total surface area of a lattu (top) as shown in the Fig. 12.5 is the sum of total surface area of hemisphere and the total surface area of cone.

Fig. 12.5

Solution : False. Total surface area of the lattu is the sum of the curved surface area of the hemisphere and curved surface area of the cone.

Sample Question 5 : Actual capacity of a vessel as shown in the Fig. 12.6 is equal to the difference of volume of the cylinder and volume of the hemisphere.

Fig. 12.6

Solution : True. Actual capacity of the vessel is the empty space inside the glass that can accomodate something when poured in it.

EXERCISE 12.2

Write ‘True’ or ‘False’ and justify your answer in the following:

~~ 1. Two identical solid hemispheres of equal base radius rcm are stuck together along their bases. The total surface area of the combination is 6πr2.

~~ 2. A solid cylinder of radius r and height h is placed over other cylinder of same height and radius. The total surface area of the shape so formed is 4πrh+4πr2.

~~ 3. A solid cone of radius r and height h is placed over a solid cylinder having same base radius and height as that of a cone. The total surface area of the combined solid is πr[r2+h2+3r+2h].

~~ 4. A solid ball is exactly fitted inside the cubical box of side a. The volume of the ball is 43πa3.

~~ 5. The volume of the frustum of a cone is 13πh[r12+r22r1r2], where h is vertical height of the frustum and r1,r2 are the radii of the ends.

~~ 6. The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Fig. 12.7 is πr23[3h2r].

Fig. 12.7

~~ 7. The curved surface area of a frustum of a cone is πl(r1+r2), where l=h2+(r1+r2)2,r1 and r2 are the radii of the two ends of the frustum and h is the vertical height.

~~ 8. An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to curved surface area of frustum of a cone + area of circular base + curved surface area of cylinder

(C) Short Answer Questions

Sample Question 1: A cone of maximum size is carved out from a cube of edge 14cm. Find the surface area of the cone and of the remaining solid left out after the cone carved out.

Solution : The cone of maximum size that is carved out from a cube of edge 14cm will be of base radius 7cm and the height 14cm.

Surface area of the cone =πrl+πr2

=227×7×72+142+227(7)2

=227×7×245+154=(1545+154)cm2=154(5+1)cm2

Surface area of the cube =6×(14)2=6×196=1176cm2

So, surface area of the remaining solid left out after the cone is carved out =(1176154+1545)cm2=(1022+1545)cm2.

Sample Question 2 : A solid metallic sphere of radius 10.5cm is melted and recast into a number of smaller cones, each of radius 3.5cm and height 3cm. Find the number of cones so formed.

Solution : The volume of the solid metallic sphere =43π(10.5)3cm3

Volume of a cone of radius 3.5cm and height 3cm=13π(3.5)2×3cm3

Number of cones so formed =43π×10.5×10.5×10.513π×3.5×3.5×3.5=126

Sample Question 3 : A canal is 300cm wide and 120cm deep. The water in the canal is flowing with a speed of 20km/h. How much area will it irrigate in 20 minutes if 8cm of standing water is desired?

Solution : Volume of water flows in the canal in one hour = width of the canal × depth of the canal × speed of the canal water =3×1.2×20×1000m3=72000m3.

In 20 minutes the volume of water =72000×2060m3=24000m3.

Area irrigated in 20 minutes, if 8cm, i.e., 0.08m standing water is required

=240000.08m2=300000m2=30 hectares.

Sample Question 4 : A cone of radius 4cm is divided into two parts by drawing a plane through the mid point of its axis and parallel to its base. Compare the volumes of the two parts.

Solution : Let h be the height of the given cone. On dividing the cone through the mid-point of its axis and parallel to its base into two parts, we obtain the following (see Fig. 12.8):

In two similar triangles OAB and DCB, we have OACD=OBBD. This implies 4r=hh2.

Therefore, r=2.

Therefore,  Volume of the smaller cone  Volume of the frustumof the cone =13π×(2)2×(h2)13π×(h2)[42+22+4×2]=17

Therefore, the ratio of volume of the smaller cone to the volume of the frustum of the cone is 1:7.

Sample Question 5 : Three cubes of a metal whose edges are in the ratio 3:4:5 are melted and converted into a single cube whose diagonal is 123cm. Find the edges of the three cubes.

Solution : Let the edges of three cubes (in cm ) be 3x,4x and 5x, respectively.

Volume of the cubes after melting is =(3x)3+(4x)3+(5x)3=216x3cm3

Let a be the side of new cube so formed after melting. Therefore, a3=216x3

So, a=6x, Diagonal =a2+a2+a2=a3

But it is given that diagonal of the new cube is 123cm. Therefore, a3=123, i.e., a=12.

This gives x=2. Therefore, edges of the three cubes are 6cm,8cm and 10cm, respectively.

EXERCISE 12.3

~~ 1. Three metallic solid cubes whose edges are 3cm,4cm and 5cm are melted and formed into a single cube.Find the edge of the cube so formed.

~~ 2. How many shots each having diameter 3cm can be made from a cuboidal lead solid of dimensions 9cm×11cm×12cm ?

~~ 3. A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The radii of the top and bottom are 28cm and 21cm, respectively. Find the height of the bucket.

~~ 4. A cone of radius 8cm and height 12cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of two parts.

~~ 5. Two identical cubes each of volume 64cm3 are joined together end to end. What is the surface area of the resulting cuboid?

~~ 6. From a solid cube of side 7cm, a conical cavity of height 7cm and radius 3cm is hollowed out. Find the volume of the remaining solid.

~~ 7. Two cones with same base radius 8cm and height 15cm are joined together along their bases. Find the surface area of the shape so formed.

~~ 8. Two solid cones A and B are placed in a cylinderical tube as shown in the Fig.12.9. The ratio of their capacities are 2:1. Find the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder.

alt text

Fig. 12.9

~~ 9. An ice cream cone full of ice cream having radius 5cm and height 10cm as shown in the Fig.12.10. Calculate the volume of ice cream, provided that its 16 part is left unfilled with ice cream.

Fig. 12.10

~~ 10 Marbles of diameter 1.4cm are dropped into a cylindrical beaker of diameter 7cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6cm.

~~ 11 How many spherical lead shots each of diameter 4.2cm can be obtained from a solid rectangular lead piece with dimensions 66cm,42cm and 21cm.

~~ 12 How many spherical lead shots of diameter 4cm can be made out of a solid cube of lead whose edge measures 44cm.

~~ 13 A wall 24m long, 0.4m thick and 6m high is constructed with the bricks each of dimensions 25cm×16cm×10cm. If the mortar occupies 110 th of the volume of the wall, then find the number of bricks used in constructing the wall.

~~ 14 Find the number of metallic circular disc with 1.5cm base diameter and of height 0.2cm to be melted to form a right circular cylinder of height 10cm and diameter 4.5cm.

(E) Long Answer Questions

Sample Question 1 : A bucket is in the form of a frustum of a cone of height 30cm with radii of its lower and upper ends as 10cm and 20cm, respectively. Find the capacity and surface area of the bucket. Also, find the cost of milk which can completely fill the container, at the rate of Rs 25 per litre ( use π=3.14 ).

Solution : Capacity (or volume) of the bucket =πh3[r12+r22+r1r2].

Here, h=30cm,r1=20cm and r2=10cm.

So, the capacity of bucket =3.14×303[202+102+20×10]cm3=21.980 litres.

Cost of 1 litre of milk = Rs 25

Cost of 21.980 litres of milk = Rs 21.980×25= Rs 549.50

Surface area of the bucket = curved surface area of the bucket + surface area of the bottom

=πl(r1+r2)+πr22,l=h2+(r1r2)2

Now, l=900+100cm=31.62cm

Therefore, surface area of the bucket =3.14×31.62(20+10)+227(10)2

=3.14[948.6+100]cm2=3.14[1048.6]cm2=3292.6cm2 (approx.) 

