ML Aggarwal Similarity MCQs Class 10 ICSE Maths Solutions Ch-13. We Provide Step by Step Answer of MCQs Questions for Ch-13 Similarity as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.
ML Aggarwal Similarity MCQs Class 10 ICSE Maths Solutions Ch-13
| Board | ICSE |
| Subject | Maths |
| Class | 10th |
| Chapter-13 | Similarity |
| Writer / Book | Understanding |
| Topics | Solutions of MCQs |
| Academic Session | 2024-2025 |
Similarity MCQs
ML Aggarwal Class 10 ICSE Maths Solutions Ch-13
Question 1 In the adjoining figure, ∆ABC ~ ∆QPR. Then ∠R is
(a) 60°
(b) 50°
(c) 70°
(d) 80°
Answer 1
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°
Page 291
Question 2 In the adjoining figure, ∆ABC ~ ∆QPR. The value of x is
(a) 2.25 cm
(b) 4 cm
(c) 4.5 cm
(d) 5.2 cm
Answer 2
In the given figure
∆ABC ~ ∆QPR
AC/QR = BC?PR
6/3 = 4.5/x
x= (3 x 4.5)/6
= 2.25
x = 2.25 cm
Question 3 In the adjoining 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°
Answer 3
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
Question 4 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
Answer 4
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 5 The adjoining figure, AB || DE. The length of CD is
(a) 2.5 cm
(b) 2.7 cm
(c) 10/ 3 cm
(d) 3.5 cm
Answer 5
In the given figure, AB || DE
∆ABC ~ ∆DCE
AB/DE = BC/CD
5/3 = 4.5/CD
CD = (3 x 4.5)/5
2.7 cm
Question 6 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
Answer 6
∆PQR ~ ∆ABC
PQ = 6 cm, AB = 8 cm
Perimeter of ∆ABC = 36 cm
Let perimeter of ∆PQR be x cm
= 27 cm
Question 7 In the adjoining figure, DE || BC and all measurements are in centimeters. The length of AE is
(a) 2 cm
(b) 2.25 cm
(c) 3.5 cm
(d) 4 cm
Answer 7
In the given figure,
DE || BC
Question 8 In the given figure, PQ || CA and all lengths are given in centimeters. The length of BC is
(a) 6.4 cm
(b) 7.5 cm
(c) 8 cm
(d) 9 cm
Answer 8
In the given figure,
PQ || CA
Let BC = x
Page 292
Question 9
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
Answer 9
In the given figure, MN || QR
PN = 3.6 cm, NR = 2.4 cm and PQ = 5 cm
Let PM = x cm
Question 10 It is given that ∆ABC ~ ∆PQR with BC/QR=1/3 area of ΔPQR/area of ΔABC then equal to
(a) 9
(b) 3
(c) 1/3
(d) 1/9
Answer 10
∆ABC ~ ∆PQR
BC/QR=1/3
Question 11 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
Answer 11
Ratio in the areas of two similar triangles = 4 : 9
Ratio in their corresponding sides
= √4 : √9
= 2 : 3 (c)
Question 12 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
Answer 12
∆ABC ~ ∆PQR, BC = 8 cm, QR = 6 cm
area of ΔABC/area of ΔPQR=BC²/QR²
8²/6² = 64/36 = 16/9 (d)
Question 13 If ∆ABC ~ ∆QRP area of ΔABC/area of ΔPQR= 9/4 AB = 18 cm and BC = 15 cm, then the length of PR is equal to
(a) 10 cm
(b) 12 cm
(c) 20/3
(d) 8 cm
Answer 13
∆ABC ~ ∆QRP
area of ΔABC/area of ΔPQR
Question 14 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
Answer 14
∆ABC ~ ∆PQR,
area of ∆ABC = 81 cm² and area of ∆PQR = 144 cm²,
QR = 6 cm, BC = ?
Question 15 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
Answer 15
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
Question 16 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
Answer 16
∆ABC and ∆BDE are an equilateral triangles
Question 17 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
Answer 17
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
Question 18 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) 49/3 cm
(b) 20/3 cm
(c) 15 cm
(d) 22.5 cm
Answer 18
∆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
— : End of ML Aggarwal Similarity MCQs Class 10 ICSE Maths Solutions Ch-13 :–
Return to: ML Aggarwal Solutions for ICSE Class-10
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