Solution

(c) In $\mathrm{△}ADB$ and $\mathrm{△}ADC$,

$\begin{array}{rl}& \mathrm{\angle }D=\mathrm{\angle }D={90}^{\circ }\\ & \mathrm{\angle }DBA=\mathrm{\angle }DAC& [\text{ each equal to }{90}^{\circ }-<\mathrm{C}]\\ & \therefore \mathrm{△}ADB\sim \mathrm{△}ADC& \text{ [by AAA similarity criterion] }\\ & \therefore {\displaystyle \frac{BD}{AD}}={\displaystyle \frac{AD}{CD}}\\ & \Rightarrow BD\cdot CD=A{D}^2\\ & \end{array}$

Solution

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

Given, $AC=16cm$ and $BD=12cm$

$\therefore$ $AO=8cm,BO=6cm$

and $\mathrm{\angle }AOB={90}^{\circ }$

In right angled $\mathrm{△}AOB$,

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

Solution

(c) Given,

$\mathrm{△}ABC\sim \mathrm{△}EDF$

$\therefore {\displaystyle \frac{AB}{ED}}={\displaystyle \frac{BC}{DF}}={\displaystyle \frac{AC}{EF}}$

Taking first two terms, we get

$\begin{array}{rlrl}& & {\displaystyle \frac{AB}{ED}}& ={\displaystyle \frac{BC}{DF}}\\ \Rightarrow & & AB\cdot DF& =ED\cdot BC\\ \text{ or }& & BC\cdot DE& =AB\cdot DF\end{array}$

So, option (d) is true.

Taking last two terms, we get

${\displaystyle \frac{BC}{DF}}={\displaystyle \frac{AC}{EF}}$

$\Rightarrow BC\cdot EF=AC\cdot DF$

So, option (a) is also true.

Taking first and last terms, we get

${\displaystyle \frac{AB}{ED}}={\displaystyle \frac{AC}{EF}}$

$\Rightarrow$

$AB\cdot EF=ED\cdot AC$

Hence, option (b) is true.

Solution

(a) Given, in two $\mathrm{△}ABC$ and $\mathrm{△}PQR,{\displaystyle \frac{AB}{QR}}={\displaystyle \frac{BC}{PR}}={\displaystyle \frac{CA}{PQ}}$

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.,

$\mathrm{△}CAB\sim \mathrm{△}PQR$

Solution

(d) In $\mathrm{△}APB$ and $\mathrm{△}CPD$,

$\mathrm{\angle }APB=\mathrm{\angle }CPD={50}^{\circ }$

[vertically opposite angles]

and

$\begin{array}{rlr}& {\displaystyle \frac{AP}{PD}}={\displaystyle \frac{6}{5}}& \dots \text{(i)}\\ & {\displaystyle \frac{BP}{CP}}={\displaystyle \frac{3}{2.5}}={\displaystyle \frac{6}{5}}& \dots \text{(ii)}\end{array}$

From Eqs. (i) and (ii)

$\begin{array}{rl}& {\displaystyle \frac{AP}{PD}}={\displaystyle \frac{BP}{CP}}\\ & \therefore \mathrm{△}APB\sim \mathrm{△}DPC\text{ [by SAS similarity criterion] }\\ & \therefore \mathrm{\angle }A=\mathrm{\angle }D={30}^{\circ }\text{ [corresponding angles of similar triangles] }\\ & \text{ In }\mathrm{△}APB,\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }APB={180}^{\circ }\text{ [sum of angles of a triangle }={180}^{\circ }\text{ ] }\\ & \Rightarrow {30}^{\circ }+\mathrm{\angle }B+{50}^{\circ }={180}^{\circ }\\ & \therefore \mathrm{\angle }B={180}^{\circ }-({50}^{\circ }+{30}^{\circ })={100}^{\circ }\\ & \text{ i.e., }\mathrm{\angle }PBA={100}^{\circ }\end{array}$

Solution

(b) Given, in $\mathrm{△}DEF$ and $\mathrm{△}PQR,\mathrm{\angle }D=\mathrm{\angle }Q,\mathrm{\angle }R=\mathrm{\angle }E$

