## RD Sharma Class 10 Solutions Chapter 7 Triangles MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles MCQS

**Other Exercises**

- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1
- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.2
- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.3
- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4
- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5
- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6
- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7
- RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise
- RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS
- RD Sharma Class 10 Solutions Chapter 7 Triangles MCQS

**Mark the correct alternative in each of the following.**

**Question 1.**

Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio

(a) 2 : 3

(b) 4 : 9

(c) 81 : 16

(d) 16 : 81

**Solution:**

**(d)** Triangles are similar and the ratio of their sides is 4 : 9

The ratio of the areas of two similar triangles are proportion to the square oT their corresponding sides

Ratio in their areas = (4)² : (9)² = 16 : 81

**Question 2.**

The areas of two similar triangles are in respectively 9 cm² and 16 cm². The ratio of their corresponding sides is

(a) 3 : 4

(b) 4 : 3

(c) 2 : 3

(d) 4 : 5

**Solution:**

**(a)** Ratio in the areas of two similar triangles = 9 cm² : 16 cm² = 9 : 16

The areas of similar triangles are proportional to the squares of their corresponding sides

Ratio in their corresponding sides = √\(\frac { 9 }{ 16 }\) = \(\frac { 3 }{ 4 }\) = 3 : 4

**Question 3.**

The areas of two similar triangles ∆ABC and ∆DEF are 144 cm² and 81 cm² respectively. If the longest side of larger ∆ABC be 36 cm, then. The longest side of the smaller triangle ∆DEF is :

(a) 20 cm

(b) 26 cm

(c) 27 cm

(d) 30 cm

**Solution:**

**(c)** Area of the larger triangle ABC = 144 cm²

and area of smaller ∆DEF = 81 cm²

Longest side of larger triangle = 36 cm

Let the longest side of smaller triangle = x cm

The ratio of the areas of two similar triangles is proportional to the squares of their corresponding sides

**Question 4.**

∆ABC and ∆BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangles ABC and BDE is :

(a) 2 : 1

(b) 1 : 2

(c) 4 : 1

(d) 1 : 4

**Solution:**

**(c)** ∆ABC and ∆BDE are equilateral triangles and D is the mid-point of PC

∆ABC and ∆BDE are both equilateral triangles

**Question 5.**

If ∆ABC and ∆DEF are similar such that 2AB = DE and BC = 8 cm, then EF =

(a) 16 cm

(b) 12 cm

(c) 8 cm

(d) 4 cm

**Solution:**

**(a)**

**Question 6.**

(a) 2 : 5

(b) 4 : 25

(c) 4 : 15

(d) 8 : 125

**Solution:**

**(b)**

**Question 7.**

XY is drawn parallel to the base BC of a ∆ABC cutting AB at X and AC at Y. If AB = 4 BX and YC = 2 cm, then AY =

(a) 2 cm

(b) 4 cm

(c) 6 cm

(d) 8 cm

**Solution:**

**(c)** In ∆ABC, XY || BC

AB = 4BX, YC = 2 cm

**Question 8.**

Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their foot is 12 m, the distance between their tops is

(a) 12 m

(b) 14 m

(c) 13 m

(d) 11 m

**Solution:**

**(c)** Let length of pole AB = 6 m

and of pole CD = 11 m

and distance between their foot = 12 m

i.e., BD = 12 m

**Question 9.**

In ∆ABC, D and E are points on side AB and AC respectively such that DE || BC and AD : DB = 3 : 1. If EA = 3.3 cm, then AC =

(a) 1.1 cm

(b) 4 cm

(c) 4.4 cm

(d) 5.5 cm

**Solution:**

**(c)** In ∆ABC, DE || BC

AD : DB = 3 : 1, EA = 3.3 cm

**Question 10.**

In triangles ABC and DEF, ∠A = ∠E = 40°, AB : ED = AC : EF and ∠F = 65°, then ∠B =

