S Chand Class 10 CHAPTER 12 Similar triangles Exercise 12 C

 Exercise 12 C

Question 1 

Ans: $\triangle A B C, A B=6 \mathrm{~cm}, A C=3 \mathrm{~cm}$ 

M is mid point of $A B$

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So $A M=\frac{1}{2} A B=\frac{1}{2} \times 6=3 \mathrm{~cm}$

$M N \| B C$ is drown

So $N$ is mid-point of $A C$

So $A N=\frac{1}{2} A C=\frac{1}{2} \times 3=\frac{3}{2}=1.5 \mathrm{~cm}$

Question 2

Ans: Parallelogram $A B C D A B=6 \mathrm{~cm} A D=5 \mathrm{~cm}$

Diagonals $A C$ and $B D$ intersect each other at $O$ 

$E$ is mid-point as $B C$, $O E$ is Joined

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To prove:

(i)$D E \| A B \quad$ (ii) $D E=\frac{1}{2} A B$

Proof: In $\triangle A B C$

O is mid-point of $A C$ 

and $E$ is the mid-point of $B C$

So $O E \angle A B$ and $O E=\frac{1}{2} A B$

Hence proved

Question 3

Ans: ABC is a triangle , D is mid point of AE and E is mid point of AC 

(i) If BC = 6cm find DE 

(ii) IF  $\angle D B C=140^{\circ}$ find $\angle A B E$

If D and E are the mid point of $\triangle A B C$

(i) SO $D E \| B C$ and $D E=\frac{1}{2} B C$ 

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

So $\begin{aligned} & \frac{A D}{A B}=\frac{D E}{B C} \Rightarrow \frac{1}{2} \frac{A B}{A B}=\frac{D E}{6} \\ \Rightarrow & \frac{D E}{6}=\frac{1}{2} \\ \Rightarrow & D E=\frac{6}{2}=3 \mathrm{~cm} \end{aligned}$

(ii) $\angle D B C=140^{\circ} .$

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

$\begin{aligned}&=140^{\circ} \\&\text { So } \angle A D E^{\circ}=140^{\circ}\end{aligned}$

Question 4

Ans: In trapezium $A B C D$

$A B \| D C$

$A D=B C=4 \mathrm{~cm}$

$\angle D A B=\angle C B A=60^{\circ}$

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(i) the length of $C D$

(ii) distance between $A B$ and $C D$

Draw DL and CM perpendicular on AB in $\triangle A D L$ and $\triangle B C M$

$\angle L=\angle M$

$\angle A=\angle B$

$D A=C B$

$\begin{aligned}\text { So } \triangle A D L & \cong M B C m \\\text { So } A L &=M B\end{aligned}$

Now in right $\triangle A D L$

since $\frac{D L}{A D} \Rightarrow \sin 60^{\circ}=\frac{D L}{4}$

$\Rightarrow \frac{3}{2}=\frac{D L}{4} \Rightarrow D L=\frac{4 \times \sqrt{3}}{2}=2 \sqrt{3} \mathrm{~cm}$

and $\tan 60^{\circ}=\frac{D L}{A L}=\sqrt{3}=\frac{2 \sqrt{3}}{A L}$

$\Rightarrow A L=\frac{2 \sqrt{3}}{\sqrt{3}}=2$

So MB = AL = 2cm

But CD = LM= AB -AL-MB

$=10-2-2=10-4=6 \mathrm{~cm}$

Question 5

Ans: (i) In ||gm ABCD , AB||CD 

Diagonals AC and BD bisects each other at K now in Triangle ABC

if E is mid point of AD and K is mid point of AD 

So EK or EKF ||CD or AB 

$\begin{aligned} & \frac{A E}{A D}=\frac{A K}{A C}=\frac{E K}{C D} \\ \Rightarrow & \frac{6}{6+6}=\frac{8}{8+8}=\frac{5}{y} \\ \Rightarrow & \frac{5}{y}=\frac{6}{12} \Rightarrow \frac{5}{y}=\frac{1}{2} \Rightarrow y=5 \times 2=10 \end{aligned}$

So $y=10 \mathrm{~cm}$

similarly in $\triangle A C B$.