Sample Question 2 : A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 4cm and the diameter of the base is 8cm. Determine the volume of the toy. If a cube circumscribes the toy, then find the difference of the volumes of cube and the toy. Also, find the total surface area of the toy.

Solution : Let r be the radius of the hemisphere and the cone and h be the height of the cone (see Fig. 12.11).

Volume of the toy = Volume of the hemisphere + Volume of the cone

Fig. 12.11

=23πr3+13πr2h=(23×227×43+13×227×42×4)cm3=14087cm3.

A cube circumscribes the given solid. Therefore, edge of the cube should be 8cm.

Volume of the cube =83cm3=512cm3.

Difference in the volumes of the cube and the toy =(51214087)cm3=310.86cm3

Total surface area of the toy = Curved surface area of cone + curved surface area of hemisphere

=πrl+2πr2, where l=h2+r2=πr(l+2r)=227×442+42+2×4cm2=227×442+8cm2=88×472+2cm2=171.68cm2

Sample Question 3 : A building is in the form of a cylinder surmounted by a hemispherical dome (see Fig. 12.12). The base diameter of the dome is equal to 23 of the total height of the building. Find the height of the building, if it contains 67121m3 of air.

Fig. 12.12

Solution : Let the radius of the hemispherical dome be r metres and the total height of the building be h metres.

Since the base diameter of the dome is equal to 23 of the total height, therefore 2r=23h. This implies r=h3. Let H metres be the height of the cylindrical portion. Therefore, H=hh3=23h metres.

Volume of the air inside the building = Volume of air inside the dome + Volume of the air inside the cylinder =23πr3+πr2H, where H is the height of the cylindrical portion

=23πh33+πh3223h=881πh3 cu. metres 

Volume of the air inside the building is 67121m3. Therefore, 881πh3=140821. This gives h=6m.

EXERCISE 12.4

~~ 1. A solid metallic hemisphere of radius 8cm is melted and recasted into a right circular cone of base radius 6cm. Determine the height of the cone.

~~ 2. A rectangular water tank of base 11m×6m contains water upto a height of 5m. If the water in the tank is transferred to a cylindrical tank of radius 3.5m, find the height of the water level in the tank.

~~ 3. How many cubic centimetres of iron is required to construct an open box whose external dimensions are 36cm,25cm and 16.5cm provided the thickness of the iron is 1.5cm. If one cubic cm of iron weighs 7.5g, find the weight of the box.

~~ 4. The barrel of a fountain pen, cylindrical in shape, is 7cm long and 5mm in diameter. A full barrel of ink in the pen is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one fifth of a litre?

~~ 5. Water flows at the rate of 10m/ minute through a cylindrical pipe 5mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40cm and depth 24cm ?

~~ 6. A heap of rice is in the form of a cone of diameter 9m and height 3.5m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

~~ 7. A factory manufactures 120000 pencils daily. The pencils are cylindrical in shape each of length 25cm and circumference of base as 1.5cm. Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at Rs 0.05 per dm2.

~~ 8. Water is flowing at the rate of 15km/h through a pipe of diameter 14cm into a cuboidal pond which is 50m long and 44m wide. In what time will the level of water in pond rise by 21cm ?

~~ 9. A solid iron cuboidal block of dimensions 4.4m×2.6m×1m is recast into a hollow cylindrical pipe of internal radius 30cm and thickness 5cm. Find the length of the pipe.

~~ 10 500 persons are taking a dip into a cuboidal pond which is 80m long and 50m broad. What is the rise of water level in the pond, if the average displacement of the water by a person is 0.04m3 ?

~~ 11 16 glass spheres each of radius 2cm are packed into a cuboidal box of internal dimensions 16cm×8cm×8cm and then the box is filled with water. Find the volume of water filled in the box.

~~ 12 A milk container of height 16cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8cm and 20cm respectively. Find the cost of milk at the rate of Rs. 22 per litre which the container can hold.

~~ 13 A cylindrical bucket of height 32cm and base radius 18cm is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24cm, find the radius and slant height of the heap.

~~ 14 A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6cm and 12cm, respectively. If the the slant height of the conical portion is 5cm, find the total surface area and volume of the rocket [Use π=3.14].

~~ 15 A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains 411921m3 of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building?

~~ 16 A hemispherical bowl of internal radius 9cm is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius 1.5cm and height 4cm. How many bottles are needed to empty the bowl?

~~ 17 A solid right circular cone of height 120cm and radius 60cm is placed in a right circular cylinder full of water of height 180cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone.

~~ 18 Water flows through a cylindrical pipe, whose inner radius is 1cm, at the rate of 80cm/sec in an empty cylindrical tank, the radius of whose base is 40cm. What is the rise of water level in tank in half an hour?

~~ 19 The rain water from a roof of dimensions 22m×20m drains into a cylindrical vessel having diameter of base 2m and height 3.5m. If the rain water collected from the roof just fill the cylindrical vessel, then find the rainfall in cm.

~~ 20 A pen stand made of wood is in the shape of a cuboid with four conical depressions and a cubical depression to hold the pens and pins, respectively. The dimension of the cuboid are 10cm,5cm and 4cm. The radius of each of the conical depressions is 0.5cm and the depth is 2.1cm. The edge of the cubical depression is 3cm. Find the volume of the wood in the entire stand.

Chapter 13

STATISTICS AND PROBABILITY

(A) Main Concepts and Results

Statistics

Measures of Central Tendency

(a) Mean of Grouped Data

(i) To find the mean of grouped data, it is assumed that the frequency of each class interval is centred around its mid-point.

(ii) Direct Method

Mean (x¯)=fixifi,

where the xi (class mark) is the mid-point of the i th class interval and fi is the corresponding frequency.

(iii) Assumed Mean Method

Mean (x¯)=a+fidifi,

a is the assumed mean and di=xia are the deviations of xi from a for each i.

(iv) Step-deviation Method

Mean(x¯)=a+hfiuifi,

where a is the assumed mean, h is the class size and ui=xiah.

(v) If the class sizes are unequal, the formula in (iv) can still be applied by taking h to be a suitable divisor of all the di ’s.

(b) Mode of Grouped Data

(i) In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies. To find the mode of grouped data, locate the class with the maximum frequency. This class is known as the modal class. The mode of the data is a value inside the modal class.

(ii) Mode of the grouped data can be calculated by using the formula

Mode =l+f1f02f1f0f2×h,

where l is the lower limit of the modal class, h is the size of the class, f1 is frequency of the modal class and f0 and f2 are the frequencies of the classes preceding and succeeding the modal class, respectively.

(c) Median of Grouped Data

(i) Cumulative frequency table - the less than type and the more than type of the grouped frequency distribution.

(ii) If n is the total number of observations, locate the class whose cumulative frequency is greater than (and nearest to) n2. This class is called the median class.

(iii) Median of the grouped data can be calculated by using the formula :

Median =l+(n2cff)h, where l is the lower limit of the median class, n is the number of observations, h is the class size, cf is the cumulative frequency of the class preceding the median class and f is the frequency of the median class.

(d) Graphical Representation of Cumulative Frequency Distribution (Ogive)

- Less than type and more than type.

(i) To find median from the graph of cumulative frequency distribution (less than type) of a grouped data.

(ii) To find median from the graphs of cumulative frequency distributions (of less than type and more than type) as the abscissa of the point of intersection of the graphs.

Probability

  • Random experiment, outcome of an experiment, event, elementary events.

  • Equally likely outcomes.

  • The theoretical (or classical) probability of an event E [denoted by P(E)] is given by

P(E)= Number of outcomes favourable to E Number of all possible outcomes of the experiment 

where the outcomes of the experiment are equally likely.

  • The probability of an event can be any number between 0 and 1 . It can also be 0 or 1 in some special cases.

  • The sum of the probabilities of all the elementary events of an experiment is 1 .

  • For an event E,P(E)+P(E)=1,

where E is the event ’not E ‘. E is called the complement of the event E.

