Solution

(c) In $\triangle A D B$ and $\triangle A D C$,

$\begin{aligned} & \angle D=\angle D=90^{\circ} \\ & \angle D B A=\angle D A C & [\text { each equal to } 90^{\circ}-<\mathrm{C} ]\\ & \therefore \quad \triangle A D B \sim \triangle A D C & \text { [by AAA similarity criterion] } \\ & \therefore \quad \dfrac{B D}{A D}=\dfrac{A D}{C D} \\ & \Rightarrow \quad B D \cdot C D=A D^2 \\ & \end{aligned}$

Solution

(b) We know that, the diagonals of a rhombus are perpendicular bisector of each other.

Given, $A C=16 cm$ and $B D=12 cm$

$\therefore$ $A O=8 cm, B O=6 cm$

and $\angle A O B=90^{\circ}$

In right angled $\triangle A O B$,

$ \begin{aligned} & A B^2 =A O^2+O B^2 & \text{[by Pythagoras theorem]}\\ \Rightarrow & A B^2 =8^2+6^2=64+36=100 \\ \therefore & A B =10 \mathrm{~cm} \end{aligned} $

Solution

(c) Given,

$ \triangle A B C \sim \triangle E D F $

$ \therefore \quad \dfrac{A B}{E D}=\dfrac{B C}{D F}=\dfrac{A C}{E F} $

Taking first two terms, we get

$ \begin{aligned} & & \dfrac{A B}{E D} & =\dfrac{B C}{D F} \\ \Rightarrow & & A B \cdot D F & =E D \cdot B C \\ \text{ or } & & B C \cdot D E & =A B \cdot D F \end{aligned} $

So, option (d) is true.

Taking last two terms, we get

$ \dfrac{B C}{D F}=\dfrac{A C}{E F} $

$ \Rightarrow \quad B C \cdot E F=A C \cdot D F $

So, option (a) is also true.

Taking first and last terms, we get

$ \dfrac{A B}{E D}=\dfrac{A C}{E F} $

$\Rightarrow$

$A B \cdot E F=E D \cdot A C$

Hence, option (b) is true.

Solution

(a) Given, in two $\triangle A B C$ and $\triangle P Q R, \dfrac{A B}{Q R}=\dfrac{B C}{P R}=\dfrac{C A}{P Q}$

which shows that sides of one triangle are proportional to the side of the other triangle, then their corresponding angles are also equal, so by SSS similarity, triangles are similar.

i.e.,

$\triangle C A B \sim \triangle P Q R$

Solution

(d) In $\triangle A P B$ and $\triangle C P D$,

$\angle A P B=\angle C P D=50^{\circ}$

[vertically opposite angles]

and

$ \begin{aligned} & \dfrac{A P}{P D}=\dfrac{6}{5} & \ldots \text{(i)}\\ & \dfrac{B P}{C P}=\dfrac{3}{2.5}=\dfrac{6}{5} & \ldots \text{(ii)} \end{aligned} $

From Eqs. (i) and (ii)

$ \begin{aligned} & \dfrac{A P}{P D}=\dfrac{B P}{C P} \\ & \therefore \quad \triangle A P B \sim \triangle D P C \quad \text{ [by SAS similarity criterion] } \\ & \therefore \quad \angle A=\angle D=30^{\circ} \text{ [corresponding angles of similar triangles] } \\ & \text{ In } \triangle A P B, \quad \angle A+\angle B+\angle A P B=180^{\circ} \quad \text{ [sum of angles of a triangle }=180^{\circ} \text{ ] } \\ & \Rightarrow \quad 30^{\circ}+\angle B+50^{\circ}=180^{\circ} \\ & \therefore \quad \angle B=180^{\circ}-(50^{\circ}+30^{\circ})=100^{\circ} \\ & \text{ i.e., } \quad \angle P B A=100^{\circ} \end{aligned} $

Solution

(b) Given, in $\triangle D E F$ and $\triangle P Q R, \angle D=\angle Q, \angle R=\angle E$

$ \begin{matrix} \therefore & \Delta D E F \sim \triangle Q R P \quad \text{ [by AAA similarity criterion] } \\ \therefore & \angle F=\angle P \quad \text{ [corresponding angles of similar triangles] } \\ \therefore & \dfrac{D F}{Q P}=\dfrac{E D}{R Q}=\dfrac{F E}{P R} \end{matrix} $

Solution

(b) In $\triangle A B C$ and $\triangle D E F, \angle B=\angle E, \angle F=\angle C$ and $A B=3 D E$

We know that, if in two triangles corresponding two angles are same, then they are similar by AAA similarity criterion. Also, $\triangle A B C$ and $\triangle D E F$ do not satisfy any rule of congruency, (SAS, ASA, SSS), so both are not congruent.

Thinking Process

Use the property of area of similar triangle.

Solution

(a) Given, $\triangle A B C \sim \triangle P Q R$ and $\dfrac{B C}{Q R}=\dfrac{1}{3}$

We know that, the ratio of the areas of two similar triangles is equal to square of the ratio of their corresponding sides.

$ \therefore \quad \dfrac{ar(\triangle P R Q)}{ar(\triangle B C A)}=\dfrac{(Q R)^{2}}{(B C)^{2}}=\dfrac{Q R}{B C}^{2}=\dfrac{3}1^{2}=\dfrac{9}{1}=9 $

Solution

(b) Given, $\triangle A B C \sim \triangle D F E$, then $\angle A=\angle D=30^{\circ}, \angle C=\angle E=50^{\circ}$

$ \begin{matrix} \therefore & \angle B=\angle F=180^{\circ}-(30^{\circ}+50^{\circ})=100^{\circ} \\ \text{ Also, } & A B=5 cm, A C=8 cm \text{ and } D F=7.5 cm \\ \therefore & \dfrac{A B}{D F}=\dfrac{A C}{D E} \\ \Rightarrow & \dfrac{5}{7.5}=\dfrac{8}{D E} \\ \therefore & D E=\dfrac{8 \times 7.5}{5}=12 cm \\ \text{ Hence, } & D E=12 cm, \angle F=100^{\circ} \end{matrix} $

Solution

(c) Given, in $\triangle A B C$ and $\triangle E D F$,

$ \dfrac{A B}{D E}=\dfrac{B C}{F D} $

By converse of basic proportionality theorem,

$ \begin{aligned} \triangle A B C & \sim \triangle E D F \\ \angle B & =\angle D, \angle A=\angle E \\ \angle C & =\angle F \end{aligned} $

Then,

and

Solution

(a) Given, $\triangle A B C \sim \triangle Q R P, A B=18 cm$ and $B C=15 cm$

We know that, the ratio of area of two similar triangles is equal to the ratio of square of their corresponding sides.

