ML Aggarwal Solution Class 10 Chapter 13 Similarity MCQs

 MCQs

Question 1

In the given figure, ∆ABC ~ ∆QPR. Then ∠R is

(a) 60°

(b) 50°

(c) 70°

(d) 80°

Sol :

In the given figure,

∆ABC ~ ∆QPR

∴ ∠A = ∠Q, ∠B = ∠P and ∠C = ∠R

But ∠A = 70°, ∠B = 50°

∴ ∠C = 180° – (70° + 500) = 180° – 120° = 60°

∠C = ∠R

∴ ∠R = ∠C = 60°

Question 2

In the given figure, ∆ABC ~ ∆QPR. The value of x is

(a) 2.25 cm

(b) 4 cm

(c) 4.5 cm

(d) 5.2 cm

Sol :

In the given figure

∆ABC ~ ∆QPR

$\therefore \frac{\mathrm{AC}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}$

$\Rightarrow \frac{6}{3}=\frac{4.5}{x}$

$\Rightarrow x=\frac{3 \times 4.5}{6}=2.25$

$\therefore x=2.25 \mathrm{~cm}$

Ans (a)

Question 3

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

(a) 50°

(b) 30°

(c) 60°

(d) 100°

Sol :

In the given figure two line segments AC and BD intersect each other at P and PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°

In ∆APB and ∆CPD

$\frac{A P}{P D}=\frac{6}{5}$ and $\frac{B P}{C P}=\frac{3}{2.5}=\frac{6}{5}$

$\because \frac{\mathrm{AP}}{\mathrm{PD}}=\frac{\mathrm{BP}}{\mathrm{CP}}$ and

$\Delta \mathrm{APB}=\angle \mathrm{CPD}$

(Vertically opposite angles)

$\therefore \Delta \mathrm{APB} \sim \Delta \mathrm{CPD}$

$\therefore \angle \mathrm{A}=\angle \mathrm{D}=30$

and third $\angle \mathrm{PBA}=180^{\circ}-\left(50^{\circ}+30^{\circ}\right)$

=180°-80°=100°

Ans (d)

Question 4

If in two triangles ABC and PQR,

$\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}$

then

(a) ∆PQR ~ ∆CAB

(b) ∆PQR ~ ∆ABC

(c) ∆CBA ~ ∆PQR

(d) ∆BCA ~ ∆PQR

Sol :

In two ∆ABC and ∆PQR

$\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}}$

$\Rightarrow \frac{\mathrm{PQ}}{\mathrm{CA}}=\frac{\mathrm{QR}}{\mathrm{AB}}=\frac{\mathrm{RP}}{\mathrm{BC}}$

Ans (a)

Question 5

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

Sol :

In ∆ABC and ∆DEF

∠B = ∠E, ∠F = ∠C

AB = 3DE

∵ Two angles of the one triangles are equal

to corresponding two angles of the other

But sides are not equal

∵ Triangles are similar but not congruent, (b)

Question 6

In the given figure, if D, E and F are midpoints of the sides BC, CA and AB respectively, then the two triangles ABC and DEF are

(a) similar

(b) congruent

(c) both similar and congruent

(d) neither similar nor congruent

Sol :

D, E and F are the mid-points of the sides BC, CA and AB of ∆ABC

then two triangles ABC and DEF are similar.

Ans (a)

Question 7

The given figure, AB || DE. The length of CD is

(a) 2.5 cm

(b) 2.7 cm

(c) $\frac{10}{3} \mathrm{~cm}$

(d) 3.5 cm

Sol :

In the given figure, AB || DE

∆ABC ~ ∆DCE

$\therefore \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{CD}}$

$\Rightarrow \frac{5}{3}=\frac{4.5}{\mathrm{CD}}$

$\Rightarrow \mathrm{CD}=\frac{3 \times 4.5}{5}=2.7 \mathrm{~cm}$

Ans (b)

Question 8

If in triangles ABC and DEF, $\frac{A B}{D E}=\frac{B C}{F D}$  ,then they will be similar when

(a) ∠B = ∠E

(b) ∠A = ∠D

(c) ∠B = ∠D

(d) ∠A = ∠F

Solution:

In two triangles ABC and DEF

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

They will be similar if their included angles are equal

∠B = ∠D 

Ans (c)

Question 9

If ∆PQR ~ ∆ABC, PQ = 6 cm, AB = 8 cm and perimeter of ∆ABC is 36 cm, then perimeter of ∆PQR is

(a) 20.25 cm

(b) 27 cm

(c) 48 cm

(d) 64 cm

Sol :

∆PQR ~ ∆ABC

PQ = 6 cm, AB = 8 cm

Perimeter of ∆ABC = 36 cm

Let perimeter of ∆PQR be x cm

$\therefore \frac{\mathrm{PQ}}{\mathrm{AB}}=\frac{\text { Perimeter of } \Delta \mathrm{PQR}}{\text { Perimeter of } \Delta \mathrm{ABC}}$

$\Rightarrow \frac{6}{8}=\frac{x}{36}$

$\Rightarrow x=\frac{6 \times 36}{8}=27 \mathrm{~cm}$

Perimeter of ΔPQR=27 cm

Ans (b)

Question 10

In the given figure, DE || BC and all measurements are in centimetres. The length of AE is

(a) 2 cm

(b) 2.25 cm

(c) 3.5 cm

(d) 4 cm

Sol :

In the given figure,

DE || BC

$\therefore \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$

$\Rightarrow \frac{3}{4}=\frac{x}{3}$

$\Rightarrow x=\frac{3 \times 3}{4}=\frac{9}{4}=2.25 \mathrm{~cm}$

Ans (b)

Question 11

In the given figure, PQ || CA and all lengths are given in centimetres. The length of BC is

(a) 6.4 cm

(b) 7.5 cm

(c) 8 cm

(d) 9 cm

Sol :

In the given figure,

PQ || CA

Let BC = x

$\therefore \frac{B Q}{Q C}=\frac{B P}{P A}$

$\Rightarrow \frac{5}{x-5}=\frac{4}{2.4}$

$\Rightarrow 4 x-20=12$

$\Rightarrow 4 x=12+20=32$

$\Rightarrow x=\frac{32}{4}=8$

∴BC=8 cm 

Ans (c)

Question 12

In the given figure, MN || QR. If PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm, then PM is

(a) 4 cm

(b) 3.6 cm

(c) 2 cm

(d) 3 cm

Sol :

In the given figure, MN || QR

PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm

Let PM = x cm

$\therefore \frac{\mathrm{PM}}{\mathrm{MQ}}=\frac{\mathrm{PN}}{\mathrm{NR}}$

$\Rightarrow \frac{x}{5-x}=\frac{3.6}{2.4}$

$\Rightarrow \frac{x}{5-x}=\frac{36}{24}$

$\Rightarrow 36(5-x)=24 x$

$\Rightarrow 180-36 x=24 x$

$\Rightarrow 180=24 x+36 x$

$\Rightarrow 60 x=180$

$\Rightarrow x=\frac{180}{60}=3$

∴PM=3 cm

Ans (d)

Question 13

D and E are respectively the points on the sides AB and AC of a ∆ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then the length of DE is

(a) 2.5 cm

(b) 3 cm

(c) 5 cm

(d) 6 cm

Sol :

D and E are the points on sides AB and AC of ∆ABC,

AD = 2 cm, BD = 3 cm,

BC = 7.5 cm

$\mathrm{DE} \| \mathrm{BC}$

Let DE $=x \mathrm{~cm}$

In $\Delta \mathrm{ABC}, \mathrm{DE} \| \mathrm{BC}$

$\therefore \Delta \mathrm{ADE} \sim \Delta \mathrm{ABC}$

$\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{DE}}{\mathrm{BC}}$

$\Rightarrow \frac{2}{2+3}=\frac{x}{7.5}$

$\Rightarrow \frac{2}{5}=\frac{x}{7.5}$

$\Rightarrow x=\frac{2 \times 7.5}{5}=\frac{15}{5}=3$

$\therefore \mathrm{DE}=3 \mathrm{~cm}$

Ans (b)