  • Impossible event, sure or a certain event

(B) Multiple Choice Questions

Choose the correct answer from the given four options:

Sample Question 1 : Construction of a cumulative frequency table is useful in determining the

(A) mean

(B) median

(C) mode

(D) all the above three measures

Solution : Answer (B)

Sample Question 2 : In the following distribution :

Monthly income range (in Rs) Number of families
Income more than Rs 10000 100
Income more than Rs 13000 85
Income more than Rs 16000 69
Income more than Rs 19000 50
Income more than Rs 22000 33
Income more than Rs 25000 15

the number of families having income range (in Rs) 16000 - 19000 is

(A) 15

(B) 16

(C) 17

(D) 19

Solution : Answer (D)

Sample Question 3 : Consider the following frequency distribution of the heights of 60 students of a class :

Height (in cm) Number of students
150155 15
155160 13
160165 10
165170 8
170175 9
175180 5

The sum of the lower limit of the modal class and upper limit of the median class is

(A) 310

(B) 315

(C) 320

(D) 330

Solution : Answer (B)

Sample Question 4 : Which of the the following can be the probability of an event?

(A) -0.04

(B) 1.004

(C) 1823

(D) 87

Solution : Answer (C)

Sample Question 5 : A card is selected at random from a well shuffled deck of 52 playing cards. The probability of its being a face card is

(A) 313

(B) 413

(C) 613

(D) 913

Solution : Answer (A)

Sample Question 6 : A bag contains 3 red balls, 5 white balls and 7 black balls. What is the probability that a ball drawn from the bag at random will be neither red nor black?

(A) 15

(B) 13

(C) 715

(D) 815

Solution : Answer (B)

EXERCISE 13.1

Choose the correct answer from the given four options:

~~ 1. In the formula

x¯=a+fidifi,

for finding the mean of grouped data dis are deviations from a of

(A) lower limits of the classes

(B) upper limits of the classes

(C) mid points of the classes

(D) frequencies of the class marks

~~ 2. While computing mean of grouped data, we assume that the frequencies are

(A) evenly distributed over all the classes

(B) centred at the classmarks of the classes

(C) centred at the upper limits of the classes

(D) centred at the lower limits of the classes

~~ 3. If xi, are the mid points of the class intervals of the grouped data , fi,s are the corresponding frequencies and x¯ is the mean, then (fixix¯) is equal to

(A) 0

(B) -1

(C) 1

(D) 2

~~ 4. In the formula x¯=a+hfiuifi, for finding the mean of grouped frequency distribution, ui=

(A) xi+ah

(B) h(xia)

(C) xiah

(D) axih

~~ 5. The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its

(A) mean

(B) median

(C) mode

(D) all the three above

~~ 6. For the following distribution :

Class 05 510 1015 1520 2025
Frequency 10 15 12 20 9

the sum of lower limits of the median class and modal class is

(A) 15

(B) 25

(C) 30

(D) 35

~~ 7. Consider the following frequency distribution :

Class 05 611 1217 1823 2429
Frequency 13 10 15 8 11

The upper limit of the median class is

(A) 17

(B) 17.5

(C) 18

(D) 18.5

~~ 8. For the following distribution :

Marks Number of students
Below 10 3
Below 20 12
Below 30 27
Below 40 57
Below 50 75
Below 60 80

the modal class is

(A) 1020

(B) 2030

(C) 3040

(D) 5060

~~ 9. Consider the data :

Class 6585 85105 105125 125145 145165 165185 185205
Frequency 4 5 13 20 14 7 4

The difference of the upper limit of the median class and the lower limit of the modal class is

(A) 0

(B) 19

(C) 20

(D) 38

~~ 10 The times, in seconds, taken by 150 atheletes to run a 110m hurdle race are tabulated below :

Class 13.814 1414.2 14.214.4 14.414.6 14.614.8 14.815
Frequency 2 4 5 71 48 20

The number of atheletes who completed the race in less then 14.6 seconds is :

(A) 11

(B) 71

(C) 82

(D) 130

~~ 11 Consider the following distribution :

Marks obtained Number of students
More than or equal to 0 63
More than or equal to 10 58
More than or equal to 20 55
More than or equal to 30 51
More than or equal to 40 48
More than or equal to 50 42

the frequency of the class 3040 is

(A) 3

(B) 4

(C) 48

(D) 51

~~ 12 If an event cannot occur, then its probability is

(A) 1

(B) 34

(C) 12

(D) 0

~~ 13 Which of the following cannot be the probability of an event?

(A) 13

(B) 0.1

(C) 3 %

(D) 1716

~~ 14 An event is very unlikely to happen. Its probability is closest to

(A) 0.0001

(B) 0.001

(C) 0.01

(D) 0.1

~~ 15 If the probability of an event is p, the probability of its complementary event will be

(A) p1

(B) p

(C) 1p

(D) 11p

~~ 16 The probability expressed as a percentage of a particular occurrence can never be

(A) less than 100

(B) less than 0

(C) greater than 1

(D) anything but a whole number

~~ 17 If P(A) denotes the probability of an event A, then

(A) P(A)<0

(B) P(A)>1

(C) 0P(A)1

(D) 1P(A)1

~~ 18 A card is selected from a deck of 52 cards. The probability of its being a red face card is

(A) 326

(B) 313

(C) 213

(D) 12

~~ 19 The probability that a non leap year selected at random will contain 53 sundays is

(A) 17

(B) 27

(C) 37

(D) 57

~~ 20 When a die is thrown, the probability of getting an odd number less than 3 is

(A) 16

(B) 13

(C) 12

(D) 0

~~ 21 A card is drawn from a deck of 52 cards. The event E is that card is not an ace of hearts. The number of outcomes favourable to E is

(A) 4

(B) 13

(C) 48

(D) 51

~~ 22 The probability of getting a bad egg in a lot of 400 is 0.035 . The number of bad eggs in the lot is

(A) 7

(B) 14

(C) 21

(D) 28

~~ 23 A girl calculates that the probability of her winning the first prize in a lottery is 0.08 . If 6000 tickets are sold, how many tickets has she bought?

(A) 40

(B) 240

(C) 480

(D) 750

~~ 24 One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is

(A) 15

(B) 35

(C) 45

(D) 13

~~ 25 Someone is asked to take a number from 1 to 100 . The probability that it is a prime is

(A) 15

(B) 625

(C) 14

(D) 1350

~~ 26 A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. A single student is selected at random to be the class monitor. The probability that the selected student is not from A,B and C is

(A) 423

(B) 623

(C) 823

(D) 1723

(C) Short Answer Questions with Reasoning

Sample Question 1: The mean of ungrouped data and the mean calculated when the same data is grouped are always the same. Do you agree with this statement? Give reason for your answer.

Solution : The statement is not true. The reason is that when we calculated mean of a grouped data, it is assumed that frequency of each class is centred at the mid-point of the class. Because of this, two values of the mean, namely, those from ungrouped and grouped data are rarely the same.

Sample Question 2 : Is it correct to say that an ogive is a graphical representation of a frequency distribution? Give reason.

Solution : Graphical representation of a frequency distribution may not be an ogive. It may be a histogram. An ogive is a graphical representation of cumulative frequency distribution.

Sample Question 3: In any situation that has only two possible outcomes, each outcome will have probability 12. True or false? Why?

Solution : False, because the probability of each outcome will be 12 only when the two outcomes are equally likely otherwise not.

EXERCISE 13.2

~~ 1. The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason.

~~ 2. In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula

x¯=a+fidifi

where a is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.

~~ 3. Is it true to say that the mean, mode and median of grouped data will always be different? Justify your answer.

~~ 4. Will the median class and modal class of grouped data always be different? Justify your answer.

~~ 5. In a family having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is 14. Is this correct? Justify your answer.

~~ 6. A game consists of spinning an \to which comes to rest pointing at one of the regions (1, 2 or 3 ) (Fig. 13.1). Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons.

Fig. 13.1

~~ 7. Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why?