$ \begin{matrix} \therefore & \dfrac{ar(\triangle A B C)}{ar(\triangle Q R P)} & =\dfrac{(B C)^{2}}{(R P)^{2}} \\ \text{ But given, } & \dfrac{ar(\triangle A B C)}{ar(\triangle P Q R)} & =\dfrac{9}{4} \\ \Rightarrow & \dfrac{(15)^{2}}{(R P)^{2}} & =\dfrac{9}{4} \\ \Rightarrow & (R P)^{2} & =\dfrac{225 \times 4}{9}=100 \\ \therefore & R P & =10 cm \end{matrix} \quad[\because B C=15 cm, \text{ given] } $

Solution

(c) Given, in $\triangle P Q R$,

[using Eqs. (ii) and (iii)]

$ \begin{aligned} & & PS=QS=RS & & \ldots \text{(i)} \\ \text{In } \Delta PSR,& & PS=RS & & \text{[From Eq.(i)]} \\ \Rightarrow & & \angle 1= \angle 2 & & \ldots \text{(ii)} \end{aligned} $

Similarly, in $\triangle R S Q$,

$\Rightarrow \quad \angle 3=\angle 4$

[corresponding angles of equal sides are equal] Now, in $\triangle P Q R$, sum of angles $=180^{\circ}$

$ \begin{matrix} \Rightarrow & \angle P+\angle Q+\angle R=180^{\circ} \\ \Rightarrow & \angle 2+\angle 4+\angle 1+\angle 3=180^{\circ} \\ \Rightarrow & \angle 1+\angle 3+\angle 1+\angle 3=180^{\circ} \\ \Rightarrow & 2(\angle 1+\angle 3)=180^{\circ} \\ \Rightarrow & \angle 1+\angle 3=\dfrac{180^{\circ}}{2}=90^{\circ} \\ \therefore & \angle R=90^{\circ} \end{matrix} $

In $\triangle P Q R$, by Pythagoras theorem,

$ P R^{2}+Q R^{2}=P Q^{2} $

Solution

False

Let $a=25 cm, b=5 cm$ and $c=24 cm$

Now,

$ \begin{aligned} b^{2}+c^{2} & =(5)^{2}+(24)^{2} \\ & =25+576=601 \neq(25)^{2} \end{aligned} $

Hence, given sides do not make a right triangle because it does not satisfy the property of Pythagoras theorem.

Solution

False

We know that, if two triangles are similar, then their corresponding angles are equal.

$\therefore \quad \angle D=\angle R, \angle F=\angle P$ and $\angle F=Q$

Solution

False

Given, $P Q=12.5 cm, P A=5 cm, B R=6 cm$ and $P B=4 cm$

Then, $\quad Q A=Q P-P A=12.5-5=7.5 cm$

Now, $\quad \dfrac{P A}{A Q}=\dfrac{5}{7.5}=\dfrac{50}{75}=\dfrac{2}{3}$

and $\quad \dfrac{P B}{B R}=\dfrac{4}{6}=\dfrac{2}{3}$

From Eqs. (i) and (ii), $\quad \dfrac{P A}{A Q}=\dfrac{P B}{B R}$

By converse of basic proportionality theorem,

$A B | Q R$

Solution

True

In $\triangle P B C$ and $\triangle P D E$,

Since, one angle of $\triangle P B C$ is equal to one angle of $\triangle P D E$ and the sides including these angles are proportional, so both triangles are similar.

Hence, $\triangle P B C \sim \triangle P D E$, by SAS similarity criterion.

Solution

False

We know that, the sum of three angles of a triangle is $180^{\circ}$.

$ \begin{matrix} \text{ In } \triangle P Q R, & \angle P+\angle Q+\angle R =180^{\circ} \\ \Rightarrow & 55^{\circ}+25^{\circ}+\angle R =180^{\circ} \\ \Rightarrow & \angle R=180^{\circ}-(55^{\circ}+25^{\circ}) =180^{\circ}-80^{\circ}=100^{\circ} \\ \ln \triangle T S M, & L T+\angle S+\angle M =180^{\circ} \\ \Rightarrow & \angle T+\angle 25^{\circ}+100^{\circ} =180^{\circ} \\ \Rightarrow & \angle T =180^{\circ}-(25^{\circ}+100^{\circ}) \\ \Rightarrow & =180^{\circ}-125^{\circ}=55^{\circ} \end{matrix} $

In $\triangle P Q R$ and $\triangle T S M$,

and

$ \begin{aligned} & \angle P=\angle T, \angle Q=\angle S \\ & \angle R=\angle M \end{aligned} $

$\therefore$

$\triangle P Q R \sim \triangle T S M$ [since, all corresponding angles are equal]

Hence, $\triangle Q P R$ is not similar to $\triangle T S M$, since correct correspondence is $P \rightarrow T, Q \rightarrow S$ and $R \rightarrow M$.

Solution

False

Two quadrilaterals are similar, if their corresponding angles are equal and corresponding sides must also be proportional.

Solution

True

Here, the corresponding two sides and the perimeters of two triangles are proportional, then third side of both triangles will also in proportion.

Solution

True

Let two right angled triangles be $\triangle A B C$ and $\triangle P Q R$.

In which

$\angle A=\angle P=90^{\circ}$

and $\angle B=\angle Q=$ acute angle

Then, by AAA similarity criterion, $\triangle A B C \sim \triangle P Q R$

Solution

False

By the property of area of two similar triangles,

$ \Rightarrow \quad \begin{aligned} \dfrac{\text{ Area }_1}{\text{ Area }_2} & =\dfrac{\text{ Altitude }_1^{2}}{\text{ Altitude }_2} \\ \dfrac{\text{ Area }_1}{\text{ Area }_2} & =\dfrac{3}{5} \\ & =\dfrac{9}{25} \neq \dfrac{6}{5} \end{aligned} $

So, given statement is not correct.

Solution

False

In $\triangle P Q D$ and $\triangle R P D$,

$ \begin{aligned} P D & =P D \\ \angle P D Q & =\angle P D R \end{aligned} $

[common side]

[each $90^{\circ}$ ]

Here, no other sides or angles are equal, so we can say that $\triangle P Q D$ is not similar to $\triangle R P D$.

But, if $\angle P=90^{\circ}$, then $\angle D P Q=\angle P R D$

[each equal to $90^{\circ}-\angle \theta$ and by ASA similarity criterion, $\triangle P Q D \sim \triangle R P D$ ]

Solution

True

In $\triangle A D E$ and $\triangle A C B$,

$ \begin{matrix} \angle A & =\angle A & \text{ [common angle] } \\ \angle D & =\angle C & \text{ [given] } \\ \triangle A D E & \sim \triangle A C B & \text{ [by AAA similarity criterion] } \end{matrix} $

Solution

False

Because, according to SAS similarity criterion, if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

Here, one angle and two sides of two triangles are equal but these sides not including equal angle, so given statement is not correct.

Solution

Given $\ln \triangle P Q R$,

$ \begin{aligned} & & P R^{2}-P Q^{2} =Q R^{2} \text{ and } Q M \perp P R \\ \text{To prove} & & Q M^{2} =P M \times M R \\ \text{Proof Since, } & & P R^{2}-P Q^{2} =Q R^{2} \\ \Rightarrow & & P R^{2} =P Q^{2}+Q R^{2} \end{aligned} $

So, $\triangle P Q R$ is right angled triangle at $Q$.