$\begin{array}{cc}\therefore & \mathrm{\Delta }DEF\sim \mathrm{△}QRP\text{ [by AAA similarity criterion] }\\ \therefore & \mathrm{\angle }F=\mathrm{\angle }P\text{ [corresponding angles of similar triangles] }\\ \therefore & {\displaystyle \frac{DF}{QP}}={\displaystyle \frac{ED}{RQ}}={\displaystyle \frac{FE}{PR}}\end{array}$

Solution

(b) In $\mathrm{△}ABC$ and $\mathrm{△}DEF,\mathrm{\angle }B=\mathrm{\angle }E,\mathrm{\angle }F=\mathrm{\angle }C$ and $AB=3DE$

We know that, if in two triangles corresponding two angles are same, then they are similar by AAA similarity criterion. Also, $\mathrm{△}ABC$ and $\mathrm{△}DEF$ 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, $\mathrm{△}ABC\sim \mathrm{△}PQR$ and ${\displaystyle \frac{BC}{QR}}={\displaystyle \frac{1}{3}}$

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

$\therefore {\displaystyle \frac{ar(\mathrm{△}PRQ)}{ar(\mathrm{△}BCA)}}={\displaystyle \frac{(QR{)}^2}{(BC{)}^2}}={{\displaystyle \frac{QR}{BC}}}^2={{\displaystyle \frac{3}{1}}}^2={\displaystyle \frac{9}{1}}=9$

Solution

(b) Given, $\mathrm{△}ABC\sim \mathrm{△}DFE$, then $\mathrm{\angle }A=\mathrm{\angle }D={30}^{\circ },\mathrm{\angle }C=\mathrm{\angle }E={50}^{\circ }$

$\begin{array}{cc}\therefore & \mathrm{\angle }B=\mathrm{\angle }F={180}^{\circ }-({30}^{\circ }+{50}^{\circ })={100}^{\circ }\\ \text{ Also, }& AB=5cm,AC=8cm\text{ and }DF=7.5cm\\ \therefore & {\displaystyle \frac{AB}{DF}}={\displaystyle \frac{AC}{DE}}\\ \Rightarrow & {\displaystyle \frac{5}{7.5}}={\displaystyle \frac{8}{DE}}\\ \therefore & DE={\displaystyle \frac{8\times 7.5}{5}}=12cm\\ \text{ Hence, }& DE=12cm,\mathrm{\angle }F={100}^{\circ }\end{array}$

Solution

(c) Given, in $\mathrm{△}ABC$ and $\mathrm{△}EDF$,

${\displaystyle \frac{AB}{DE}}={\displaystyle \frac{BC}{FD}}$

By converse of basic proportionality theorem,

$\begin{array}{rl}\mathrm{△}ABC& \sim \mathrm{△}EDF\\ \mathrm{\angle }B& =\mathrm{\angle }D,\mathrm{\angle }A=\mathrm{\angle }E\\ \mathrm{\angle }C& =\mathrm{\angle }F\end{array}$

Then,

and

Solution

(a) Given, $\mathrm{△}ABC\sim \mathrm{△}QRP,AB=18cm$ and $BC=15cm$

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

$\begin{array}{ccc}\therefore & {\displaystyle \frac{ar(\mathrm{△}ABC)}{ar(\mathrm{△}QRP)}}& ={\displaystyle \frac{(BC{)}^2}{(RP{)}^2}}\\ \text{ But given, }& {\displaystyle \frac{ar(\mathrm{△}ABC)}{ar(\mathrm{△}PQR)}}& ={\displaystyle \frac{9}{4}}\\ \Rightarrow & {\displaystyle \frac{(15{)}^2}{(RP{)}^2}}& ={\displaystyle \frac{9}{4}}\\ \Rightarrow & (RP{)}^2& ={\displaystyle \frac{225\times 4}{9}}=100\\ \therefore & RP& =10cm\end{array}[\because BC=15cm,\text{ given] }$

Solution

(c) Given, in $\mathrm{△}PQR$,

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

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

Similarly, in $\mathrm{△}RSQ$,

$\Rightarrow \mathrm{\angle }3=\mathrm{\angle }4$

[corresponding angles of equal sides are equal] Now, in $\mathrm{△}PQR$, sum of angles $={180}^{\circ }$