Question 14

It is given that ∆ABC ~ ∆PQR with $\frac{B O}{Q R}=\frac{1}{3}$ then $\begin{array}{lll}\text { area } & \text { of } & \Delta P Q R \\\hline \text { area } & \text { of } & \Delta A B C\end{array}$ equal to

(a) 9

(b) 3

(c) $\frac{1}{3}$

(d) $\frac{1}{9}$

Sol :

∆ABC ~ ∆PQR

$\frac{B C}{Q R}=\frac{1}{3}$

$\therefore \frac{\text { area of } \Delta \mathrm{PQR}}{\text { area of } \Delta \mathrm{ABC}}=\frac{\mathrm{QR}^{2}}{\mathrm{BC}^{2}}$

$=\frac{(3)^{2}}{(1)^{2}}=\frac{9}{1}=9$

Ans (a)

Question 15

If the areas of two similar triangles are in the ratio 4 : 9, then their corresponding sides are in the ratio

(a) 9 : 4

(b) 3 : 2

(c) 2 : 3

(d) 16 : 81

Sol :

Ratio in the areas of two similar triangles = 4 : 9

Ratio in their corresponding sides

= √4 : √9

= 2 : 3 (c)

Question 16

If ∆ABC ~ ∆PQR, BC = 8 cm and QR = 6 cm, then the ratio of the areas of ∆ABC and ∆PQR is

(a) 8 : 6

(b) 3 : 4

(c) 9 : 16

(d) 16 : 9

Sol :

∆ABC ~ ∆PQR, BC = 8 cm, QR = 6 cm

$\begin{array}{ll}\text { area } & \text { of } \quad \Delta A B C \\\hline \text { area } & \text { of } \quad \Delta P Q R\end{array}=\frac{B C^{2}}{Q R^{2}}$

$=\frac{8^{2}}{6^{2}}=\frac{64}{36}=\frac{16}{9}$

Ans (d)

Question 17

If ∆ABC ~ ∆QRP $\begin{array}{lll}\text { area } & \text { of } & \Delta A B C \\\hline \text { area } & \text { of } & \Delta P Q R\end{array}$AB = 18 cm and BC = 15 cm, then the length of PR is equal to

(a) 10 cm

(b) 12 cm

(c) $\frac{20}{3}$

(d) 8 cm

Sol :

∆ABC ~ ∆QRP

$\begin{array}{ccc}\text { area } & \text { of } & \Delta A B C \\\hline \text { area } & \text { of } & \Delta P Q R\end{array}$

$A B=18 \mathrm{~cm}, B C=15 \mathrm{~cm}^{2}$

Length of PR

$\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{9}{4}=\frac{\mathrm{BC}^{2}}{\mathrm{PR}^{2}}$

$\Rightarrow \frac{9}{4}=\frac{15^{2}}{\mathrm{PR}^{2}}$

$\Rightarrow \frac{3}{2}=\frac{15}{\mathrm{PR}}$

(Taking square root)

$\therefore P R=\frac{15 \times 2}{3}=10 \mathrm{~cm}$

Ans (a)

Question 18

If ∆ABC ~ ∆PQR, area of ∆ABC = 81 cm², area of ∆PQR = 144 cm² and QR = 6 cm, then length of BC is

(a) 4 cm

(b) 4.5 cm

(c) 9 cm

(d) 12 cm

Sol :

∆ABC ~ ∆PQR,

area of ∆ABC = 81 cm² and area of ∆PQR = 144 cm²,

QR = 6 cm, BC = ?

$\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{\mathrm{BC}^{2}}{\mathrm{QR}^{2}}=\frac{\mathrm{BC}^{2}}{(6)^{2}}=\frac{81}{144}$

(given)

$\Rightarrow \frac{\mathrm{BC}^{2}}{(6)^{2}}=\left(\frac{9}{12}\right)^{2}$

$\Rightarrow \frac{\mathrm{BC}}{6}=\frac{9}{12}$

$\therefore \mathrm{BC}=\frac{6 \times 9}{12}=\frac{9}{2}=4.5 \mathrm{~cm}$

Ans (b)

Question 19

In the given figure, DE || CA and D is a point on BD such that BD : DC = 2 : 1. The ratio of area of ∆ABC to area of ∆BDE is