In $\triangle Q M R$ and $\triangle P M Q$,

$ \begin{aligned} & & \angle M =\angle M & & [\text{each }90 ^{\circ}]\\ & & \angle M Q R =\angle Q P M & & [\text{each equal to }90 ^{\circ}- \angle R]\\ & & \triangle Q M R \sim \triangle P M Q & & [\text{by AAA similarilty criterion }] \end{aligned} $

Now, using property of area of similar triangles, we get

$ \begin{matrix} \dfrac{ar(\triangle Q M R)}{ar(\triangle P M Q)} & =\dfrac{(Q M)^{2}}{(P M)^{2}} \\ \Rightarrow & \dfrac{\dfrac{1}{2} \times R M \times Q M}{\dfrac{1}{2} \times P M \times Q M} & =\dfrac{(Q M)^{2}}{(P M)^{2}} \quad[\because \text{ area of triangle }=\dfrac{1}{2} \times \text{ base } \times \text{ height }] \\ \Rightarrow \quad Q M^{2} & =P M \times R M \quad \text{ Hence proved. } \end{matrix} $

Solution

Given,

$ \begin{aligned} & D E | A B \\ & \dfrac{C D}{A D}=\dfrac{C E}{B E} \end{aligned} $

$ \begin{matrix} \Rightarrow & & \dfrac{x+3}{3 x+19} & =\dfrac{x}{3 x+4} \\ \Rightarrow & (x+3)(3 x+4) & =x(3 x+19) \\ \Rightarrow & 3 x^{2}+4 x+9 x+12 & =3 x^{2}+19 x \\ \Rightarrow & 19 x-13 x & =12 \\ \Rightarrow & 6 x & =12 \\ \therefore & & x & =\dfrac{12}{6}=2 \end{matrix} $

[by basic proportionality theorem]

Hence, the required value of $x$ is 2 .

Thinking Process

Firstly, show that $S T | Q R$ with the help of given information, then use AAA similarity criterion To prove required result.

Solution

Given $\triangle NSQ \cong \triangle M T R$ and $\angle 1=\angle 2$

To prove $\triangle P T S \sim \triangle P R Q$

Proof Since,

$ \triangle NSQ \cong \triangle MTR $

So,

$$ \begin{equation*} S Q=T R \tag{i} \end{equation*} $$

Also,

$$ \begin{equation*} \angle 1=\angle 2 \Rightarrow P T=P S \tag{ii} \end{equation*} $$

From Eqs. (i) and (ii),

[since, sides opposite to equal angles are also equal]

$\Rightarrow$ $\dfrac{P S}{S Q}=\dfrac{P T}{T R}$

$\therefore$ $S T | Q R \quad$ [by convense of basic proportionality theorem]

and

$\angle 1=\angle P Q R$

In $\triangle P T S$ and $\triangle P R Q$,

$ \angle 2=\angle P R Q $

$$ \begin{aligned}\angle P & =\angle P \\ \angle 1 & =\angle P Q R \\ \angle \quad \angle 2 & =\angle P R Q \\ \therefore \quad \triangle P T S & \sim \triangle P R Q\end{aligned} $$

[by AAA similarity criterion]

Hence proved.

Thinking Process

Firstly, show that $\triangle P O Q$ and $\triangle R O S$ are similar by AAA similarity, then use property of area of similar triangle to get required ratio.

Solution

Given $P Q R S$ is a trapezium in which $P Q | R S$ and $P Q=3 R S$

$\Rightarrow$

$$ \begin{equation*} \dfrac{P Q}{R S}=\dfrac{3}{1} \tag{i} \end{equation*} $$

In $\triangle P O Q$ and $\triangle R O S$,

$ \begin{aligned} \angle S O R & =\angle Q O P & \text{[vertically opposite angles]}\\ \angle S R P & =\angle R P Q & \text{[alternate angles]} \\ \therefore \triangle P O Q & \sim \triangle R O S & \text{[by AAA similarity criterion]} \end{aligned} $

By property of area of similar triangle,

$ \begin{matrix} & \dfrac{ar(\triangle P O Q)}{ar(\triangle S O R)}=\dfrac{(P Q)^{2}}{(R S)^{2}}=\dfrac{P Q}{R S}{ }^{2}=\dfrac{3}{1} \\ \Rightarrow \quad & \dfrac{ar(\triangle P O Q)}{ar(\triangle S O R)}=\dfrac{9}{1} \end{matrix} $

Hence, the required ratio is $9: 1$.

Solution

Given $A C$ and $P Q$ intersect each other at the point $O$ and $A B | D C$.

To prove $O A \cdot C Q=O C \cdot A P$

Proof $\ln \triangle A O P$ and $\triangle C O Q, \quad \angle A O P=\angle C O Q \quad$ [vertically opposite angles] $\angle A P O=\angle C Q O$

[since, $A B | D C$ and $P Q$ is transversal, so alternate angles]

$\therefore \quad \triangle A O P \sim \triangle C O Q \quad$ [by AAA similarity criterion]

Then,

$\Rightarrow \quad O A \cdot C Q=O C \cdot A P$

Hence proved.

Solution

Let $A B C$ be an equilateral triangle of side $8 cm$ i.e., $A B=B C=C A=8 cm$.

Draw altitude $A D$ which is perpendicular to $B C$. Then, $D$ is the mid-point of $B C$.

$ \begin{aligned} & \therefore \quad B D=C D=\dfrac{1}{2} B C=\dfrac{8}{2}=4 cm \\ & \text{ Now, } \quad A B^{2}=A D^{2}+B D^{2} \\ & \text{ [by Pythagoras theorem] } \\ & \Rightarrow \quad(8)^{2}=A D^{2}+(4)^{2} \\ & \Rightarrow \quad 64=A D^{2}+16 \\ & \Rightarrow \quad A D^{2}=64-16=48 \\ & \Rightarrow \quad A D=\sqrt{48}=4 \sqrt{3} cm . \end{aligned} $

Hence, altitude of an equilateral triangle is $4 \sqrt{3} cm$.

Solution

Given $A B=4 cm, D E=6 cm$ and $E F=9 cm$ and $F D=12 cm$

Also,

$ \begin{aligned} \triangle A B C & \sim \triangle D E F \\ \dfrac{A B}{E D} & =\dfrac{B C}{E F}=\dfrac{A C}{D F} \\ \dfrac{4}{6} & =\dfrac{B C}{9}=\dfrac{A C}{12} \end{aligned} $

$ \begin{aligned} & \therefore \\ & \Rightarrow \end{aligned} $

On taking first two terms, we get

$ \begin{aligned} \dfrac{4}{6} & =\dfrac{B C}{9} \\ B C & =\dfrac{4 \times 9}{6}=6 cm \\ & =A C=\dfrac{6 \times 12}{9}=8 cm \end{aligned} $

Now,

$ \text{ perimeter of } \triangle A B C=A B+B C+A C $

$ =4+6+8=18 cm $

Solution

Given, $D E | B C, D E=6 cm$ and $B C=12 cm$

In $\triangle A B C$ and $\triangle A D E$,

and

$ \begin{aligned} \angle A B C & =\angle A D E \\ \angle A C B & =\angle A E D \\ \angle A & =\angle A \end{aligned} $

$\therefore$

Then, $\quad \dfrac{ar(\triangle A D E)}{ar(\triangle A B C)}=\dfrac{(D E)^{2}}{(B C)^{2}}$

[corresponding angle] [corresponding angle]

[common side]

[by AAA similarity criterion]

$ \begin{aligned} & =\dfrac{(6)^{2}}{(12)^{2}}=\dfrac{1}2^{2} \\ \Rightarrow \quad \dfrac{ar(\triangle A D E)}{ar(\triangle A B C)} & =\dfrac{1}2^{2}=\dfrac{1}{4} \end{aligned} $

Let $ar(\triangle A D E)=k$, then $ar(\triangle A B C)=4 k$

Now, ar $(D E C B)=ar(A B C)-ar(A D E)=4 k-k=3 k$

$\therefore$ Required ratio $=ar(A D E): ar(D E C B)=k: 3 k=1: 3$

Solution

Given, a trapezium $A B C D$ in which $A B | D C . P$ and $Q$ are points on $A D$ and $B C$, respectively such that $P Q | D C$. Thus, $A B|P Q| D C$.