(a) 35°

(b) 65°

(c) 75°

(d) 85°

**Solution:**

**(c)** In ∆ABC and ∆DEF,

∠A = ∠E = 40°

AB : ED = AC : EF, ∠F = 65°

**Question 11.**

If ABC and DEF are similar triangles such that ∠A = 47° and ∠E = 83°, then ∠C =

(a) 50°

(b) 60°

(c) 70°

(d) 80°

**Solution:**

**(a)** ∆ABC ~ ∆DEF

∠A = 47°, ∠E = 83°

∆ABC and ∆DEF are similar

∠A = ∠D, ∠B = ∠E and ∠C = ∠F

∠A = 47°

∠B = ∠E = 83°

But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)

47° + 83° + ∠C = 180°

=> 130° + ∠C = 180°

=> ∠C = 180° – 130°

=> ∠C = 50°

**Question 12.**

If D, E, F are the mid-points of sides BC, CA and AB respectively of ∆ABC, then the ratio of the areas of triangles DEF and ABC is

(a) 1 : 4

(b) 1 : 2

(c) 2 : 3

(d) 4 : 5

**Solution:**

**(a)** D, E and F are the mid points of the sides. BC, CA and AB respectively of ∆ABC

DE, EF and FD are joined

**Question 13.**

In an equilateral triangle ABC, if AD ⊥ BC, then

(a) 2AB² = 3AD²

(b) 4AB² = 3AD²

(c) 3AB² = 4AD²

(d) 3AB² = 2AD²

**Solution:**

**(c)** In equilateral ∆ABC,

AD ⊥ BC

**Question 14.**

If ∆ABC is an equilateral triangle such that AD ⊥ BC, then AD² =

(a) \(\frac { 3 }{ 2 }\) DC²

(b) 2 DC²

(c) 3 CD²

(d) 4 DC²

**Solution:**

**(c)** In equilateral ∆ABC, AD ⊥ BC

AD bisects BC at D

**Question 15.**

In a ∆ABC, AD is the bisector of ∠BAC. If AB = 6 cm, AC = 5 cm and BD = 3 cm, then DC =

(a) 11.3 cm

(b) 2.5 cm

(c) 3.5 cm

(d) None of these

**Solution:**

**(b)** In ∆ABC, AD is the bisector of ∠BAC

AB = 6 cm, AC = 5 cm, BD = 3 cm

**Question 16.**

In a ∆ABC, AD is the bisector of ∠BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm. Find AC

(a) 4 cm

(b) 6 cm

(c) 3 cm

(d) 8 cm

**Solution:**

**(a)** In ∆ABC, AD is the bisector of ∠BAC

AB = 8 cm, BD = 6 cm and DC = 3 cm

**Question 17.**

ABCD is a trapezium such that BC || AD and AB = 4 cm. If the diagonals AC and BD intersect at O such that \(\frac { AO }{ OC }\) = \(\frac { DO }{ OB }\) = \(\frac { 1 }{ 2 }\) , then BC =

(a) 7 cm

(b) 8 cm

(c) 9 cm

(d) 6 cm

**Solution:**

**(b)** In trapezium ABCD, BC || AD

AD = 4 cm, diagonals AC and BD intersect

=> x = 8

BC = 8 cm

**Question 18.**

If ABC is a right triangle right-angled at B and M, N are the mid-points of AB and BC respectively, then 4 (AN² + CM²) =

(a) 4 AC²

(b) 5 AC²

(c) \(\frac { 5 }{ 4 }\) AC²

(d) 6 AC²

**Solution:**

**(b)** In right ∆ABC, ∠B = 90°

M and N are the mid points of AB and BC respectively

**Question 19.**

If in ∆ABC and ∆DEF, \(\frac { AB }{ DE }\) = \(\frac { BC }{ FD }\), then ∆ABC ~ ∆DEF when