$K E \| A B$

So $\begin{aligned} \frac{C K}{C A} &=\frac{K F}{A B} \Rightarrow \frac{8}{16}=\frac{x}{10} \\ \Rightarrow x &=\frac{8 \times 10}{16}=5 \end{aligned}$

Hence $x=5 \mathrm{~cm}, y=10 \mathrm{~cm}$

(ii)In the figure, PQRS is a parallelogram

$\begin{aligned}&O L=L T=6, S T=4, \angle S=3 \\&P Q=y \text { and } P L=x\end{aligned}$

we have to find the value of $x$ and $y$ in $\triangle Q T R$

L is mid point of QT  So QL = LT =6cm and PLS|| QR 

So S is mid point of TR 

So TS= SR = SR = 4

But PQ = SR

So PQ = 4 or y = 4

In $\triangle P L Q$ and $\triangle S L T$

$\begin{aligned}L Q &=L T \\P Q &=T S \\\angle P L Q &=\angle S L T \\\text { So } \triangle P L Q & \cong \triangle S L T \\\text { So } P L &=L S \\\Rightarrow x &=3\end{aligned}$

Hence $x=3, y=4$

(iii)In $\triangle A B D$.

$A L=L D=3$

So $L$ is mid-point of $A D$ 

if $\quad L M \| B D$

So $m$ is mid -point of $A B$ 

So $\triangle A L M \backsim \triangle A D B$

So $\begin{aligned} & \frac{A L}{A D}=\frac{L m}{B D} \Rightarrow \frac{3}{6}=\frac{9}{p} \\ \Rightarrow & \frac{q}{p}=\frac{1}{2} \end{aligned}$

$\Rightarrow \quad 2 9=p$..........(i)

Similarly in $\triangle B C D$

Question 6

Ans: $\triangle A B C$ is an isoscelies triangle in which $A B=A C=10 \mathrm{~cm}$ $B C=12 \mathrm{~cm}$

$P Q R S$ is a Rectangle inside the isosceles triangle $P Q=58$ $=y \mathrm{~cm}$ and $P S=Q R=2 x \mathrm{~cm}$

To prove:

$x=6-\frac{3 y}{4}$

Construction: From A, draw AL ⊥ BC 

Which Bisects BC at L 

Proof: If L is mid point of BC 

So $B L=L C=\frac{12}{2}=6 \mathrm{~cm}$ Now in rignt $\triangle A B L$

$\begin{aligned}& A B^{2}=A L^{2}+B L^{2} \\\Rightarrow &(10)^{2}=A L^{2}+(6)^{2} \\\Rightarrow & A L^{2}=(10)^{2}-(6)^{2} \\\Rightarrow & A L^{2}=100-36=64=8^{2}\end{aligned}$

So $A L=8 \mathrm{~cm}$

Now in $\triangle B P Q$ and $\triangle B A L$

$\begin{aligned}&\angle Q=\angle L \\&\angle B=\angle B\end{aligned}$

So $\triangle B P Q$∽ $\triangle B A L$

$\begin{aligned} & \frac{P Q}{A L}=\frac{B Q}{B C} \Rightarrow \frac{y}{8}=\frac{6-x}{6} \\ \Rightarrow & 6 y=48-8 x \\ \Rightarrow & 8 x=48-6 y \end{aligned}$

$x=\frac{48-6 y}{8}=6-\frac{3 y}{4}$

Hence $x=6-\frac{3 y}{4}$

Question 7

Ans: In the figure PQ|| ST 

To prove: 

(i) $\triangle$ PQR and STR are similar

(ii) PR.RT = QR.RS

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proof :

(a) In $\triangle P Q R$ and $\triangle$ STR

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

So $\triangle P Q R ~ \triangle S T R$

(b) $\begin{aligned}& \frac{P R}{R S}=\frac{R Q}{R T} \\\Rightarrow & P R \times R T=Q R \times R S\end{aligned}$

Hence proved

Question 8

Ans: ABCD is a trapezium in which AB||DC and diagonals AC and BD intersect each other at E. 