Join $B D$.

$\begin{array}{lrr}\text { In } \triangle A B D, & P O | A B & {[\because P Q | A B]} \\ \text { By basic proportionality theorem, } & \dfrac{D P}{A P}=\dfrac{D O}{O B} & \ldots \text { (i) }\end{array}$

In $\triangle B D C$,

$O Q | D C$

By basic proportionality theorem,

$ \begin{matrix} & \dfrac{B Q}{Q C}=\dfrac{O B}{O D} \\ \Rightarrow & \dfrac{Q C}{B Q}=\dfrac{O D}{O B} & \ldots \text{(ii)}\\ \text{ From Eqs. (i) and (ii), } & \dfrac{D P}{A P}=\dfrac{Q C}{B Q} \\ \Rightarrow & \dfrac{18}{A P}=\dfrac{15}{35} \\ \Rightarrow & A P=\dfrac{18 \times 35}{15}=42 \\ \therefore & A D=A P+D P=42+18=60 cm \end{matrix} $

Thinking Process

Use the property area of similar triangle to get required area.

Solution

Given, ratio of corresponding sides of two similar triangles $=2: 3$ or $\dfrac{2}{3}$

Area of smaller triangle $=48 cm^{2}$

By the property of area of two similar triangle,

Ratio of area of both riangles $=(\text{ Ratio of their corresponding sides })^{2}$

$ \begin{matrix} \text{ i.e., } & \dfrac{ar \text{ (smaller triangle })}{ar(\text{ larger triangle })}=\dfrac{2}{3} \\ \Rightarrow & \dfrac{48}{ar(\text{ larger triangle })}=\dfrac{4}{9} \\ \Rightarrow & \text{ ar (larger triangle })=\dfrac{48 \times 9}{4}=12 \times 9=108 cm^{2} \end{matrix} $

Thinking Process

Firstly, show that $\triangle Q N P \sim \triangle R N Q$, by SAS similarity criterion and then use the property that sum of all angles of a triangle is $180^{\circ}$.

Solution

Given $\triangle P Q R, N$ is a point on $P R$, such that $Q N \perp P R$

and

To prove Proof We have, $\Rightarrow$ $$ \begin{aligned} \angle P Q R & =90^{\circ} \\ P N \cdot N R & =Q N^2 \\ P N \cdot N R & =Q N \cdot Q N \\ \dfrac{P N}{Q N} & =\dfrac{Q N}{N R} \end{aligned} $$

In $\triangle Q N P$ and $\triangle R N Q$,

and

$\therefore$

Then, $\triangle Q N P$ and $\triangle R N Q$ are equiangulars.

i.e.,

$ \begin{aligned} & \angle P Q N=\angle Q R N \\ & \angle R Q N=\angle Q P N \end{aligned} $

On adding both sides, we get

$ \begin{aligned} & \angle P Q N+\angle R Q N=\angle Q R N+\angle Q P N \\ & \Rightarrow \quad \angle P Q R=\angle Q R N+\angle Q P N \end{aligned} $

We know that, sum of angles of a triangle $=180^{\circ}$

In $\triangle P Q R, \quad \angle P Q R+\angle Q P R+\angle Q R P=180^{\circ}$

$ \begin{matrix} \Rightarrow & \angle P Q R+\angle Q P N+\angle Q R N=180^{\circ} & {[\because \angle Q P R=\angle Q P N \text{ and } \angle Q R P=\angle Q R N]} \\ \Rightarrow & \angle P Q R+\angle P Q R=180^{\circ} & \\ \Rightarrow & 2 \angle P Q R=180^{\circ} \\ \Rightarrow & \angle P Q R=\dfrac{180^{\circ}}{2}=90^{\circ} & \end{matrix} $

$ \therefore \quad \angle P Q R=90^{\circ} \quad \text{ Hence proved. } $

Solution

Given, area of smaller triangle $=36 cm^{2}$ and area of larger triangle $=100 cm^{2}$

Also, length of a side of the larger triangle $=20 cm$

Let length of the corresponding side of the smaller triangle $=x cm$

By property of area of similar triangle,

$ \begin{matrix} & \dfrac{\text{ ar (larger triangle })}{\text{ ar (smaller triangle })} & =\dfrac{(\text{ Side of larger triangle })^{2}}{\text{ Side of smaller triangle }{ }^{2}} \\ \Rightarrow & \dfrac{100}{36} & =\dfrac{(20)^{2}}{x^{2}} \Rightarrow x^{2}=\dfrac{(20)^{2} \times 36}{100} \\ & \therefore & x^{2} & =\dfrac{400 \times 36}{100}=144 \\ & x & =\sqrt{144}=12 cm \end{matrix} $

Hence, the length of corresponding side of the smaller triangle is $12 cm$.

Solution

Given, $A C=8 cm, A D=3 cm$ and $\angle A C B=\angle C D A$

From figure,

$\angle C D A=90^{\circ}$

$\therefore$ $\angle A C B=\angle C D A=90^{\circ}$

In right angled $\triangle A D C$,

$A C^{2}=A D^{2}+C D^{2}$

$\Rightarrow$ $(8)^{2}=(3)^{2}+(C D)^{2}$

$\Rightarrow$ $64-9=C D^{2}$

$\Rightarrow$ $C D=\sqrt{55} cm$

In $\triangle C D B$ and $\triangle A D C$,

$\angle B D C=\angle A D C$

[each $90^{\circ}$ ]

$\begin{matrix} \therefore & \angle D B C=\angle D C A \\ \triangle C D B \sim \triangle A D C\end{matrix} $

[each equal to $90^{\circ}-\angle A$ ]

Then,

$\dfrac{C D}{B D}=\dfrac{A D}{C D}$

$\Rightarrow$

$C D^{2}=A D \times B D$

$\therefore \quad B D=\dfrac{C D^{2}}{A D}=\dfrac{(\sqrt{55})^{2}}{3}=\dfrac{55}{3} cm$

Thinking Process

Firstly, draw the figure according to given conditions, then show that both triangles are similar by AAA similarity criterion and then use ratio of sides of both triangles to get required length.

Solution

Let $B C=15 m$ be the tower and its shadow $A B$ is $24 m$. At that time $\angle C A B=\theta$. Again, let $E F=h$ be a telephone pole and its shadow $D E=16 m$. At the same time $\angle E D F=\theta$. Here, $\triangle A B C$ and $\triangle D E F$ both are right angled triangles.

In $\triangle A B C$ and $\triangle D E F$,

$\therefore$

$ \angle C A B=\angle E D F=\theta $

$ \angle B=\angle E $

$\triangle A B C \sim \triangle D E F$ [each $90^{\circ}$ ]

[by AAA similarity criterion]

Then,

$ \begin{aligned} \dfrac{A B}{D E} & =\dfrac{B C}{E F} \\ \dfrac{24}{16} & =\dfrac{15}{h} \\ h & =\dfrac{15 \times 16}{24}=10 \end{aligned} $

Hence, the height of the telephone pole is $10 m$.

Solution

Let $A B$ be a vertical wall and $A C=10 m$ is a ladder. The top of the ladder reaches to $A$ and distance of ladder from the base of the wall $B C$ is $6 m$.