~~ 8. When we toss a coin, there are two possible outcomes - Head or Tail. Therefore, the probability of each outcome is 12. Justify your answer.

~~ 9. A student says that if you throw a die, it will show up 1 or not 1 . Therefore, the probability of getting 1 and the probability of getting ’not 1 ’ each is equal to 12. Is this correct? Give reasons.

~~ 10 I toss three coins together. The possible outcomes are no heads, 1 head, 2 heads and 3 heads. So, I say that probability of no heads is 14. What is wrong with this conclusion?

~~ 11 If you toss a coin 6 times and it comes down heads on each occasion. Can you say that the probability of getting a head is 1 ? Give reasons.

~~ 12 Sushma tosses a coin 3 times and gets tail each time. Do you think that the outcome of next toss will be a tail? Give reasons.

~~ 13 If I toss a coin 3 times and get head each time, should I expect a tail to have a higher chance in the 4th  toss? Give reason in support of your answer.

~~ 14 A bag contains slips numbered from 1 to 100. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since this situation has only two possible outcomes, so, the probability of each is 12. Justify.

(D) Short Answer Questions

Sample Question 1 : Construct the cumulative frequency distribution of the following distribution :

Class 12.517.5 17.522.5 22.527.5 27.532.5 32.537.5
Frequency 2 22 19 14 13

Solution : The required cumulative frequency distribution of the given distribution is given below :

Class Frequency Cumulative frequency
12.5-17.5 2 2
17.522.5 22 24
22.527.5 19 43
27.532.5 14 57
32.537.5 13 70

Sample Question 2 : Daily wages of 110 workers, obtained in a survey, are tabulated below :

Daily wages (in Rs) Number of workers
100120 10
120140 15
140160 20
160180 22
180200 18
200220 12
220240 13

Compute the mean daily wages of these workers.

Solution : We first find the classmark, xi, of each class and then proceed as follows:

Daily wages
(in Rs)
Classes
Class marks
(xi)
Number of workers
(fi)
fixi
100120 110 10 1100
120140 130 15 1950
140160 150 20 3000
160180 170 22 3740
180200 190 18 3420
200220 210 12 2520
220240 230 13 2990
fi=110 fixi=18720

Therefore, Mean =x¯=fixifi=18720110=170.20

Hence, the mean daily wages of the workers is Rs 170.20 .

Note : Mean daily wages can also be calculated by the assumed mean method or step deviation method.

Sample Question 3 : The percentage of marks obtained by 100 students in an examination are given below:

Marks 3035 3540 4045 4550 5055 5560 6065
Frequency 14 16 18 23 18 8 3

Determine the median percentage of marks.

Solution :

Marks
(Class)
Number of Students
(Frequency)
Cumulative frequency
3035 14 14
3540 16 30
4045 18 48
4550 23 71 Median Class
5055 18 89
5560 8 97
6065 3 100

Here, n=100.

Therefore, n2=50, This observation lies in the class 45-50.

l (the lower limit of the median class) =45

cf (the cumulative frequency of the class preceding the median class) =48

f (the frequency of the median class) =23

h (the class size) =5

 Median =l+(n2cff)h=45+(504823)×5=45+1023=45.4

So, the median percentage of marks is 45.4.

Sample Question 4 : The frequency distribution table of agricultural holdings in a village is given below :

Area of land
(in hectares)
13 35 57 79 911 1113
Number of
families
20 45 80 55 40 12

Find the modal agricultural holdings of the village.

Solution : Here the maximum class frequency is 80 , and the class corresponding to this frequency is 5-7.

So, the modal class is 5-7.

l( lower limit of modal class )=5

fl (frequency of the modal class) =80 f0 (frequency of the class preceding the modal class) =45

f2 (frequency of the class succeeding the modal class) =55

h( class size )=2

Mode =l+f1f02f1f0f2×h

=5+80452(80)4555×2=5+3560×2=5+3530=5+1.2=6.2

Hence, the modal agricultural holdings of the village is 6.2 hectares.

EXERCISE 13.3

~~ 1. Find the mean of the distribution :

Class 13 35 57 710
Frequency 9 22 27 17

~~ 2. Calculate the mean of the scores of 20 students in a mathematics test :

Marks 1020 2030 3040 4050 5060
Number of
students
2 4 7 6 1

~~ 3. Calculate the mean of the following data :

Class 47 811 1215 1619
Frequency 5 4 9 10

~~ 4. The following table gives the number of pages written by Sarika for completing her own book for 30 days :

Number of pages
written per day
1618 1921 2224 2527 2830
Number of
days
1 3 4 9 13

Find the mean number of pages written per day.

~~ 5. The daily income of a sample of 50 employees are tabulated as follows :

Income
(in Rs)
1200 201400 401600 601800
Number of
employees
14 15 14 7

Find the mean daily income of employees.

~~ 6. An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table :

Number of seats 100104 104108 108112 112116 116120
Frequency 15 20 32 18 15

Determine the mean number of seats occupied over the flights.

~~ 7. The weights (in kg ) of 50 wrestlers are recorded in the following table :

Weight (in kg) 100110 110120 120130 130140 140150
Number of
wrestlers
4 14 21 8 3

Find the mean weight of the wrestlers.

~~ 8. The mileage ( km per litre) of 50 cars of the same model was tested by a manufacturer and details are tabulated as given below :

Mileage
(km/l)
1012 1214 1416 1618
Number of cars 7 12 18 13

Find the mean mileage.

The manufacturer claimed that the mileage of the model was 16km/litre. Do you agree with this claim?

~~ 9. The following is the distribution of weights (in kg ) of 40 persons :

Weight (in kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75 75-80
Number of
persons
4 4 13 5 6 5 2 1

Construct a cumulative frequency distribution (of the less than type) table for the data above.

~~ 10 The following table shows the cumulative frequency distribution of marks of 800 students in an examination:

Marks Number of students
Below 10 10
Below 20 50
Below 30 130
Below 40 270
Below 50 440
Below 60 570
Below 70 670
Below 80 740
Below 90 780
Below 100 800

Construct a frequency distribution table for the data above.

~~ 11 Form the frequency distribution table from the following data:

Marks (out of 90) Number of candidates
More than or equal to 80 4
More than or equal to 70 6
More than or equal to 60 11
More than or equal to 50 17
More than or equal to 40 23
More than or equal to 30 27
More than or equal to 20 30
More than or equal to 10 32
More than or equal to 0 34

~~ 12 Find the unknown entries a,b,c,d,e,f in the following distribution of heights of students in a class :

Height
(in cm)
Frequency Cumulative frequency
150155 12 a
155160 b 25
160165 10 c
165170 d 43
170175 e 48
175180 2 f
Total 50

~~ 13 The following are the ages of 300 patients getting medical treatment in a hospital on a particular day :

Age (in years) 1020 2030 3040 4050 5060 6070
Number of
patients
60 42 55 70 53 20

Form:

(i) Less than type cumulative frequency distribution.

(ii) More than type cumulative frequency distribution.

~~ 14 Given below is a cumulative frequency distribution showing the marks secured by 50 students of a class :

Marks Below 20 Below 40 Below 60 Below 80 Below 100
Number of
students
17 22 29 37 50

Form the frequency distribution table for the data.

~~ 15 Weekly income of 600 families is tabulated below :

Weekly income
(in Rs)
Number of families
01000 250
10002000 190
20003000 100
30004000 40
40005000 15
50006000 5
Total 600

Compute the median income.

~~ 16 The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows :

Speed (km/h) 85100 100115 115130 130145
Number of
players
11 9 8 5

Calculate the median bowling speed.

~~ 17 The monthly income of 100 families are given as below :

Income (in Rs) Number of families
05000 8
500010000 26
1000015000 41
1500020000 16
2000025000 3
2500030000 3
3000035000 2
3500040000 1

Calculate the modal income.

~~ 18 The weight of coffee in 70 packets are shown in the following table :

Weight (in g) Number of packets
200201 12
201202 26
202203 20
203204 9
204205 2
205206 1

Determine the modal weight.