(a) ∠A = ∠F

(b) ∠A = ∠D

(c) ∠B = ∠D

(d) ∠B = ∠E

**Solution:**

**(c)**

**Question 20.**

**Solution:
**

**(a)**

**Question 21.**

∆ABC ~ ∆DEF, ar (∆ABC) = 9 cm², ar (∆DEF) = 16 cm². If BC = 2.1 cm, then the measure of EF is

(a) 2.8 cm

(b) 4.2 cm

(c) 2.5 cm

(d) 4.1 cm

**Solution:**

**(a)** ∆ABC ~ ∆DEF

ar (∆ABC) = 9 cm², ar (∆DEF) =16 cm²,

BC = 2.1 cm

∆ABC ~ ∆DEF

**Question 22.**

The length of the hypotenuse of an isosceles right triangle whose one side is 4√2 cm is

(a) 12 cm

(b) 8 cm

(c) 8√2 cm

(d) 12√2 cm

**Solution:**

**(b)** In isosceles right ∆ABC

∠B = 90°, AB = BC = 4√2

AC = √2

equal side = √2 x 4√2 = 8 cm

**Question 23.**

A man goes 24 m due west and then 7 m due north. How far is he from the starting point ?

(a) 31 m

(b) 17 m

(c) 25 m

(d) 26 m

**Solution:**

**(c)** In the figure, O is starting point OA = 24 m and AB = 7 m

By Pythagoras Theorem

OB² = OA² + AB²

= (24)² + (7)² = 576 + 49 = 625 = (25)²

OB = 25 m

**Question 24.**

∆ABC ~ ∆DEF. If BC = 3 cm, EF = 4 cm and ar (∆ABC) = 54 cm², then ar (∆DEF)

(a) 108 cm²

(b) 96 cm²

(c) 48 cm²

(d) 100 cm²

**Solution:**

**(b)**

**Question 25.**

∆ABC ~ ∆PQR such that ar (∆ABC) = 4 ar (∆PQR). If BC = 12 cm, then QR =

(a) 9 cm

(b) 10 cm

(c) 6 cm

(d) 8 cm

**Solution:**

**(c)** ∆ABC ~ ∆PQR

ar (∆ABC) = 4ar (∆PQR), BC = 12 cm

∆ABC ~ ∆PQR

**Question 26.**

The areas of two similar triangles are 121 cm² and 64 cm² respectively. If the median of the first triangle is 12.1 cm, then the corresponding median of the other triangles is

(a) 11 cm

(b) 8.8 cm

(c) 11.1 cm

(d) 8.1 cm

**Solution:**

**(b)** Areas of two similar triangles are 121 cm² and 64 cm²

Median of first triangle = 12.1 cm

In ∆ABC, area ∆DEF

AD and PS are their corresponding median

∆ABC ~ ∆DEF

**Question 27.**

In an equilateral triangle ABC if AD ⊥ BC, then AD² =

(a) CD²

(b) 2 CD²

(c) 3 CD²

(d) 4 CD²

**Solution:**

**(c)** In equilateral ∆ABC, AD ⊥ BC

**Question 28.**

In an equilateral triangle ABC if AD ⊥ BC, then

(a) 5 AB² = 4 AD²

(b) 3 AB² = 4 AD²

(c) 4 AB² = 3 AD²

(d) 2 AB² = 3 AD²

**Solution:**

**(b)** In equilateral ∆ABC, AD ⊥ BC

**Question 29.**

In an isosceles triangle ABC if AC = BC and AB² = 2 AC², then ∠C =

(a) 30°

(b) 45°

(c) 90°

(d) 60°

**Solution:**

**(c)** In isosceles ∆ABC, AC = BC

and AB2 = 2 AC² = AC² + AC²

= AC² + BC² ( AC = BC)

By converse of Pythagoras Theorem,

∠C = 90°

**Question 30.**

∆ABC is an isosceles triangle in which ∠C = 90°. If AC = 6 cm, then AB =

(a) 6√2 cm

(b) 6 cm

(c) 2√6 cm

(d) 4√2 cm

**Solution:**

**(a)** ∆ABC is an isosceles with ∠C= 90°

AC = BC

AC = 6 cm

AB² = AC² + BC² (Pythagoras Theorem)

(6)² + (6)² = 36 + 36 = 72 (AC = BC)

AB = √72 = √(36 x 2) = 6√2 cm

**Question 31.**

If in two triangles ABC and DEF, ∠A = ∠E, ∠B = ∠F, then which of the following is not true ?