In right angled $\triangle A B C$,

$ \begin{aligned} A C^{2} & =A B^{2}+B C^{2} \\ & (10)^{2}=A B^{2}+(6)^{2} \end{aligned} $

$ \begin{matrix} \Rightarrow & (10)^{2}=A B^{2}+(6)^{2} \\ \Rightarrow & 100=A B^{2}+36 \\ \Rightarrow & A B^{2}=100-36=64 \\ \therefore & A B=\sqrt{64}=8 cm \end{matrix} $

Hence, the height of the point on the wall where the top of the ladder reaches is $8 cm$.

Solution

Given, $\angle A=\angle C, A B=6 cm, B P=15 cm, A P=12 cm$ and $C P=4 cm$ In $\triangle A P B$ and $\triangle C P D$,

$ \begin{aligned} \angle A & =\angle C \\ \angle A P B & =\angle C P D \end{aligned} $

[by Pythagoras theorem]

Triangles

$$ \begin{array}{ll} \therefore & \Delta A P D \sim \triangle C P D \\ \Rightarrow & \dfrac{A P}{C P}=\dfrac{P B}{P D}=\dfrac{A B}{C D} \\ \Rightarrow & \dfrac{12}{4}=\dfrac{15}{P D}=\dfrac{6}{C D} \end{array} $$ [by AAA similarity criterion]

On taking first two terms, we get

$ \begin{aligned} & \dfrac{12}{4}=\dfrac{15}{P D} \\ \Rightarrow \quad & P D=\dfrac{15 \times 4}{12}=5 cm \end{aligned} $

On taking first and last term, we get

$ \begin{aligned} & \dfrac{12}{4}=\dfrac{6}{C D} \\ \Rightarrow \quad C D & =\dfrac{6 \times 4}{12}=2 cm \end{aligned} $

Hence, length of $P D=5 cm$ and length of $C D=2 cm$

Thinking Process

Use the property of similar triangles i.e., the corresponding sides are in the same ratio and then simplify.

Solution

Given, $\triangle A B C \sim \triangle E D F$, so the corresponding sides of $\triangle A B C$ and $\triangle E D F$ are in the same ratio. i.e.,

$$ \begin{equation*} \dfrac{A B}{E D}=\dfrac{A C}{E F}=\dfrac{B C}{D F} \tag{i} \end{equation*} $$

Also,

$ \begin{aligned} & A B=5 cm, A C=7 cm \\ & D F=15 cm \text{ and } D E=12 cm \end{aligned} $

On putting these values in Eq. (i), we get

On taking first and second terms, we get

$ \dfrac{5}{12}=\dfrac{7}{E F}=\dfrac{B C}{15} $

$ \begin{aligned} & \dfrac{5}{12}=\dfrac{7}{E F} \\ & \Rightarrow \quad E F=\dfrac{7 \times 12}{5}=16.8 cm \end{aligned} $

On taking first and third terms, we get

$ \Rightarrow \quad \begin{aligned} \dfrac{5}{12} & =\dfrac{B C}{15} \\ \Rightarrow \quad B C & =\dfrac{5 \times 15}{12}=6.25 cm \end{aligned} $

Hence, lengths of the remaining sides of the triangles are $E F=16.8 cm$ and $B C=6.25 cm$.

Solution

Let a $\triangle A B C$ in which a line $D E$ parallel to $B C$ intersects $A B$ at $D$ and $A C$ at $E$.

To prove $D E$ divides the two sides in the same ratio.

i.e.,

Construction Join $B E, C D$ and draw $E F \perp A B$ and $D G \perp A C$.

Proof Here, $\quad \dfrac{ar(\triangle A D E)}{ar(\triangle B D E)}=\dfrac{\dfrac{1}{2} \times A D \times E F}{\dfrac{1}{2} \times D B \times E F} \quad[\because.$ area of triangle $=\dfrac{1}{2} \times$ base $\times$ height $]$

$$ \begin{equation*} =\dfrac{A D}{D B} \tag{i} \end{equation*} $$

similarly,

$$ \begin{equation*} \dfrac{ar(\triangle A D E)}{ar(\triangle D E C)}=\dfrac{\dfrac{1}{2} \times A E \times G D}{\dfrac{1}{2} \times E C \times G D}=\dfrac{A E}{E C} \tag{ii} \end{equation*} $$

Now, since, $\triangle B D E$ and $\triangle D E C$ lie between the same parallel $D E$ and $B C$ and on the same base $D E$.

So,

$$ \begin{equation*} ar(\triangle B D E)=ar(\triangle D E C) \tag{iii} \end{equation*} $$

From Eqs. (i), (ii) and (iii),

$ \dfrac{A D}{D B}=\dfrac{A E}{E C} $

Hence proved.

Solution

Given $P Q R S$ is a parallelogram, so $P Q | S R$ and $P S | Q R$. Also, $A B | P S$.

To prove $O C | S R$

Proof in $\triangle O P S$ and $\triangle O A B$,

$ P S | A B $

$\angle P O S=\angle A O B$ $\angle O S P=\angle O B A$

$\therefore \quad \triangle O P S \sim \triangle O A B$

Then,

$$ \begin{equation*} \dfrac{P S}{A B}=\dfrac{O S}{O B} \tag{i} \end{equation*} $$

[common angle] [corresponding angles] [by AAA similarity criterion]

In $\triangle C Q R$ and $\triangle C A B$,

$ \begin{aligned} & & \angle Q C R & =\angle A C B \\ \therefore & & \angle C R Q & =\angle C B A \\ Then, & & \Delta C Q R & \sim \triangle C A B \\ \Rightarrow & & \dfrac{Q R}{A B} & =\dfrac{C R}{C B} \\ & & \dfrac{P S}{A B} & =\dfrac{C R}{C B}\end{aligned} $

[since, $P Q R S$ is a parallelogram, so $P S \equiv Q R$ ]

From Eqs. (i) and (ii),

$ \dfrac{O S}{O B}=\dfrac{C R}{C B} \text{ or } \dfrac{O B}{O S}=\dfrac{C B}{C R} $

On subtracting from both sides, we get

$ \begin{aligned} \Rightarrow & \dfrac{O B}{O S}-1 & =\dfrac{C B}{C R}-1 \\ \Rightarrow & \dfrac{O B-O S}{O S} & =\dfrac{C B-C R}{C R} \\ \Rightarrow & \dfrac{B S}{O S} & =\dfrac{B R}{C R} \end{aligned} $

By converse of basic proportionality theorem,

$S R | O C$

Hence proved.

Solution

Let $A C$ be the ladder of length $5 m$ and $B C=4 m$ be the height of the wall, which ladder is placed. If the foot of the ladder is moved $1.6 m$ towards the wall i.e, $A D=1.6 m$, then the ladder is slide upward i.e., $C E=x m$. In right angled $\triangle A B C$,

$ \begin{aligned} & A C^{2}=A B^{2}+B C^{2} \quad \text{ [by Pythagoras theorem] } \\ & \Rightarrow \quad(5)^{2}=(A B)^{2}+(4)^{2} \\ & \Rightarrow \quad A B^{2}=25-16=9 \Rightarrow A B=3 m \\ & \therefore \quad D B=A B-A D=3-1.6=1.4 m \\ & E D^{2}=E B^{2}+B D^{2} \\ & (5)^{2}=(E B)^{2}+(1.4)^{2} \\ & 25=(E B)^{2}+1.96 \\ & \Rightarrow \\ & (E B)^{2}=25-1.96=23.04 \\ & E B=\sqrt{23.04}=4.8 \\ & \begin{matrix} \Rightarrow \\ \text{ Now } \end{matrix} \\ & E C=E B-B C=4.8-4=0.8 \end{aligned} $

Hence, the top of the ladder would slide upwards on the wall at distance $0.8 m$.