$\begin{array}{cc}\Rightarrow & \mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R={180}^{\circ }\\ \Rightarrow & \mathrm{\angle }2+\mathrm{\angle }4+\mathrm{\angle }1+\mathrm{\angle }3={180}^{\circ }\\ \Rightarrow & \mathrm{\angle }1+\mathrm{\angle }3+\mathrm{\angle }1+\mathrm{\angle }3={180}^{\circ }\\ \Rightarrow & 2(\mathrm{\angle }1+\mathrm{\angle }3)={180}^{\circ }\\ \Rightarrow & \mathrm{\angle }1+\mathrm{\angle }3={\displaystyle \frac{{180}^{\circ }}{2}}={90}^{\circ }\\ \therefore & \mathrm{\angle }R={90}^{\circ }\end{array}$

In $\mathrm{△}PQR$, by Pythagoras theorem,

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

Solution

False

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

Now,

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

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.

$\therefore \mathrm{\angle }D=\mathrm{\angle }R,\mathrm{\angle }F=\mathrm{\angle }P$ and $\mathrm{\angle }F=Q$

Solution

False

Given, $PQ=12.5cm,PA=5cm,BR=6cm$ and $PB=4cm$

Then, $QA=QP-PA=12.5-5=7.5cm$

Now, ${\displaystyle \frac{PA}{AQ}}={\displaystyle \frac{5}{7.5}}={\displaystyle \frac{50}{75}}={\displaystyle \frac{2}{3}}$

and ${\displaystyle \frac{PB}{BR}}={\displaystyle \frac{4}{6}}={\displaystyle \frac{2}{3}}$

From Eqs. (i) and (ii), ${\displaystyle \frac{PA}{AQ}}={\displaystyle \frac{PB}{BR}}$

By converse of basic proportionality theorem,

$AB|QR$

Solution

True

In $\mathrm{△}PBC$ and $\mathrm{△}PDE$,

Since, one angle of $\mathrm{△}PBC$ is equal to one angle of $\mathrm{△}PDE$ and the sides including these angles are proportional, so both triangles are similar.

Hence, $\mathrm{△}PBC\sim \mathrm{△}PDE$, by SAS similarity criterion.

Solution

False

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

$\begin{array}{cc}\text{ In }\mathrm{△}PQR,& \mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R={180}^{\circ }\\ \Rightarrow & {55}^{\circ }+{25}^{\circ }+\mathrm{\angle }R={180}^{\circ }\\ \Rightarrow & \mathrm{\angle }R={180}^{\circ }-({55}^{\circ }+{25}^{\circ })={180}^{\circ }-{80}^{\circ }={100}^{\circ }\\ \ln \mathrm{△}TSM,& LT+\mathrm{\angle }S+\mathrm{\angle }M={180}^{\circ }\\ \Rightarrow & \mathrm{\angle }T+\mathrm{\angle }{25}^{\circ }+{100}^{\circ }={180}^{\circ }\\ \Rightarrow & \mathrm{\angle }T={180}^{\circ }-({25}^{\circ }+{100}^{\circ })\\ \Rightarrow & ={180}^{\circ }-{125}^{\circ }={55}^{\circ }\end{array}$

In $\mathrm{△}PQR$ and $\mathrm{△}TSM$,

and

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

$\therefore$

$\mathrm{△}PQR\sim \mathrm{△}TSM$ [since, all corresponding angles are equal]

Hence, $\mathrm{△}QPR$ is not similar to $\mathrm{△}TSM$, since correct correspondence is $P\to T,Q\to S$ and $R\to M$.

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 $\mathrm{△}ABC$ and $\mathrm{△}PQR$.

In which

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

and $\mathrm{\angle }B=\mathrm{\angle }Q=$ acute angle

Then, by AAA similarity criterion, $\mathrm{△}ABC\sim \mathrm{△}PQR$

Solution

False

By the property of area of two similar triangles,

$\Rightarrow \begin{array}{rl}{\displaystyle \frac{{\text{ Area }}_1}{{\text{ Area }}_2}}& ={\displaystyle \frac{{\text{ Altitude }}_1^2}{{\text{ Altitude }}_2}}\\ {\displaystyle \frac{{\text{ Area }}_1}{{\text{ Area }}_2}}& ={\displaystyle \frac{3}{5}}\\ & ={\displaystyle \frac{9}{25}}\ne {\displaystyle \frac{6}{5}}\end{array}$

So, given statement is not correct.