(a) \(\frac { BC }{ DF }\) = \(\frac { AC }{ DE }\)

(b) \(\frac { AB }{ DE }\) = \(\frac { BC }{ DF }\)

(c) \(\frac { AB }{ EF }\) = \(\frac { AC }{ DE }\)

(d) \(\frac { BC }{ DF }\) = \(\frac { AB }{ EF }\)

**Solution:**

**(b)** In two triangles ABC and DEF

∠A = ∠E, ∠B = ∠F

**Question 32.**

In the figure, the measures of ∠D and ∠F are respectively

(a) 50°, 40°

(b) 20°, 30°

(c) 40°, 50°

(d) 30°, 20°

**Solution:**

**(b)** In ∆ABC

∠A = 180° – (∠B + ∠C)

**Question 33.**

In the figure, the value of x for which DE || AB is

(a) 4

(b) 1

(c) 3

(d) 2

**Solution:**

**(b)**

**Question 34.**

In the figure, if ∠ADE = ∠ABC, then CE =

(a) 2

(b) 5

(c) \(\frac { 9 }{ 2 }\)

(c) 3

**Solution:**

**(c)** In the figure ∠ADE = ∠ABC

AB = 2, DB = 3, AE = 3

Let EC = x

∠ADE = ∠ABC

But these are corresponding angles DE || BC

∆ADE ~ ∆ABC

**Question 35.**

In the figure, RS || DB || PQ. If CP = PD = 11 cm and DR = RA = 3 cm. Then the values of x and y are respectively

(a) 12, 10

(b) 14, 6

(c) 10, 7

(d) 16, 8

**Solution:**

**(d)** In the figure RS || DB || PQ

CP = PD = 11 cm DR = RA = 3 cm

In ∆ABD

RS || BD and AR = RD

RS = \(\frac { 1 }{ 2 }\) BD

y = \(\frac { 1 }{ 2 }\) x or x = 2y

Only 16, 8 is possible

**Question 36.**

In the figure, if PB || CF and DP || EF, then \(\frac { AD }{ DE }\) =

(a) \(\frac { 3 }{ 4 }\)

(b) \(\frac { 1 }{ 3 }\)

(c) \(\frac { 1 }{ 4 }\)

(d) \(\frac { 2 }{ 3 }\)

**Solution:**

**(b)** In the figure, PB || CF, DP || EF

AB = 2 cm, AC = 8 cm

BC = AC – AB = 8 – 2 = 6 cm

In ∆ACF, BP || CF

\(\frac { AD }{ DE }\) = \(\frac { 1 }{ 3 }\)

**Question 37.**

A chord of a circle of radius 10 cm subtends a right angle at the centre. The length of the chord (in cm) is

(a) 5√2

(b) 10√2

(c) \(\frac { 5 }{ \surd 2 }\)

(d) 10√3 **[ICSE 2014]**

**Solution:**

**(b)**

AB² = OA² + OB² (Pythagoras Theorem)

AB² = 10² + 10²

AB² = 2 (10)²

AB = 10√2

**Question 38.**

A vertical stick 20 m long casts a shadow 10 m long on the ground. At the same time, a tower casts a shadow 50 m long on the ground. The height of the tower is