Thinking Process

Firstly draw the figure according to the given conditions and use Pythagoras theorem to find the value of $x$. Then, required saved distance will be equal to the difference of $(A C+B C)$ and 26 .

Solution

Given, $A C \perp C B, km, C B=2(x+7) km$ and $A B=26 km$

On drawing the figure, we get the right angled $\triangle A C B$ right angled at $C$.

Now, In $\triangle A C B$, by Pythagoras theorem,

$ \begin{matrix} & A B^{2}=A C^{2}+B C^{2} \\ \Rightarrow & (26)^{2}=(2 x)^{2}+{2(x+7)}^{2} \\ \Rightarrow & 676=4 x^{2}+4(x^{2}+49+14 x) \\ \Rightarrow & 676=4 x^{2}+4 x^{2}+196+56 x \\ \Rightarrow & 676=8 x^{2}+56 x+196 \\ \Rightarrow & 8 x^{2}+56 x-480=0 \\ \text{ On dividing by 8, we get } & x^{2}+7 x-60=0 \\ \Rightarrow & x^{2}+12 x-5 x-60=0 \\ \Rightarrow & x(x+12)-5(x+12)=0 \\ \Rightarrow & (x+12)(x-5)=0 \\ \therefore & x=-12, x=5 \end{matrix} $

Since, distance cannot be negative.

$ \begin{aligned} & \therefore \quad x=5 \quad[\because x \neq-12] \\ & \text{ Now, } \quad A C=2 x=10 km \\ & \text{ and } \quad B C=2(x+7)=2(5+7)=24 km \end{aligned} $

The distance covered to reach city $B$ from city $A$ via city $C$

$ \begin{aligned} & =A C+B C \\ & =10+24 \\ & =34 km \end{aligned} $

Distance covered to reach city $B$ from city $A$ after the construction of the highway $=B A=26 km$

Hence, the required saved distance is $34-26$ i.e., $8 km$.

Solution

Let $B C=18 m$ be the flag pole and its shadow be $A B=9.6 m$. The distance of the top of the pole, $C$ from the far end i.e., $A$ of the shadow is $A C$.

$ \begin{matrix} \text{ In right angled } \triangle A B C, & A C^{2}=A B^{2}+B C^{2} \\ \Rightarrow & A C^{2}=(9.6)^{2}+(18)^{2} \\ \Rightarrow & A C^{2}=92.16+324 \\ \therefore & A C^{2}=416.16 \\ & A C=\sqrt{416.16}=20.4 m \\ \text{ Hence, the required distance is } 20.4 m . \end{matrix} $

Thinking Process

Firstly, draw the figure according to the question and get two triangles. Then, show both triangles are similar by AAA similarity criterion and then calculate the required distance.

Solution

Let $A$ be the position of the street bulb fixed on a pole $A B=6 m$ and $C D=1.5 m$ be the height of a woman and her shadow be $E D=3 m$. Let distance between pole and woman be $x m$.

Here, woman and pole both are standing vertically.

So,

In $\triangle C D E$ and $\triangle A B E$,

$ \therefore $

Then,

$ C D | A B $

$ \begin{aligned} \angle E & =\angle E \\ \angle A B E & =\angle C D E \\ \triangle C D E & \sim \triangle A B E \end{aligned} $

$ \dfrac{E D}{E B}=\dfrac{C D}{A B} $

$ \Rightarrow \dfrac{3}{3+x}=\dfrac{1.5}{6} $

$ \begin{aligned} \Rightarrow & 3 \times 6 =1.5(3+x) \\ \Rightarrow & 18 =1.5 \times 3+1.5 x \\ \Rightarrow & 1.5 x =18-4.5 \\ \therefore & x =\dfrac{13.5}{1.5}=9 m \end{aligned} $

[common angle] [each equal to $90^{\circ}$ ] [by AAA similarity criterion]

Hence, she is at the distance of $9 m$ from the base of the pole.

Solution

Given, $\triangle A B C$ in which $\angle B=90^{\circ}$ and $B D \perp A C$

Also,

In $\triangle A D B$ and $\triangle C D B$,

and

Then,

In right angled $\triangle B D C$,

Again,

$\Rightarrow \quad \dfrac{2 \sqrt{5}}{5}=\dfrac{B A}{3 \sqrt{5}}$

$\therefore \quad B A=\dfrac{2 \sqrt{5} \times 3 \sqrt{5}}{5}=6 cm$

Hence, $B D=2 \sqrt{5} cm$ and $A B=6 cm$ $A D=4 cm$ and $C D=5 cm$

$\angle A D B=\angle C D B$

$\angle B A D=\angle D B C$

$\triangle D B A \sim \triangle D C B$

$\dfrac{D B}{D A}=\dfrac{D C}{D B}$

$D B^{2}=D A \times D C$

$D B^{2}=4 \times 5$

$D B=2 \sqrt{5} cm$

$B C^{2}=B D^{2}+C D^{2}$

$B C^{2}=(2 \sqrt{5})^{2}+(5)^{2}$

$ =20+25=45 $

$ B C=\sqrt{45}=3 \sqrt{5} $

$\triangle D B A \sim \triangle D C B$,

$\dfrac{D B}{D C}=\dfrac{B A}{B C}$

Solution

Given, $\triangle P Q R$ in which $\angle Q=90^{\circ}, Q S \perp P R$ and $P Q=6 cm, P S=4 cm$

$\angle P S Q=\angle R S Q$ [each equal to $90^{\circ}$ ]

$\therefore$ $\triangle S Q P \sim \triangle S R Q$

Then, $\dfrac{S Q}{P S}=\dfrac{S R}{S Q}$

In $\triangle S Q P$ and $\triangle S R Q$, $\Rightarrow$

In right angled $\triangle P S Q$,

$$ \begin{align*} \Rightarrow & S Q^{2} =P S \times S R & \ldots \text{(i)}\\ P Q^{2} =P S^{2}+Q S^{2} & \text{[by Pythagoras theorem]}\\ \Rightarrow & (6)^{2} =(4)^{2}+Q S^{2} \\ \Rightarrow & 36 =16+Q S^{2} \\ \Rightarrow & Q S^{2} =36-16=20 \\ \therefore & Q S =\sqrt{20}=2 \sqrt{5} cm \end{align*} $$

On putting the value of $Q S$ in $E q$. (i), we get

$ \begin{matrix} & (2 \sqrt{5})^{2} =4 \times S R \\ \Rightarrow & S R =\dfrac{4 \times 5}{4}=5 cm \\ \text{ In right angled } \triangle Q S R, & Q R^{2} =Q S^{2}+S R^{2} \\ \Rightarrow & Q R^{2} =(2 \sqrt{5})^{2}+(5)^{2} \\ \Rightarrow & Q R^{2} =20+25 \\ & Q R =\sqrt{45}=3 \sqrt{5} cm \\ & \text{ Hence, } Q S=2 \sqrt{5} cm, R S=5 cm \text{ and } Q R=3 \sqrt{5} cm \end{matrix} $

Thinking Process

Apply the Pythagoras theorem in both $\triangle P D Q$ and $\triangle P D R$ to get two equations and then equate them To prove required result.