To prove: AE.DF = BE.CE 

Proof: In $\triangle A E B$ and $\triangle C E D$

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

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

and $\angle E P A=\angle E D C$

So $\triangle A E B \backsim \triangle C E D$

So $\frac{A E}{C E}=\frac{B E}{D E}$

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

Question 9

Ans: In $\triangle A B C$

$A L \perp B C$ and $B M \perp A C$ are drawn To prove:

To prove:

(i) $\triangle A L C \sim \triangle B M C$

(ii) $\mathrm{Cm} \cdot \mathrm{CA}=\mathrm{Cl} \cdot \mathrm{CB}$

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proof: In $\triangle A L C$ and $\triangle B M C$

$\begin{aligned}\angle L &=\angle m \\\angle C &=\angle C\end{aligned}$

So $\triangle A L C ~ \triangle B M C$

So $\frac{\mathrm{CM}}{\mathrm{CL}}=\frac{\mathrm{CB}}{\mathrm{CA}}$

$\begin{aligned}&\Rightarrow \mathrm{Cm} \cdot \mathrm{CA}=\mathrm{CB} \cdot \mathrm{CL} \\&\Rightarrow \mathrm{CM} \cdot \mathrm{CA}=\mathrm{CL} \cdot \mathrm{CB}\end{aligned}$

Hence proved 

Question 10

Ans: $\triangle A B C$ is a right angled Triangle $A D \perp B C$ is drawn

To prove:

(i) $\triangle A B D \backsim \triangle A B C$

(ii) $\triangle A C D=\triangle A B C$

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proof :

(i) In $\triangle A B D$ and $\triangle B C$

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

(ii) Similarly in $\triangle A C D$ and $\triangle A B C$

$\angle C=\angle C$

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

So $\triangle A C D ∽ \triangle A B C$

Hence proved 

Question 11

Ans: 

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

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To prove:

(i) $\frac{A B}{A D}=\frac{A C}{A E}$

(ii) $\angle A D B=\angle A E C$

Proof: $\angle B A D=\angle C A E$

Adding $\angle C A D$ both sides.

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

Now in $\triangle A B C$ and $\triangle A D E$

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

(i) So $\triangle A B C \backsim \triangle A D E$

So $\frac{A B}{A D}=\frac{A C}{A E}$

or $\frac{A B}{A C}=\frac{A D}{A E}$

(ii) Now in $\triangle A B D$ and $\triangle A E C$

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

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

So $\angle A D B=\angle A E C$ Hence proved 

Question 12

Ans:  In the  figure $A B C D$ and $A E F G$ are two squares 

To prove 

(i)  AF : $A C=A C: A D$

(ii) triangles ACF and ADG are similar

proof : If AC and AF are the diagonals of two squares ABCD and AEFG respectively 

So $\triangle A D C \sim \triangle A G F$

Now $\angle B A F=\angle B A C-\angle F A C$

 $\angle G A C=\angle G A F-\angle F A C$ 

But $\angle B A C=\angle G A F=45^{\circ}$

(Diagonals bisects the opposite angles of a square)

So $\angle B A F=45^{\circ}-\angle F A C$ .........(i)

and $\angle G A C=45^{\circ}-\angle F A C$...........(ii)

From (i) and (ii)

So $\angle B A F=\angle G A C$

Similarly $\angle D A G=\angle D A C-\angle G A C=45^{\circ}-\angle G A C$

and $\angle F A C=\angle B A C-\angle B A F=45^{\circ}-\angle B A F$ 

But $\angle G A C=\angle B A F$

So $\angle D A G=\angle F A C$

Now in $\triangle A D G$ and $\triangle A C F$

$\angle D A G=\angle F A C$

So From (A)