~~ 19 Two dice are thrown at the same time. Find the probability of getting

(i) same number on both dice.

(ii) different numbers on both dice.

~~ 20 Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is

(i) 7 ?

(ii) a prime number?

(iii) 1 ?

~~ 21 Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is

(i) 6

(ii) 12

(iii) 7

~~ 22 Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.

~~ 23 Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.

~~ 24 A coin is tossed two times. Find the probability of getting at most one head.

~~ 25 A coin is tossed 3 times. List the possible outcomes. Find the probability of getting

(i) all heads

(ii) at least 2 heads

~~ 26 Two dice are thrown at the same time. Determine the probabiity that the difference of the numbers on the two dice is 2 .

~~ 27 A bag contains 10 red, 5 blue and 7 green balls. A ball is drawn at random. Find the probability of this ball being a

(i) red ball

(ii) green ball

(iii) not a blue ball

~~ 28 The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is

(i) a heart

(ii) a king

~~ 29 Refer to Q.28. What is the probability that the card is

(i) a club

(ii) 10 of hearts

~~ 30 All the jacks, queens and kings are removed from a deck of 52 playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value 1 similar value for other cards, find the probability that the card has a value

(i) 7

(ii) greater than 7

(iii) less than 7

~~ 31 An integer is chosen between 0 and 100 . What is the probability that it is

(i) divisible by 7 ?

(ii) not divisible by 7 ?

~~ 32 Cards with numbers 2 to 101 are placed in a box. A card is selected at random. Find the probability that the card has

(i) an even number

(ii) a square number

~~ 33 A letter of English alphabets is chosen at random. Determine the probability that the letter is a consonant.

~~ 34 There are 1000 sealed envelopes in a box, 10 of them contain a cash prize of Rs 100 each, 100 of them contain a cash prize of Rs 50 each and 200 of them contain a cash prize of Rs 10 each and rest do not contain any cash prize. If they are well shuffled and an envelope is picked up out, what is the probability that it contains no cash prize?

~~ 35 Box A contains 25 slips of which 19 are marked Re 1 and other are marked Rs 5 each. Box B contains 50 slips of which 45 are marked Re 1 each and others are marked Rs 13 each. Slips of both boxes are poured into a third box and resuffled. A slip is drawn at random. What is the probability that it is marked other than Re 1 ?

~~ 36 A carton of 24 bulbs contain 6 defective bulbs. One bulbs is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?

~~ 37 A child’s game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6 are blue and rest are red. One piece is lost at random. Find the probability that it is a

(i) triangle

(ii) square

(iii) square of blue colour

(iv) triangle of red colour

~~ 38 In a game, the entry fee is Rs 5 . The game consists of a tossing a coin 3 times. If one or two heads show, Sweta gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise she will lose. For tossing a coin three times, find the probability that she

(i) loses the entry fee.

(ii) gets double entry fee.

(iii) just gets her entry fee.

~~ 39 A die has its six faces marked 0,1,1,1,6,6. Two such dice are thrown together and the total score is recorded.

(i) How many different scores are possible?

(ii) What is the probability of getting a total of 7 ?

~~ 40 A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is

(i) acceptable to Varnika?

(ii) acceptable to the trader?

~~ 41 A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is selected at random. What is the probability that it is

(i) not red?

(ii) white?

~~ 42 At a fete, cards bearing numbers 1 to 1000 , one number on one card, are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square greater than 500 , the player wins a prize. What is the probability that

(i) the first player wins a prize?

(ii) the second player wins a prize, if the first has won?

(E) Long Answer Questions

Sample Question 1: The following is the cumulative frequency distribution (of less than type) of 1000 persons each of age 20 years and above. Determine the mean age.

Age below
(in years)
30 40 50 60 70 80
Number of
persons
100 220 350 750 950 1000

Solution : First, we make the frequency distribution of the given data and then proceed to calculate mean by computing class marks (xi),ui ’s and fiui ’s as follows :

Class Frequency Class mark ui=xi4510 fiui
(fi) (xi)
2030 100 25 -2 -200
3040 120 35 -1 -120
4050 130 45 0 0
5060 400 55 1 400
6070 200 65 2 400
7080 50 75 3 150
fi=1000 fiui=630

ANSWERS

EXERCISE 1.1

~~ 1. (C)

~~ 2. (D)

~~ 3. (C)

~~ 4. (B)

~~ 5. (A)

~~ 6. (B)

~~ 7. (C)

~~ 8. (A)

~~ 9. (D)

~~ 10 (D)

EXERCISE 1.2

~~ 1. No, because an integer can be written in the form 4q,4q+1,4q+2,4q+3.

~~ 2. True, because n(n+1) will always be even, as one out of n or (n+1) must be even.

~~ 3. True, because n(n+1)(n+2) will always be divisible by 6 , as atleast one of the factors will be divisible by 2 and atleast one of the factors will be divisible by 3 .

~~ 4. No. Since any positive integer can be written as 3q,3q+1,3q+2,

therefore, square will be 9q2=3m,9q2+6q+1=3(3q2+2q)+1=3m+1, 9q2+12q+3+1=3m+1.

~~ 5. No. (3q+1)2=9q2+6q+1=3(3q2+2q)=3m+1.

~~ 6. HCF=75, as HCF is the highest common factor.

~~ 7. 3×5×7+7=7(3×5+1)=7(16), which has more than two factors.

~~ 8. No, because HCF (18) does not divide LCM (380).

~~ 9. Terminating decimal expansion, because 9871050047500 and 5005322

[98710500=3293500=32922537=472253=.094.]

~~ 10 Since 327.7081 is a terminating decimal number, so q must be of the form 2m5n; m,n are natural numbers.

EXERCISE 1.3

~~ 8. 63

~~ 9. 625

~~ 10 2520cm

~~ 11 23.54,0.0514

EXERCISE 2.1

~~ 1. (A)

~~ 2. (C)

~~ 3. (D)

~~ 4. (A)

~~ 5. (D)

~~ 6. (B)

~~ 7. (A)

~~ 8. (B)

~~ 9. (C)

~~ 10 (A)

EXERCISE 2.2

~~ 1. (i) No

(ii) 0,ax2+bx+c

(iii) degp(x)<degg(x)

(iv) degg(x)degp(x)

(v) No

~~ 2. (i) False

(vi) False

(ii) False

(vii) False

(iii) True

(iv) True

(v) True

EXERCISE 2.3

~~ 1. 1,14

~~ 2. 23,2

~~ 3. 1,75

~~ 4. 0,3,5

~~ 5. 32,14

~~ 6. 24,322

~~ 7. 12,2

~~ 8. 3,53

~~ 9. 25,52

~~ 10 23,17

EXERCISE 2.4

~~ 1. (i) 2,23

(ii) 52,18

(iii) 33,3

(iv) 55,52

~~ 2. a=1 and b=3 or a=5,b=3. Zeroes are 1,2,5

~~ 3. 22,223

~~ 4. k=-3

Zeroes of 2x4+x314x2+5x+6 are 1,3,2,12

Zeroes of x2+2x3 are 1,3

~~ 5. 5,52,52

~~ 6. a=1,b=2

1 and 2 are the zeroes of q(x) which are not the zeroes of p(x).

EXERCISE 3.1

~~ 1. (D)

~~ 2. (D)

~~ 3. (C)

~~ 4. (D)

~~ 5. (D)

~~ 6. (C)

~~ 7. (C)

~~ 8. (D)

~~ 9. (D)

~~ 10 (D)

~~ 11 (C)

~~ 12 (D)

~~ 13 (C)

EXERCISE 3.2

~~ 1. (i) Yes

(ii) No

(iii) No

~~ 2. (i) No

(ii) Yes

(iii) No

~~ 3. (i) No

(ii) Yes

(iii) Yes

(iv) No

~~ 4. No

~~ 5. False

~~ 6. Not true

EXERCISE 3.3

~~ 1. (i) λ=1

(ii) λ=1

(iii) All real values of λ except ±1.