Solution

False

In $\mathrm{△}PQD$ and $\mathrm{△}RPD$,

$\begin{array}{rl}PD& =PD\\ \mathrm{\angle }PDQ& =\mathrm{\angle }PDR\end{array}$

[common side]

[each ${90}^{\circ }$ ]

Here, no other sides or angles are equal, so we can say that $\mathrm{△}PQD$ is not similar to $\mathrm{△}RPD$.

But, if $\mathrm{\angle }P={90}^{\circ }$, then $\mathrm{\angle }DPQ=\mathrm{\angle }PRD$

[each equal to ${90}^{\circ }-\mathrm{\angle }\theta$ and by ASA similarity criterion, $\mathrm{△}PQD\sim \mathrm{△}RPD$ ]

Solution

True

In $\mathrm{△}ADE$ and $\mathrm{△}ACB$,

$\begin{array}{ccc}\mathrm{\angle }A& =\mathrm{\angle }A& \text{ [common angle] }\\ \mathrm{\angle }D& =\mathrm{\angle }C& \text{ [given] }\\ \mathrm{△}ADE& \sim \mathrm{△}ACB& \text{ [by AAA similarity criterion] }\end{array}$

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 $\ln \mathrm{△}PQR$,

$\begin{array}{rlr}& & P{R}^2-P{Q}^2=Q{R}^2\text{ and }QM\perp PR\\ \text{To prove}& & Q{M}^2=PM\times MR\\ \text{Proof Since, }& & P{R}^2-P{Q}^2=Q{R}^2\\ \Rightarrow & & P{R}^2=P{Q}^2+Q{R}^2\end{array}$

So, $\mathrm{△}PQR$ is right angled triangle at $Q$.

In $\mathrm{△}QMR$ and $\mathrm{△}PMQ$,

$\begin{array}{rlrlr}& & \mathrm{\angle }M=\mathrm{\angle }M& & [\text{each }{90}^{\circ }]\\ & & \mathrm{\angle }MQR=\mathrm{\angle }QPM& & [\text{each equal to }{90}^{\circ }-\mathrm{\angle }R]\\ & & \mathrm{△}QMR\sim \mathrm{△}PMQ& & [\text{by AAA similarilty criterion }]\end{array}$

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

$\begin{array}{cc}{\displaystyle \frac{ar(\mathrm{△}QMR)}{ar(\mathrm{△}PMQ)}}& ={\displaystyle \frac{(QM{)}^2}{(PM{)}^2}}\\ \Rightarrow & {\displaystyle \frac{{\displaystyle \frac{1}{2}}\times RM\times QM}{{\displaystyle \frac{1}{2}}\times PM\times QM}}& ={\displaystyle \frac{(QM{)}^2}{(PM{)}^2}}[\because \text{ area of triangle }={\displaystyle \frac{1}{2}}\times \text{ base }\times \text{ height }]\\ \Rightarrow Q{M}^2& =PM\times RM\text{ Hence proved. }\end{array}$

Solution

Given,

$\begin{array}{rl}& DE|AB\\ & {\displaystyle \frac{CD}{AD}}={\displaystyle \frac{CE}{BE}}\end{array}$

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

[by basic proportionality theorem]

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

Thinking Process

Firstly, show that $ST|QR$ with the help of given information, then use AAA similarity criterion To prove required result.

Solution

Given $\mathrm{△}NSQ\cong \mathrm{△}MTR$ and $\mathrm{\angle }1=\mathrm{\angle }2$

To prove $\mathrm{△}PTS\sim \mathrm{△}PRQ$

Proof Since,

$\mathrm{△}NSQ\cong \mathrm{△}MTR$

So,

$$\begin{array}{}\text{(i)}& SQ=TR\end{array}$$

Also,

$$\begin{array}{}\text{(ii)}& \mathrm{\angle }1=\mathrm{\angle }2\Rightarrow PT=PS\end{array}$$