$\triangle A D G \sim \triangle A C F$

So 

$\begin{aligned} & \frac{A C}{A F}=\frac{A D}{A C} \\ \Rightarrow & \frac{A F}{A G}=\frac{A C}{A D} \end{aligned}$

$\Rightarrow A F: A G=A C: A D$   Hence proved

Question 13

Ans: In a square $A B C D$ diagonals $A C$ and $B D$ bisect each other at O

Bisector of $\angle C A B$ meets $B D$ is $X$ and $B C$

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To prove: $\triangle A CY \sim \triangle A B X$

$\angle C A Y=\angle B A Y$

$\angle A C Y=\angle A B X$

So $\triangle A C Y \sim \triangle A B X$ Hence proved 

Question 14

Ans: In rectangle PQRS , X is a point on PQ such that $SX^{2}$ 

$=P x \cdot P Q$

To prove: $\Delta P \times S \sim \triangle X S R$

Proof: If $S \times{ }^{2}=P \times \cdot P Q$

$\Rightarrow \frac{s x}{p Q}=\frac{p x}{s x} \Rightarrow \frac{s x}{s R}=\frac{p x}{s x}$

Now In $\Delta$ PXS and $\Delta X S R$

$\frac{S X}{S R}=\frac{P X}{S X}$ 

$\angle PX S=\angle R S X$

So $\triangle$ PXS $\sim$ △XSR Hence proved 

Question 15

Ans: $\triangle A B C, D$ is a point on BC such that $\angle A D C=\angle B A C$

 To prove $: \frac{B C}{C A}=\frac{C A}{C D}$

Proof: In ISABC and $\triangle A D C$

$\begin{aligned}&\angle B A C=\angle A D C \\&\angle C=\angle C \\&\text { SO } \triangle A B C \sim \triangle A D C \\&\text { SO } \frac{B C}{A C}=\frac{A C}{C D} \\&\Rightarrow \frac{B C}{C A}=\frac{C A}{C D}\end{aligned}$

Question 16

Ans: Given: In $\triangle P Q R, P Q=P R$

Z is a point on PR such that Q R^{2}=P R \times R Z

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To prove: $Q Z=Q R$

Proof: $\quad Q R^{2}=P R \times R Z$

$\Rightarrow \frac{P R}{Q R}=\frac{Q R}{R Z}$

Now In $\triangle B Q R$ and $\triangle Q R Z$

$\frac{P R}{Q R}=\frac{Q R}{R Z}$

$\angle R=\angle R$

So $\triangle P O R \sim \triangle Q R Z$

$\frac{Q R}{R Z}=\frac{P R}{Q R}=\frac{P Q}{Q Z}$

But $\frac{P Q}{Q Z}=\frac{P R}{Q Z}$

So $\frac{P R}{Q R}=\frac{P R}{Q Z} \Rightarrow Q R=Q Z$

or QZ = QR Hence proved 

Question 17

Ans: In $\triangle A B C, A D$ and $B E$ are perpendicular to the sides $B C$ and $A C$ respectively

To prove:

(i) $\triangle A D C \backsim \triangle B E C$.

(ii) $C A \cdot C E=C B \cdot C D$

(iii) $\triangle A B C \backsim \triangle D E C$

(iv) $C D \cdot A B=C A \cdot D E$

Proof: 

(i) In $\triangle A D C$ and $\triangle B E C$

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

So $\triangle A D C \backsim \triangle B E C$

So $\frac{C A}{C B}=\frac{C D}{C E}$

(ii) So $C A \cdot C E=C B \cdot C D$

Again in $\triangle A B C$ and $\triangle D E C$

$\angle C=\angle C$

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

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

SO $\frac{C B}{C D}=\frac{A B}{D E}$

(iv) $\begin{aligned} \Rightarrow & C A \cdot D E=C D \cdot A B \\ & \operatorname{or} C D \cdot A B=C A \cdot D E \end{aligned}$