~~ 2. k=6

~~ 3. a=3,b=1

~~ 4. (i) All real values of p except 10.

(ii) p=1

(iii) All real values of p except 910.

(iv) All real values of p except - 4 .

(v) p=4,q=8

~~ 5. Do not cross each other.

~~ 6. xy=4

2x+3y=7; infinitely many pairs.

~~ 7. 31,57

~~ 8. x=1,y=4

~~ 9. (i) x=1.2,y=2.1

(ii) x=6,y=8

(iii) x=3,y=2

(iv) x=16,y=14

(v) x=1,y=1

(vi) x=a2,y=b2

(vii) x=12,y=32

~~ 10 x=340,y=165;12

~~ 11 (i) consistent; x=1,y=1

(ii) inconsistent

(iii) consistent. The solution is given by y=3x, where x can take any value, i.e., there are infinitely many solutions.

~~ 12 (2,0),(0,4),(0,4);8 sq. units.

~~ 13 x=y; Infinitely many lines.

~~ 14 a=5,b=2.

~~ 15 55,85.

~~ 16 Salim’s age =38 years, Daughter’s age =14 years.

~~ 17 40 years.

~~ 18 40,48 .

~~ 19 100 students in hall A,80 students in hall B.

~~ 20 Rs 10, Rs 3 .

~~ 21 100 .

~~ 22 x=20,y=30,A=130,B=100,C=50,D=80

EXERCISE 3.4

~~ 1. x=1,y=4;4:1

~~ 2. (0,0),(4,4),(6,2)

~~ 3. 8 sq. units

~~ 4. 4x+4y=100,3x=y+15, where Rs x and Rs y are the costs of a pen and a pencil box respectively; Rs 10 , Rs 15 5. (1,0),(2,3),(4,2)6.10 km/h, 40 km/h

~~ 5. 2.5km/h

~~ 6. Rs 2500 , Rs 30

~~ 7. Rs 12000 in scheme A, Rs 10000 in scheme B

~~ 8. 10km/h,4km/h

~~ 9. Rs 600 , Rs 400

~~ 10 83

~~ 11 500

EXERCISE 4.1

~~ 1. (D)

~~ 2. (C)

~~ 3. (C)

~~ 4. (A)

~~ 5. (B)

~~ 6. (D)

~~ 7. (B)

~~ 8. (C)

~~ 9. (B)

~~ 10 (A)

~~ 11 (C)

EXERCISE 4.2

~~ 1. (i) No, because discriminant =7<0.

(ii) Yes, because discriminant =9>0.

(iii) No, because discriminant =0.

(iv) Yes, because discriminant =4>0.

(v) No, because discriminant =64<0.

(vi) Yes, because discriminant =(222)20.

(vii) Yes, because discriminant =1>0.

(viii) No, because discriminant =7<0.

(ix) Yes, because discriminant =1>0.

(x) Yes, because discriminant =8>0.

~~ 2. (i) False, for example: x2=1 is a quadratic equation with two roots.

(ii) False, for example x2+1=0 has no real root.

(iii) False, for example: x2+1=0 is a quadratic equation which has no real roots.

(iv) True, because every quadratic polynomial has almost two zeroes.

(v) True, because if in ax2+bx+c=0,a and c have opposite signs, then ac<0 and so b24ac>0.

(vi) True, because if in ax2+bx+c=0,a and c have same sign and b=0, then b24ac=4ac<0.

~~ 3. x23x+1=0 is an equation with integral coefficients but its roots are not integers.

~~ 4. x26x70, which has roots 32,32

~~ 5. Yes. 3x273x1230, which has roots 3,4

~~ 6. No.

~~ 7. Yes

EXERCISE 4.3

~~ 1. (i) 52,1

(ii) 1,85

(iii) 43,3

(iv) 5,2

(v) 32,2

(vi) 5,25

(vii) 113,113

~~ 2. (i) 32,23

(ii) 12,3

(iii) 2,26

(iv) 53,25

(v) 121,121

EXERCISE 4.4

~~ 1. (i) Real roots exist; roots are 12,34

(ii) Real roots exist; roots are 2,12

(iii) Real roots exist; roots are 15515,15515

(iv) Real roots exist; roots are 4+322,4322

(v) Real roots exist; roots are 75,25

~~ 2. The natural number is 12

~~ 3. The natural number is 8

~~ 4. Original speed of the train is 45km/h

~~ 5. Zeba’s age now is 14 years

~~ 6. Nisha’s age is 5 years and Asha’s age is 27 years

~~ 7. Length of the pond is 34m and breadth is 24m

~~ 8. 14

EXERCISE 5.1

~~ 1. (D)

~~ 2. (B)

~~ 3. (B)

~~ 4. (B)

~~ 5. (C)

~~ 6. (B)

~~ 7. (B)

~~ 8. (B)

~~ 9. (C)

~~ 10 (A)

~~ 11 (C)

~~ 12 (D)

~~ 13 (B)

~~ 14 (C)

~~ 15 (A)

~~ 16 (A)

~~ 17 (C)

~~ 18 (A)

EXERCISE 5.2

~~ 1. (i), (iv) and (vii) form an AP as in each of these Double subscripts: use braces to clarify is the same for different values of k.

~~ 2. False, as a4a3a3a2.

~~ 3. Yes, a30a203020d10d40.

~~ 4. The difference between any two corresponding terms of such APs is the same as the difference between their first terms.

~~ 5. No.

~~ 6. No, as the total fare (in Rs) after each km is 15,23,31,39, —

~~ 7. (i), (ii) and (iii) form an AP as in the list of numbers formed every succeeding term is obtained by adding a fixed number.

~~ 8. (i) Yes

(ii) No

(iii) No

EXERCISE 5.3

~~ 1. (A1)(B4)

(A2)(B5)

(A3)(B1)

(A4)(B2)

~~ 2. (i) 1,54,32

(ii) 113,103,3

(iii) 43,53,63

(iv) (a+2)+(b+1),(a+2)+(b+2),(a+3)+(b+2)

(v) 5a4,6a5,7a6

~~ 3. (i) 12,13,16

(ii) 5,8,11

(iii) 2,32,42

~~ 4. a1,b15,c31

~~ 5. 3,7,11,15,

~~ 6. d15,n27

~~ 7. 1,6,11,16,

~~ 8. 126

~~ 10. Yes, 17th  term.

~~ 11 k=0

~~ 12 67,69,71

~~ 13 40,60,80

~~ 14 16th  term; -21

~~ 15 -1

~~ 16 -78

~~ 17 12th term

~~ 18 73

~~ 19 3

~~ 20 n=6,d=10

~~ 21 (i) -9400 (ii) 7n12 (iii) 1111a6bab

~~ 22 16th  term; -632

~~ 23 -780

~~ 24 5,13,21,

~~ 25 k=27

~~ 26 -510

~~ 27 100

~~ 28 330

~~ 29 1170

~~ 30 504

~~ 31 n=5,11

~~ 32 11

~~ 33 Rs 800

~~ 34 25 months.

EXERCISE 5.4

~~ 1. 970

~~ 2. (i) 12250

(ii) 12750

(iii) 75250

~~ 3. 3

~~ 4. 3,7,11,15, —

~~ 5. (i) 1683

(ii) 13167

~~ 6. 1:3;5:49

~~ 8. 50

~~ 9. Rs 3900; Rs 44500

~~ 10. 728m;26m.

EXERCISE 6.1

~~ 1. (C)

~~ 2. (B)

~~ 3. (C)

~~ 4. (A)

~~ 5. (D)

~~ 6. (B)

~~ 7. (B)

~~ 8. (A)

~~ 9. (B)

~~ 10 (C)

~~ 11 (A)

~~ 12 (C)

EXERCISE 6.2

~~ 1. No,25252242

~~ 2. No, D=R but FP.

~~ 3. Yes, because PAQAPBBR

~~ 4. Yes, SAS criterion.