From Eqs. (i) and (ii),

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

$\Rightarrow$ ${\displaystyle \frac{PS}{SQ}}={\displaystyle \frac{PT}{TR}}$

$\therefore$ $ST|QR$ [by convense of basic proportionality theorem]

and

$\mathrm{\angle }1=\mathrm{\angle }PQR$

In $\mathrm{△}PTS$ and $\mathrm{△}PRQ$,

$\mathrm{\angle }2=\mathrm{\angle }PRQ$

$$\begin{array}{rl}\mathrm{\angle }P& =\mathrm{\angle }P\\ \mathrm{\angle }1& =\mathrm{\angle }PQR\\ \mathrm{\angle }\mathrm{\angle }2& =\mathrm{\angle }PRQ\\ \therefore \mathrm{△}PTS& \sim \mathrm{△}PRQ\end{array}$$

[by AAA similarity criterion]

Hence proved.

Thinking Process

Firstly, show that $\mathrm{△}POQ$ and $\mathrm{△}ROS$ are similar by AAA similarity, then use property of area of similar triangle to get required ratio.

Solution

Given $PQRS$ is a trapezium in which $PQ|RS$ and $PQ=3RS$

$\Rightarrow$

$$\begin{array}{}\text{(i)}& {\displaystyle \frac{PQ}{RS}}={\displaystyle \frac{3}{1}}\end{array}$$

In $\mathrm{△}POQ$ and $\mathrm{△}ROS$,

$\begin{array}{rlr}\mathrm{\angle }SOR& =\mathrm{\angle }QOP& \text{[vertically opposite angles]}\\ \mathrm{\angle }SRP& =\mathrm{\angle }RPQ& \text{[alternate angles]}\\ \therefore \mathrm{△}POQ& \sim \mathrm{△}ROS& \text{[by AAA similarity criterion]}\end{array}$

By property of area of similar triangle,

$\begin{array}{cc}& {\displaystyle \frac{ar(\mathrm{△}POQ)}{ar(\mathrm{△}SOR)}}={\displaystyle \frac{(PQ{)}^2}{(RS{)}^2}}={\displaystyle \frac{PQ}{RS}}{}^2={\displaystyle \frac{3}{1}}\\ \Rightarrow & {\displaystyle \frac{ar(\mathrm{△}POQ)}{ar(\mathrm{△}SOR)}}={\displaystyle \frac{9}{1}}\end{array}$

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

Solution

Given $AC$ and $PQ$ intersect each other at the point $O$ and $AB|DC$.

To prove $OA\cdot CQ=OC\cdot AP$

Proof $\ln \mathrm{△}AOP$ and $\mathrm{△}COQ,\mathrm{\angle }AOP=\mathrm{\angle }COQ$ [vertically opposite angles] $\mathrm{\angle }APO=\mathrm{\angle }CQO$

[since, $AB|DC$ and $PQ$ is transversal, so alternate angles]

$\therefore \mathrm{△}AOP\sim \mathrm{△}COQ$ [by AAA similarity criterion]

Then,

$\Rightarrow OA\cdot CQ=OC\cdot AP$

Hence proved.

Solution

Let $ABC$ be an equilateral triangle of side $8cm$ i.e., $AB=BC=CA=8cm$.

Draw altitude $AD$ which is perpendicular to $BC$. Then, $D$ is the mid-point of $BC$.

$\begin{array}{rl}& \therefore BD=CD={\displaystyle \frac{1}{2}}BC={\displaystyle \frac{8}{2}}=4cm\\ & \text{ Now, }A{B}^2=A{D}^2+B{D}^2\\ & \text{ [by Pythagoras theorem] }\\ & \Rightarrow (8{)}^2=A{D}^2+(4{)}^2\\ & \Rightarrow 64=A{D}^2+16\\ & \Rightarrow A{D}^2=64-16=48\\ & \Rightarrow AD=\sqrt{48}=4\sqrt{3}cm.\end{array}$

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

Solution

Given $AB=4cm,DE=6cm$ and $EF=9cm$ and $FD=12cm$

Also,

$\begin{array}{rl}\mathrm{△}ABC& \sim \mathrm{△}DEF\\ {\displaystyle \frac{AB}{ED}}& ={\displaystyle \frac{BC}{EF}}={\displaystyle \frac{AC}{DF}}\\ {\displaystyle \frac{4}{6}}& ={\displaystyle \frac{BC}{9}}={\displaystyle \frac{AC}{12}}\end{array}$