Solution

Given In $\triangle P Q R, P D \perp Q R, P Q=a, P R=b, Q D=c$ and $D R=d$

To prove $(a+b)(a-b)=(c+d)(c-d)$

Proof In right angled $\triangle P D Q$,

$ \begin{aligned} & .P Q^{2}=P D^{2}+Q D^{2} \quad \text{ [by Pythagoras theorem ]} \\ & \Rightarrow \quad a^{2}=P D^{2}+c^{2} \\ & \Rightarrow \quad P D^{2}=a^{2}-c^{2} \end{aligned} $

$ \begin{aligned} \text{In right angled }\triangle P D R, & P R^{2} =P D^{2}+D R^{2} & \text{ [by Pythagoras theorem ]}\\ \Rightarrow & b^{2} =P D^{2}+d^{2} \\ \Rightarrow & P D^{2} =b^{2}-d^{2} \end{aligned} $

From Eqs. (i) and (ii),

$ \begin{aligned} & a^{2}-c^{2} & =b^{2}-d^{2} \\ \Rightarrow & a^{2}-b^{2} & =c^{2}-d^{2} \\ \Rightarrow & (a-b)(a+b) & =(c-d)(c+d) \end{aligned} $

Thinking Process

Firstly, produce $A B$ and $D C$ and show that $\angle E=90^{\circ}$. Then, apply Pythagoras theorem in different right triangles made with $E$ and show the required result.

Solution

Given Quadrilateral $A B C D$, in which $\angle A+\angle D=90^{\circ}$ To prove $A C^{2}+B D^{2}=A D^{2}+B C^{2}$

Construct Produce $A B$ and $C D$ to meet at $E$.

Also, join $A C$ and $B D$.

Proof

$ \begin{aligned}\text{In} \triangle A E D & \angle A+\angle D =90^{\circ} & \text{[given]}\\ \therefore & \angle E =180^{\circ}-(\angle A+\angle D)=90^{\circ} \end{aligned} $

$ [\because \text{ sum of angles of a triangle }=180^{\circ}] $

Then, by Pythagoras theorem,

In $\triangle B E C$, by Pythagoras theorem, $B C^{2}=B E^{2}+E F^{2}$

On adding both equations, we get

$$ \begin{equation*} A D^{2}+B C^{2}=A E^{2}+D E^{2}+B E^{2}+C E^{2} \tag{i} \end{equation*} $$

In $\triangle A E C$, by Pythagoras theorem,

$ A C^{2}=A E^{2}+C E^{2} $

and in $\triangle B E D$, by Pythagoras theorem,

$ B D^{2}=B E^{2}+D E^{2} $

On adding both equations, we get

$$ \begin{equation*} A C^{2}+B D^{2}=A E^{2}+C E^{2}+B E^{2}+D E^{2} \tag{ii} \end{equation*} $$

From Eqs. (i) and (ii),

$ A C^{2}+B D^{2}=A D^{2}+B C^{2} $

Hence proved.

Solution

Given $l | m$ and line segments $A B, C D$ and $E F$ are concurrent at point $P$.

To prove

Proof $\ln \triangle A P C$ and $\triangle B P D$,

$\therefore \quad \triangle A P C \sim \triangle B P D$

Then,

$ \begin{aligned} & \dfrac{A E}{B F}=\dfrac{A C}{B D}=\dfrac{C E}{F D} \\ & \angle A P C=\angle B P D & \text{[vertically opposite angles] }\\ & \angle P A C=\angle P B D & \text{[alternate angles]}\\\ & \triangle A P C \sim \triangle B P D & \text{[by AAA similarity criterion]}\\\ & \dfrac{A P}{P B}=\dfrac{A C}{B D}=\dfrac{P C}{P D} & \ldots \text{(i)} \end{aligned} $

In $\triangle A P E$ and $\triangle B P F$ $ \text{[vertically opposite angles] } $

$\angle A P E=\angle B P F$
$\angle P A E=\angle P B F \text{[alternate angles]} $

$\therefore$ $\triangle A P E \sim \triangle B P F \text{[by AAA similarity criterion]} $

Then, $\dfrac{A P}{P B}=\dfrac{A E}{B F}=\dfrac{P E}{P F} \ldots \text{(ii)}$

In $\triangle P E C$ and $\triangle P F D$, $\angle E P C=\angle F P D \quad\quad \text{[vertically opposite angles] } $
$\angle P C E=\angle P D F \quad \quad \text{[alternate angles]} $

$ \begin{aligned} \therefore & \triangle P E C \sim \triangle P F D & \text{[by AAA similarity criterion]}\\ \dfrac{P E}{P F} =\dfrac{P C}{P D}=\dfrac{E C}{F D} & \ldots \text{(iii)} \end{aligned} $

from Eqs. (i), (ii) and (iii),

$ \begin{aligned} & \dfrac{A P}{P B}= & \dfrac{A C}{B D}=\dfrac{A E}{B F}=\dfrac{P E}{P F}=\dfrac{E C}{F D} \\ \therefore & & \dfrac{A E}{B F}=\dfrac{A C}{B D}=\dfrac{C E}{F D} \end{aligned} $

Hence proved.

Solution

Given, $A B=6 cm, B C=9 cm, C D=12 cm$ and $S P=36 cm$ Also, $P A, Q B, R C$ and $S D$ are all perpendiculars to line $l$.

$\therefore \quad P A||Q B|| R C || S D$

By basic proportionality theorem,

$ \begin{aligned} P Q: Q R: R S & =A B: B C: C D \\ & =6: 9: 12 \end{aligned} $

Let $ \quad \quad \quad P Q=6 x, Q R=9 x \text{ and } R S=12 x $

Since, length of $ \quad \quad \quad $ PS =36 km

$ \therefore \hspace{20 mm} P Q+Q R+R S=36 $

$ \Rightarrow \hspace{20 mm} 6 x+9 x+12 x=36 $

$\Rightarrow \hspace{20 mm}27 x=36$

$ \therefore \hspace{20 mm} x=\dfrac{36}{27}=\dfrac{4}{3} $

Now, $ \hspace{20 mm} P Q=6 x=6 \times \dfrac{4}{3}=8 cm $

$ \begin{aligned} & \hspace{20 mm} Q R=9 x=9 \times \dfrac{4}{3}=12 cm \\ & \text{and }\hspace{20 mm}R S=12 x=12 \times \dfrac{4}{3}=16 cm \end{aligned} $

Solution

Given $A B C D$ is a trapezium. Diagonals $A C$ and $B D$ are intersect at $O$.