Hence proved 

Question 18

 

Ans: In quadrilateral ABCD 

PQR and S are the points of trisection of the sides AB, BC ,CD and DA respectively and are adjacent to A and C

To Prove: PQRS is a parallelogram 

construction: Join diagonals $A C$

proof: In $\triangle A D C$

$A P=\frac{1}{3} A B$ and $C Q=\frac{1}{3} B C$ or $\frac{A P}{A B}=\frac{C Q}{B C}$ $=\frac{1}{3}$

 

So $P Q \| A C$ and $P Q=\frac{2}{3} A C$ ............(i)

Similarly in  $\triangle A D C$

 

$A S=\frac{1}{3} A D$ and $C R=\frac{1}{3} C D$ or $\frac{A S}{A D}=\frac{C R}{C D}$

$=\frac{2}{3}$

So

RS $||A C$ and $R S=\frac{2}{3} \mathrm{AC}$........(ii)

from (i) and (ii)

PQRS is a parallelogram Hence proved 

Question 19

 

Ans:  In $\triangle A B C, A D \perp B C$

$B D=4, A D=6, C D=9$

To prove : 

(i) $\triangle A D B \sim \triangle C O A$

(ii) $\angle B A C$ is a right angle

Proof : In $\triangle A D B$ and $\triangle C D A$

$\begin{aligned}\therefore & \angle A D B=\angle A D C \\& \frac{A D}{B D}=\frac{6}{4}, \frac{C D}{A D}=\frac{9}{6} \\\Rightarrow & \frac{A D}{B D}=\frac{3}{2} \text { and } \frac{C D}{A D}=\frac{3}{2}\end{aligned}$

So $\frac{A D}{B D}=\frac{C D}{A D}$

So $\triangle A D B \backsim \triangle C P A$

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

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

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

But $\angle B A C+\angle A B D+\angle A C D=180^{\circ}=\angle D A C=90^{\circ}$

Question 20

Ans: $\triangle A B C \backsim \triangle D E F$

$A L \perp B C$ and $D M \perp E F$

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To prove: $\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}=\frac{A L}{D M}$

Proof: if $\triangle A B C \sim \triangle D E F$

So $\angle B=\angle E$ and $\frac{A B}{D E}=\frac{B C}{E F}$.............(i)

Now in  $\triangle A B C$ and $\triangle D E M$

 $\angle B=\angle E$

$\angle L=\angle M$

$\triangle A B L$ and $\triangle D E M$

So $\frac{A B}{D E}=\frac{A L}{D M}$...........(ii)

From (i) and (ii)

$\frac{A B}{D E}=\frac{B C}{E F}=\frac{A L}{D M}$ Hence proved 

Question 21

Ans: In parallelogram $A B C D$

$B D$ is diagonal and $E$ is any point on $B C$

 $A E$ is Joined which intersect $B D$ at $F$

To prove: DF.EF $=F B \cdot F A$ 

proof: In$\triangle AFD and$\triangle B F E$

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

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

So $\frac{D F}{F B}=\frac{F A}{F E}$

So $D F \cdot E F=F B \cdot FA$

Question 22

Ans: X is any point in  $\triangle D E F$

XD , XE and SF are joined . P is a point on DX , PQ || DE is drawn meeting XE in Q and QR||EF is drawn meeting XF in R 

To prove: PR||DF 

Proof: If P is any point on DX and PQ||DE

So $\frac{X P}{P D}=\frac{X Q}{Q E}$......(i)

if $Q R \| E F$

So $\frac{X Q}{Q E}=\frac{X R}{R F}$

from (i)and (ii)

$\frac{X P}{P D}=\frac{X R}{R F}$

So $P R \| D F$ Hence proved