$\begin{array}{rl}& \therefore \\ & \Rightarrow \end{array}$

On taking first two terms, we get

$\begin{array}{rl}{\displaystyle \frac{4}{6}}& ={\displaystyle \frac{BC}{9}}\\ BC& ={\displaystyle \frac{4\times 9}{6}}=6cm\\ & =AC={\displaystyle \frac{6\times 12}{9}}=8cm\end{array}$

Now,

$\text{ perimeter of }\mathrm{△}ABC=AB+BC+AC$

$=4+6+8=18cm$

Solution

Given, $DE|BC,DE=6cm$ and $BC=12cm$

In $\mathrm{△}ABC$ and $\mathrm{△}ADE$,

and

$\begin{array}{rl}\mathrm{\angle }ABC& =\mathrm{\angle }ADE\\ \mathrm{\angle }ACB& =\mathrm{\angle }AED\\ \mathrm{\angle }A& =\mathrm{\angle }A\end{array}$

$\therefore$

Then, ${\displaystyle \frac{ar(\mathrm{△}ADE)}{ar(\mathrm{△}ABC)}}={\displaystyle \frac{(DE{)}^2}{(BC{)}^2}}$

[corresponding angle] [corresponding angle]

[common side]

[by AAA similarity criterion]

$\begin{array}{rl}& ={\displaystyle \frac{(6{)}^2}{(12{)}^2}}={{\displaystyle \frac{1}{2}}}^2\\ \Rightarrow {\displaystyle \frac{ar(\mathrm{△}ADE)}{ar(\mathrm{△}ABC)}}& ={{\displaystyle \frac{1}{2}}}^2={\displaystyle \frac{1}{4}}\end{array}$

Let $ar(\mathrm{△}ADE)=k$, then $ar(\mathrm{△}ABC)=4k$

Now, ar $(DECB)=ar(ABC)-ar(ADE)=4k-k=3k$

$\therefore$ Required ratio $=ar(ADE):ar(DECB)=k:3k=1:3$

Solution

Given, a trapezium $ABCD$ in which $AB|DC.P$ and $Q$ are points on $AD$ and $BC$, respectively such that $PQ|DC$. Thus, $AB|PQ|DC$.

Join $BD$.

$\begin{array}{lrr}\text{ In }\mathrm{△}ABD,& PO|AB& [\because PQ|AB]\\ \text{ By basic proportionality theorem, }& {\displaystyle \frac{DP}{AP}}={\displaystyle \frac{DO}{OB}}& \dots \text{ (i) }\end{array}$

In $\mathrm{△}BDC$,

$OQ|DC$

By basic proportionality theorem,

$\begin{array}{cc}& {\displaystyle \frac{BQ}{QC}}={\displaystyle \frac{OB}{OD}}\\ \Rightarrow & {\displaystyle \frac{QC}{BQ}}={\displaystyle \frac{OD}{OB}}& \dots \text{(ii)}\\ \text{ From Eqs. (i) and (ii), }& {\displaystyle \frac{DP}{AP}}={\displaystyle \frac{QC}{BQ}}\\ \Rightarrow & {\displaystyle \frac{18}{AP}}={\displaystyle \frac{15}{35}}\\ \Rightarrow & AP={\displaystyle \frac{18\times 35}{15}}=42\\ \therefore & AD=AP+DP=42+18=60cm\end{array}$

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 ${\displaystyle \frac{2}{3}}$

Area of smaller triangle $=48c{m}^2$

By the property of area of two similar triangle,

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

$\begin{array}{cc}\text{ i.e., }& {\displaystyle \frac{ar\text{ (smaller triangle })}{ar(\text{ larger triangle })}}={\displaystyle \frac{2}{3}}\\ \Rightarrow & {\displaystyle \frac{48}{ar(\text{ larger triangle })}}={\displaystyle \frac{4}{9}}\\ \Rightarrow & \text{ ar (larger triangle })={\displaystyle \frac{48\times 9}{4}}=12\times 9=108c{m}^2\end{array}$

Thinking Process

Firstly, show that $\mathrm{△}QNP\sim \mathrm{△}RNQ$, by SAS similarity criterion and then use the property that sum of all angles of a triangle is ${180}^{\circ }$.