~~ 5. No, ΔQPRΔSTM

~~ 6. No, Corresponding sides must also be proportional.

~~ 7. Yes, as the corresponding two sides and the perimeters are equal, their third sides will also be equal.

~~ 8. Yes, AAA criterion.

~~ 9. No, ratio will be 925.

~~ 10 No, For this, P should be 90.

~~ 11 Yes, AA criterion.

~~ 12 No, angles should be included angles between the two pairs of proportional sides.

EXERCISE 6.3

~~ 2. x=2

~~ 4. 9:1

~~ 6. 43cm

~~ 7. 18cm

~~ 8. 1:3

~~ 9. 60cm

~~ 10 108cm2

~~ 12 12cm

~~ 13 553cm

~~ 14 10m

~~ 15 8m

EXERCISE 6.4

~~ 1. 5cm,2cm

~~ 2. BC=6.25cm,EF=16.8cm.

~~ 5. 0.8m

~~ 6. 8km

~~ 7. 20.4m

~~ 8. 9m

~~ 9. 25cm,6cm

~~ 10. 25cm,5cm,35cm

~~ 14. 8cm,12cm,16cm

EXERCISE 7.1

~~ 1. (B)

~~ 2. (B)

~~ 3. (C)

~~ 4. (B)

~~ 5. (C)

~~ 6. (B)

~~ 7. (C)

~~ 8. (B)

~~ 9. (D)

~~ 10 (A)

~~ 11 (B)

~~ 12 (D)

~~ 13 (B)

~~ 14 (A)

~~ 15 (A)

~~ 16 (D)

~~ 17 (D)

~~ 18 (B)

~~ 19 (B)

~~ 20 (C)

EXERCISE 7.2

~~ 1. True. Because all three sides of both triangles are proportional.

~~ 2. True. The three points lie on the line x=4.

~~ 3. False, since two points lie on the y - axis and one point lies in quadrant I.

~~ 4. False. PA=2 and PB=10, i.e., PAPB.

~~ 5. True, since ar (ABC)=0.

~~ 6. False, since the diagonals donot bisect each other.

~~ 7. True, radius of the circle =5 and OP>5

~~ 8. False, since AP AQ

~~ 9. True, since P divides AB in the ratio 1:2

~~ 10 True, since B divides AC in the ratio 2:7

~~ 11 False, since PC=266, P will lie inside the circle.

~~ 12 True, Mid-points of both the diagonals are the same and the diagonals are of equal length.

EXERCISE 7.3

~~ 1. Scalene triangle

~~ 2. (9,0),(5,0),2 points

~~ 3. Rectangle

~~ 4. a=3

~~ 5. (3,5) the middle point of AB. Infinite number of points. In fact all points which are solutions of the equation 2x+y+1=0.

~~ 6. 12,0, isosceles triangle

~~ 7. 1914

~~ 8. y=3,5,PQ=290,132

~~ 9. 0

~~ 10. 6:7,3413,0

~~ 11. 1:5

~~ 12. a=1 b = -3

~~ 13. k=22,AB=261

~~ 14 a=5,3

~~ 15. 19

~~ 16. 11

~~ 17 a=2, Area =6 sq. unit

~~ 18 45,215

~~ 19 2,12

~~ 20. 8:1,83,19

EXERCISE 7.4

~~ 1. 0,343

~~ 2. 34 sq. units.

~~ 3. (i)

x2x32,y2y32

(ii) x1x2x33,y1y2y33

(iii) same as (ii)

(iv) same as (ii)

~~ 4. a=3,h122613

~~ 5. Yes, Jaspal should be placed at the point (7,5)

~~ 6. House to Bank =5km

Bank to school =10km

School to Office =12km

Total distance travelled =27km

Distance from house to office =24.6km

Extra distance =2.4km

EXERCISE 8.1

~~ 1. (B)

~~ 2. (A)

~~ 3. (B)

~~ 4. (C)

~~ 5. (B)

~~ 6. (B)

~~ 7. (C)

~~ 8. (A)

~~ 9. (A)

~~ 10 (D)

~~ 11 (B)

~~ 12 (C)

~~ 13 (C)

~~ 14 (B)

~~ 15 (A)

EXERCISE 8.2

~~ 1. True

~~ 2. False

~~ 3. False [sin80sin10=. positive : as θ increases, value of sinθ increases ]

~~ 4. True

~~ 5. True

~~ 6. False

~~ 7. False

~~ 8. False

~~ 9. False

~~ 10 False

~~ 11 False

~~ 12 True

EXERCISE 8.3

~~ 8. 30

~~ 9. 12

~~ 10 152m

~~ 11 1

~~ 12 90

~~ 13 45

EXERCISE 8.4

~~ 3. 1031m

~~ 7. 253m

~~ 13. 103m;10m

~~ 14. h(cotαcotβ)

~~ 16 5(3+3)m18.8m

EXERCISE 9.1

~~ 1. (B)

~~ 2. (D)

~~ 3. (C)

~~ 4. (A)

~~ 5. (D)

~~ 6. (C)

~~ 7. (A)

~~ 8. (A)

~~ 9. (D)

~~ 10 (B)

EXERCISE 9.2

~~ 1. False

~~ 2. False

~~ 3. True

~~ 4. True

~~ 5. True

~~ 6. False

~~ 7. True

~~ 8. False

~~ 9. True

~~ 10 True

EXERCISE 9.3

~~ 1. 3cm

EXERCISE 9.4

~~ 3. 20cm

~~ 5. 4.8cm

~~ 7. 30

~~ 11. 203cm

~~ 12. 70

~~ 13. 82cm2

~~ 14. 24cm

EXERCISE 10.1

~~ 1. (D)

~~ 2. (B)

~~ 3. (A)

~~ 4. (C)

~~ 5. (B)

~~ 6. (D)

EXERCISE 10.2

~~ 1. True

~~ 2. False

~~ 3. False

~~ 4. True

EXERCISE 10.3

~~ 2. Yes

~~ 3. No

EXERCISE 10.4

~~ 1. 3.25cm

~~ 2. Yes, yes

~~ 3. 4cm

~~ 6. 8cm

EXERCISE 11.1

~~ 1. (B)

~~ 2. (A)

~~ 3. (B)

~~ 4. (A)

~~ 5. (B)

~~ 6. (A)

~~ 7. (D)

~~ 8. (B)

~~ 9. (C)

~~ 10 (D)

EXERCISE 11.2

~~ 1. No, radius of the circle is a2

~~ 2. Yes, side of the square is 2acm

~~ 3. No, side of the outer square = diagonal of the inner square

~~ 4. No, it is only true for minor segment.

~~ 5. No, it is πd.

~~ 6. Yes, distance covered in one revolution =2πr

~~ 7. No, it will depend on the value of radius.

~~ 8. Yes, it will be true for the arcs of the same circle.

~~ 9. No, it will be true for the arcs of the same circle.

~~ 10 No, it will be true for arcs of the same circle.

~~ 11 Yes, radius of the circle breadth of the rectangle.

~~ 12 Yes, their radii are equal

~~ 13 Yes, their radii are equal

~~ 14 No, diagonal of the square is pcm.

EXERCISE 11.3

~~ 1. 33cm

~~ 2. (16π32)cm2

~~ 3. 308cm2

~~ 4. 500 .

~~ 5. 154m2

~~ 6. (380+25π)cm2

~~ 7. 54.5cm2

~~ 8. (32+2π)m2

~~ 9. (2484π)m2

~~ 10. 3083493cm2

~~ 11. 30.96cm2

~~ 12. 39.25cm2

~~ 13 308cm2

~~ 14 15246m2

~~ 15 1386cm2

~~ 16 60πcm

EXERCISE 11.4

~~ 1. Rs 26400

~~ 2. 560

~~ 3. 242177m2

~~ 4. 75.36363cm2

~~ 5. Rs 3061.50

~~ 6. 196 cm2

~~ 7. 1.967cm2 (approx)

~~ 8. 8.7cm2

~~ 9. 42cm2

~~ 10. 168cm

~~ 11 4.3m2

~~ 12 800cm2

~~ 13 1:3:5

~~ 14 4556cm2

~~ 15 7313cm, Areas: 1543cm2,154cm2; Arc lengths: 443cm;

Arc lengths of two sectors of two different circles may be equal, but their area need not be equal.