(a) 4 : 1

(b) 9 : 1

(c) 9 : 4

(d) 3 : 2

Sol :

In the given figure, DE || CA ,

D is a point on BC and BD : DC = 2 : 1

BD : BC = 2 : (2 + 1) = 2 : 3

In ∆ABC, DE || CA

∆ABC ~ ∆BDE

$\therefore \frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{BDE}}=\frac{\mathrm{BC}^{2}}{\mathrm{BD}^{2}}=\frac{(3)^{2}}{(2)^{2}}=\frac{9}{4}$

∴Ratio=9 : 4

Ans (c)

Question 20

If ABC and BDE are two equilateral triangles such that D is mid-point of BC, then the ratio of the areas of triangles ABC and BDE is

(a) 2 : 1

(b) 1 : 2

(c) 1 : 4

(d) 4 : 1

Sol :

∆ABC and ∆BDE are an equilateral triangles

Figure to be added

$\therefore \Delta \mathrm{ABC} \sim \Delta \mathrm{BDE}$

$\mathrm{D}$ is mid-point of $\mathrm{BC}$

$\therefore \mathrm{E}$ is mid-point of $\mathrm{AB}$

$\therefore \mathrm{DE} \| \mathrm{CA}=\frac{1}{2} \mathrm{CA}$

$\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{BDE}}=\frac{\mathrm{AC}^{2}}{\mathrm{ED}^{2}}=\frac{\mathrm{AC}^{2}}{\left(\frac{1}{2} \mathrm{AC}\right)^{2}}$

$=\frac{\mathrm{AC}^{2}}{\frac{1}{4} \mathrm{AC}^{2}}=\frac{4}{1}=4: 1$

$\therefore$ Ratio $=4: 1$

Ans (d)

Question 21

The areas of two similar triangles are 81 cm² and 49 cm² respectively. If an altitude of the smaller triangle is 3.5 cm, then the corresponding altitude of the bigger triangle is

(a) 9 cm

(b) 7 cm

(c) 6 cm

(d) 4.5 cm

Sol :

Areas of two similar triangles are 81 cm² and 49 cm²

The altitude of the smaller triangle is 3.5 cm

Let the altitude of the bigger triangle is x, then

$\frac{\text { area of bigger triangle }}{\text { area of smaller triangle }}=\frac{81}{49}$

$=\frac{(\text { bigger altitude })^{2}}{(\text { smaller altitude })^{2}}$

$\Rightarrow \frac{81}{49}=\frac{x^{2}}{(3.5)^{2}}$

$\Rightarrow \frac{9}{7}=\frac{x}{3.5}$

$\Rightarrow x=\frac{9 \times 3.5}{7}=4.5 \mathrm{~cm}$

$\therefore$ Altitude of bigger triangle $=4.5 \mathrm{~cm}$

Ans (d)

Question 22

Given ∆ABC ~ ∆PQR, area of ∆ABC = 54 cm² and area of ∆PQR = 24 cm². If AD and PM are medians of ∆’s ABC and PQR respectively, and length of PM is 10 cm, then length of AD is

(a) $\frac{49}{3} \mathrm{~cm}$

(b) $\frac{20}{3} \mathrm{~cm}$

(c) 15 cm

(d) 22.5 cm

Sol :

∆ABC ~ ∆PQR

area of ∆ABC = 54 cm²

and of ∆PQR = 24 cm²

AD and PM are their median respectively

PM = 10 cm

Let AD = x cm, then

$\frac{\text { area of } \Delta \mathrm{ABC}}{\text { area of } \Delta \mathrm{PQR}}=\frac{\mathrm{AD}^{2}}{\mathrm{PM}^{2}}=\frac{x^{2}}{(10)^{2}}=\frac{54}{24}$

$\therefore \frac{x^{2}}{(10)^{2}}=\frac{54}{24}=\frac{9}{4}=\left(\frac{3}{2}\right)^{2}$

$\therefore \frac{x}{10}=\frac{3}{2}$

$\Rightarrow x=\frac{3 \times 10}{2}=15 \mathrm{~cm}$

∴AD=15 cm