Solution

Given $\mathrm{△}PQR,N$ is a point on $PR$, such that $QN\perp PR$

and

To prove Proof We have, $\Rightarrow$ $$\begin{array}{rl}\mathrm{\angle }PQR& ={90}^{\circ }\\ PN\cdot NR& =Q{N}^2\\ PN\cdot NR& =QN\cdot QN\\ {\displaystyle \frac{PN}{QN}}& ={\displaystyle \frac{QN}{NR}}\end{array}$$

In $\mathrm{△}QNP$ and $\mathrm{△}RNQ$,

and

$\therefore$

Then, $\mathrm{△}QNP$ and $\mathrm{△}RNQ$ are equiangulars.

i.e.,

$\begin{array}{rl}& \mathrm{\angle }PQN=\mathrm{\angle }QRN\\ & \mathrm{\angle }RQN=\mathrm{\angle }QPN\end{array}$

On adding both sides, we get

$\begin{array}{rl}& \mathrm{\angle }PQN+\mathrm{\angle }RQN=\mathrm{\angle }QRN+\mathrm{\angle }QPN\\ & \Rightarrow \mathrm{\angle }PQR=\mathrm{\angle }QRN+\mathrm{\angle }QPN\end{array}$

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

In $\mathrm{△}PQR,\mathrm{\angle }PQR+\mathrm{\angle }QPR+\mathrm{\angle }QRP={180}^{\circ }$

$\begin{array}{ccc}\Rightarrow & \mathrm{\angle }PQR+\mathrm{\angle }QPN+\mathrm{\angle }QRN={180}^{\circ }& [\because \mathrm{\angle }QPR=\mathrm{\angle }QPN\text{ and }\mathrm{\angle }QRP=\mathrm{\angle }QRN]\\ \Rightarrow & \mathrm{\angle }PQR+\mathrm{\angle }PQR={180}^{\circ }& \\ \Rightarrow & 2\mathrm{\angle }PQR={180}^{\circ }\\ \Rightarrow & \mathrm{\angle }PQR={\displaystyle \frac{{180}^{\circ }}{2}}={90}^{\circ }& \end{array}$

$\therefore \mathrm{\angle }PQR={90}^{\circ }\text{ Hence proved. }$

Solution

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

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

Let length of the corresponding side of the smaller triangle $=xcm$

By property of area of similar triangle,

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

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

Solution

Given, $AC=8cm,AD=3cm$ and $\mathrm{\angle }ACB=\mathrm{\angle }CDA$

From figure,

$\mathrm{\angle }CDA={90}^{\circ }$

$\therefore$ $\mathrm{\angle }ACB=\mathrm{\angle }CDA={90}^{\circ }$

In right angled $\mathrm{△}ADC$,

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

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

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

$\Rightarrow$ $CD=\sqrt{55}cm$

In $\mathrm{△}CDB$ and $\mathrm{△}ADC$,

$\mathrm{\angle }BDC=\mathrm{\angle }ADC$

[each ${90}^{\circ }$ ]

$\begin{array}{cc}\therefore & \mathrm{\angle }DBC=\mathrm{\angle }DCA\\ \mathrm{△}CDB\sim \mathrm{△}ADC\end{array}$

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

Then,

${\displaystyle \frac{CD}{BD}}={\displaystyle \frac{AD}{CD}}$

$\Rightarrow$

$C{D}^2=AD\times BD$

$\therefore BD={\displaystyle \frac{C{D}^2}{AD}}={\displaystyle \frac{(\sqrt{55}{)}^2}{3}}={\displaystyle \frac{55}{3}}cm$

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 $BC=15m$ be the tower and its shadow $AB$ is $24m$. At that time $\mathrm{\angle }CAB=\theta$. Again, let $EF=h$ be a telephone pole and its shadow $DE=16m$. At the same time $\mathrm{\angle }EDF=\theta$. Here, $\mathrm{△}ABC$ and $\mathrm{△}DEF$ both are right angled triangles.

In $\mathrm{△}ABC$ and $\mathrm{△}DEF$,

$\therefore$

$\mathrm{\angle }CAB=\mathrm{\angle }EDF=\theta$

$\mathrm{\angle }B=\mathrm{\angle }E$

$\mathrm{△}ABC\sim \mathrm{△}DEF$ [each ${90}^{\circ }$ ]