~~ 17 1808πcm2

~~ 18 40

~~ 19 25π4+252cm2

~~ 20 462cm2

EXERCISE 12.1

~~ 1. (A)

~~ 2. (A)

~~ 3. (B)

~~ 4. (B)

~~ 5. (C)

~~ 6. (D)

~~ 7. (A)

~~ 8. (A)

~~ 9. (B)

~~ 10 (A)

~~ 11 (B)

~~ 12 (C)

~~ 13 (A)

~~ 14 (A)

~~ 15 (A)

~~ 16 (B)

~~ 17 (C)

~~ 18 (A)

~~ 19 (A)

~~ 20 (D)

EXERCISE 12.2

~~ 1. False

~~ 2. False

~~ 3. False

~~ 4. False

~~ 5. False

~~ 6. True

~~ 7. False

~~ 8. True

EXERCISE 12.3

~~ 1. 6cm

~~ 2. 84

~~ 3. 15cm

~~ 4. 7:1

~~ 5. 160cm2

~~ 6. 277cm3

~~ 7. 855cm2 (approx.)

~~ 8. 14cm,7cm;132cm3,66cm3;396cm3

~~ 9. 327.4cm3

~~ 10 150

~~ 11 1500

~~ 12 2541

~~ 13 12960

~~ 14 450

EXERCISE 12.4

~~ 1 28.44cm

~~ 2 8.6m

~~ 3 3960cm3,29.7kg

~~ 4 480000 words

~~ 5 51 minutes 12sec

~~ 6 74.25m3,80.61m2

~~ 7 Rs 2250

~~ 8 2 hours

~~ 9 112m

~~ 10 0.5cm

~~ 11 487.6cm3

~~ 12 Rs 230.12

~~ 13. 13.36cm,43.27cm

~~ 14 301.44cm2,377.1cm3

~~ 15 4m

~~ 16 54

~~ 17 1.584m3

~~ 18 90cm

~~ 19 2.5cm

~~ 20 170.8cm3

EXERCISE 13.1

1. (C) 2. (B) 3. (A) 4. (C) 5. (B)
6. (B) 7. (B) 8. (C) 9. (C) 10. (C)
11. (A) 12. (D) 13. (D) 14. (A) 15. (C)
16. (B) 17. (C) 18. (A) 19. (A) 20. (A)
21. (D) 22. (B) 23. (C) 24. (A) 25. (C)
26. (B)

EXERCISE 13.2

~~ 1. Not always, because for calculating median of a grouped data, the formula used is based on the assumption that the observations in the classes are uniformly distributed (or equally spaced).

~~ 2. Not necessary, the mean of the data does not depend on the choice of a (assumed mean).

~~ 3. No, it is not always the case. The values of these three measures can be the same. It depends on the type of data.

~~ 4. Not always. It depends on the data.

~~ 5. No, the outcomes are not equally likely. For example, outcome ‘one girl’ means gbb,bgb,bbg ’three girls’ means ggg and so on.

~~ 6. No, the outcomes are not equally likely. The outcome ’ 3 ’ is more likely than the others.

~~ 7. Peehu; probability of Apoorv’s getting 36136 while probability of Peehu’s getting 3616636.

~~ 8. Yes, the probability of each outcome is 12, since the two outcomes are equally likely.

~~ 9. No, outcomes ’ 1 ’ and ’not 1 ’ are not equally likely, P(1)=16,P(not1)=56,

~~ 10 No, the outcomes are not equally likely. Outcome ’no head’ means ‘TTT’; outcome ‘one head’ means THT, HTT, TTH and so on. P( TTT )=18,P( one head )=38 and so on.

~~ 11 No, the outcomes ‘head’ and ’tail’ are equally likely every time regardless of what you get in a few tosses.

~~ 12 It could be a tail or head as both the outcomes are equally likely, in each toss.

~~ 13 No, head and tail are equally likely. So, no question of expecting a tail to have a higher chance in the 4th toss.

~~ 14 Yes, the outcomes ‘odd number’, ’even number’ are equally likely in the situation considered.

EXERCISE 13.3

~~ 1. 5.5

~~ 2. 35

~~ 3. 12.93

~~ 4. 26

~~ 5. Rs. 356.5

~~ 6. 109.92

~~ 7. 123.4kg

~~ 8. 14.48km/l; No, the manufacturer is claiming mileage 1.52km/h more than the average mileage

~~ 9.

Weight (in kg) Number of persons
Less then 45 4
Less then 50 8
Less then 55 21
Less then 60 26
Less then 65 32
Less then 70 37
Less then 75 39
Less then 80 40

~~ 10

Marks Number of students
010 10
1020 40
2030 80
3040 140
4050 170
5060 130
6070 100
7080 70
8090 40
90100 20

~~ 11

Marks Number of candidates
010 2
1020 2
2030 3
3040 4
4050 6
5060 6
6070 5
7080 2
8090 4

~~ 12 a=12,b=13,c=35,d=8,e=5,f=50

~~ 13

(i) Less than type (ii) More than type
Ages (in years) Number of
students
Ages (in years) Number of
students
Less than 10 0 More than or equal to 10 300
Less than 20 60 More than or equal to 20 240
Less than 30 102 More than or equal to 30 198
Less than 40 157 More than or equal to 40 143
Less than 50 227 More than or equal to 50 73
Less than 60 280 More than or equal to 60 60
Less than 70 300

~~ 14

Marks Number of students
020 17
2040 5
4060 7
6080 8
80100 13

~~ 15 Rs 1263.15

~~ 16 109.17km/h

~~ 17 Rs 11875

~~ 18 201.7kg

~~ 19 (i) 16 (ii) 56

~~ 20 (i) 16

(ii) 512

(iii) 0

~~ 21 (i) 19

(ii) 19

(iii) 0

~~ 22 49

~~ 23 P(2)=118,P(3)=19,P(4)=16,P(5)=16,P(6)=16,P(7)=16,P(8)=19P(9)=118

~~ 24 34

~~ 25 (i) 18 (ii) 12

~~ 26 29

~~ 27 (i) 511

(ii) 722

(iii) 1722

~~ 28 (i) 1349

(ii) 349

~~ 29 (i) 1049

(ii) 149

~~ 30 (i) 110

(ii) 310

(iii) 35

~~ 31 (i) 1499

(ii) 8599

~~ 32 (i) 12

(ii) 9100

~~ 33 2126

~~ 34 0.69

~~ 35 1175

~~ 36 P( not defective )=34,P(2 nd bulb defective )=523

~~ 37 (i) 49

(ii) 59

(iii) 13

(iv) 518

~~ 38 (i) 18

(ii) 18

(iii) 34

~~ 39 (i) 5 scores (0,1,2,6,7,12)

(ii) 13

~~ 40 (i) 78

(ii) 1516

~~ 41 (i) 56

(ii) 13

~~ 42 (i) 0.009

(ii) 8999

[Hint : (ii) After first player has won the prize the number of perfect squares greater than 500 will be reduced by 1]

EXERCISE 13.4

~~ 1. 51.75

~~ 2. 48.41

~~ 3. 31 years

~~ 4. 201.96g

~~ 7. Median salary = Rs 13420 , Modal salary = Rs 12730

~~ 8. f1=28,f2=24 9. p=5,q=7

~~ 11. Median =17.81 hectares, Mode =17.76 hectares

~~ 12. Median rainfall =21.25cm

~~ 13. average =170.3sec.

~~ 14 (i)

Distance (in m) No. of students Cummulative frequency
020 6 6
2040 11 17
4060 17 34
6080 12 46
80100 4 50

(iii) 49.41m.



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