$ P Q||A B|| D C $

To prove

$ P O =Q O$

Proof In $ \triangle A B D$ and $\triangle P O D$,

$ \begin{array}{lcr} & P O \| A B & [\because PQ \| AB] \\ \\ & \angle D=\angle D & \text{[common angle ] } \\ \\ & \angle A B D=\angle P O D & \text{[corresponding angles]}\\ \\ \therefore & \quad \triangle A B D \sim \triangle P O D & \text{[by AAA similarity criterion]}\\ \\ \text{Then, } & \dfrac{O P}{A B}=\dfrac{P D}{A D} & \ldots \text{(i)} \\ \\ \text{In} \triangle ABC \text{ and } \triangle OQC, & OQ|| AB & [\because OQ\|AB] \\ \\ & \angle C=\angle C & \text{[common angle ] } \\ \\ & \angle B A C=\angle Q O C & \text{[corresponding angles]} \\ \\ \therefore & \quad \triangle A B D \sim \triangle OQC & \text{[by AAA similarity criterion]} \\ \\ \text{ Then, }& \dfrac{O Q}{A B}=\dfrac{Q C}{B C} & \ldots \text{(ii)} \end{array} $

Then,

Now, in $\triangle A D C \hspace{20 mm} OP||DC $,

$\therefore \hspace{20 mm} \dfrac{AP}{PP}=\dfrac{OA}{OC} \hspace{13 mm} \text{[by basic proportionality theorem]} \ldots \text{(iii)}$

In $\triangle A B C $,

$\therefore \hspace{20 mm} \dfrac{BQ}{QC}=\dfrac{OA}{OC} \hspace{13 mm} \text{[by basic proportionality theorem]} \ldots \text{(iv)}$

From Eqs. (iii) and (iv),

$ \dfrac{A P}{P D}=\dfrac{B Q}{Q C} $

Adding 1 on both sides, we get

$ \begin{matrix} \Rightarrow & \dfrac{A P}{P D}+1=\dfrac{B Q}{Q C}+1 \\ \\ \Rightarrow & \dfrac{A P+P D}{P D}=\dfrac{B Q+Q C}{Q C} \\ \\ \Rightarrow & \dfrac{A D}{P D}=\dfrac{B C}{Q C} \\ \\ \Rightarrow & \dfrac{P D}{A D}=\dfrac{Q C}{B C} \\ \\ \Rightarrow & \dfrac{O P}{A B}=\dfrac{O Q}{B C} & \text{from Eq. (i) and (ii)}\\ \\ \Rightarrow & \dfrac{O P}{A B}=\dfrac{O Q}{A B} & \text{from Eq. (ii)} \\ \\ \Rightarrow & O P=O Q & \text{Hence prove} \end{matrix} $

[from Eqs. (i) and (ii)] [from Eq. (ii)] Hence proved.

Solution

Given $\triangle A B C, E$ is the mid-point of $C A$ and $\angle A E F=\angle A F E$

To prove $\dfrac{B D}{C D}=\dfrac{B F}{C E}$

Construction Take a point $G$ on $A B$ such that $C G \| E F$.

Proof Since, $E$ is the mid-point of $C A$.

$\therefore \quad C E=A E$

In $\triangle A C G, C G \| E F$ and $E$ is mid-point of $C A$.

So,

$C E=G F$

Now, in $\triangle B C G$ and $\triangle B D F$, [by mid-point theorem]

$ \begin{matrix} & \dfrac{B C}{C D}=\dfrac{B G}{G F} & \text{ [by basic proportionality theorem] } \\ \\ \Rightarrow & \dfrac{B C}{C D}=\dfrac{B F-G F}{G F} \Rightarrow & \dfrac{B C}{C D}=\dfrac{B F}{G F}-1 & \\ \\ \Rightarrow & \dfrac{B C}{C D}+1 =\dfrac{B F}{C E} & \text{ [from Eq. (ii)] }\\ \\ \Rightarrow & \dfrac{B C+C D}{C D} =\dfrac{B F}{C E} \Rightarrow \dfrac{B D}{C D}=\dfrac{B F}{C E} & \text{ Hence Proved.} \end{matrix} $

Thinking Process

Firstly, draw these semi-circles on three sides of right triangle taking each side as diameter. Then, find area of each semi-circle by using formula area of semi-circle $=\dfrac{\pi r^{2}}{2}$ and then proceed required result.

Solution

Let $A B C$ be a right triangle, right angled at $B$ and $A B=y, B C=x$.

Three semi-circles are drawn on the sides $A B, B C$ and $A C$, respectively with diameters $A B$, $B C$ and $A C$, respectively.

Again, let area of circles with diameters $A B, B C$ and $A C$ are respectively $A_1, A_2$ and $A_3$.

To prove $A_3=A_1+A_2$

Proof $\ln \triangle A B C$, by Pythagoras theorem,

$ \begin{aligned} \Rightarrow & A C^{2} & =A B^{2}+B C^{2} \\ \Rightarrow & A C^{2} & =y^{2}+x^{2} \\ \Rightarrow & A C & =\sqrt{y^{2}+x^{2}} \end{aligned} $

We know that, area of a semi-circle with radius, $r=\dfrac{\pi^{2}}{2}$

$\therefore$ Area of semi-circle drawn on $A_3 = \dfrac{\pi}{2}. (\dfrac{AC}{2})^2 = \dfrac{\pi}{2}. (\dfrac{\sqrt{y^2+x^2}}{2})^2$

$\Rightarrow \quad A_3=\dfrac{\pi(y^{2}+x^{2})}{8}$

Now, area of semi-circle drawn on $A_1 = \dfrac{\pi}{2}. (\dfrac{AB}{2})^2 = \dfrac{\pi}{2}. (\dfrac{y}{2})^2$

$\Rightarrow A_1=\dfrac{\pi y^{2}}{8}$

and area of semi-circle drawn on $B C, A_2 = \dfrac{\pi}{2}. (\dfrac{BC}{2})^2 = \dfrac{\pi}{2}. (\dfrac{x}{2})^2$

$\Rightarrow \quad A_2=\dfrac{\pi x^{2}}{8}$

On adding Eqs. (ii) and (iii), we get $A_1+A_2=\dfrac{\pi y^{2}}{8}+\dfrac{\pi x^{2}}{8}$

$ =\dfrac{\pi (y^2 + x^2 )}{8} =A _3 \text{ from Eq. (i) } $

$ \Rightarrow \quad A _1+A _2 =A _3 \text{ Hence proved. } $

Thinking Process

Firstly draw equilateral triangles on each side of right angled $\triangle A B C$ and then find the area of each equilateral triangle by using the formula, area of equilateral triangle

$=\dfrac{\sqrt{3}}{4}(\text{ Side })^{2}$ and prove the required result.

Solution

Let a right triangle $B A C$ in which $\angle A$ is right angle and $A C=y, A B=x$.

Three equilateral triangles $\triangle A E C, \triangle A F B$ and $\triangle C B D$ are drawn on the three sides of $\triangle A B C$.

Again let area of triangles made on $A C, A B$ and $B C$ are $A_1, A_2$ and $A_3$, respectively.

To prove $A_3=A_1+A_2$

Proof In $\triangle C A B$, by Pythagoras theorem,

$ \begin{aligned} \Rightarrow & B C^{2} & =A C^{2}+A B^{2} \\ \Rightarrow & B C^{2} & =y^{2}+x^{2} \\ \Rightarrow & B C & =\sqrt{y^{2}+x^{2}} \end{aligned} $

We know that, area of an equilateral triangle $=\dfrac{\sqrt{3}}{4}(\text{ Side })^{2}$

$\therefore$ Area of equilateral $\triangle A E C, A_1=\dfrac{\sqrt{3}}{4}(A C)^{2}$

$ \begin{aligned} \Rightarrow & A _1=\dfrac{\sqrt{3}}{4} y^{2} & \ldots \text{(i)} \end{aligned} $

and area of equilateral $\triangle A_2 = \dfrac{\sqrt{3}}{4}. (AB)^2 = \dfrac{\sqrt{3}}{4}. (\sqrt{y^2+x^2})^2$

$ \begin{aligned} & =\dfrac{\sqrt{3}}{4}(y^{2}+x^{2})=\dfrac{\sqrt{3}}{4} y^{2}+\dfrac{\sqrt{3}}{4} x^{2} \\ & =A_1+A_2 \end{aligned} $