(a) 100 m

(b) 120 m

(c) 25 m

(d) 200 m

**Solution:**

**(a)** Height of a stick = 20 m

and length of its shadow = 10 m

At the same time

Let height of tower = x m

and its shadow = 50 m

20 : x = 10 : 50

x x 10 = 20 x 50

=> x = \(\frac { 20 x 50 }{ 10 }\) = 100

Height of tower = 100 m

**Question 39.**

Two isosceles triangles have equal angles and their areas are in the ratio 16 : 25. The ratio of their corresponding heights is :

(a) 4 : 5

(b) 5 : 4

(c) 3 : 2

(d) 5 : 7

**Solution:**

**(a)** The corresponding angles of two isosceles triangles are equal These are similar Ratio in their areas = 16 : 25

The ratio of areas of similar triangles are proportion to the squares of their corresponding altitudes (heights)

Ratio in their altitudes = \(\surd \frac { 16 }{ 25 } =\frac { 4 }{ 5 }\)

= i.e., 4 : 5

**Question 40.**

∆ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. If ∆DEF ~ ∆ABC and EF = 4 cm, then perimeter of ∆DEF is

(a) 7.5 cm

(b) 15 cm

(c) 22.5 cm

(d) 30 cm

**Solution:**

**(b)** ∆DEF ~ ∆ABC

AB = 3 cm, BC = 2 cm, CA = 2.5 cm, EF = 4 cm

∆s are similar

**Question 41.**

In ∆ABC, a line XY parallel to BC cuts AB at X and AC at Y. If BY bisects ∠XYC, then :

(a) BC = CY

(b) BC = BY

(c) BC ≠ CY

(d) BC ≠ BY

**Solution:**

**(a)** In ∆ABC, XY || BC

BY is the bisector of ∠XYC

∠XYB = ∠CXB ….(i)

XY || BC

∠XYB = ∠XBC (Alternate angles) ……….(ii)

From (i) and (ii)

∠CYB = ∠YBC

BC = CY

**Question 42.**

In a ∆ABC, ∠A = 90°, AB = 5 cm and AC = 12 cm. If AD ⊥ BC, then AD =

(a) \(\frac { 13 }{ 2 }\) cm

(b) \(\frac { 60 }{ 13 }\) cm

(c) \(\frac { 13 }{ 60 }\) cm

(d) \(\frac { 2\surd 15 }{ 13 }\) cm

**Solution:**

**(b)** In ∆ABC,

∠A = 90°, AB = 5 cm, AC = 12 cm

**Question 43.**

In a ∆ABC, perpendicular AD from A on BC meets BC at D. If BD = 8 cm, DC = 2 cm and AD = 4 cm, then

(a) ∆ABC is isosceles

(b) ∆ABC is equilateral

(c) AC = 2 AB

(d) ∆ABC is right-angled at A

**Solution:**

**(d)** In ∆ABC, AD ⊥ BC

BD = 8 cm, DC = 2 cm, AD = 4 cm

In right ∆ACD,

AC² = AD² + CD² (Pythagoras Theorem)

= (4)² + (2)² = 16 + 4 = 20

and in right ∆ABD,

AB² = AD² + DB²

= (4)² + (8)2 = 16 + 64 = 80

and BC² = (BD + DC)² = (8 + 2 )² = (10)² = 100

AB² + AC² = 80 + 20 = 100 = BC²

∆ABC is a right triangle whose ∠A = 90°

**Question 44.**

In a ∆ABC, point D is on side AB and point G is on side AC, such that BCED is a trapezium. If DE : BC = 3:5, then Area (∆ADE) : Area (BCED) =

(a) 3 : 4

(b) 9 : 16

(c) 3 : 5

(d) 9 : 25

**Solution:**

**(b)**

In ∆ABC, D and E are points on the side AB and AC respectively, such that BCED is a trapezium DE : BC = 3 : 5

In ∆ABC, DE || BC

∆ADE ~ ∆ABC

**Question 45.**

If ABC is an isosceles triangle and D is a point on BC such that AD ⊥ BC, then

(a) AB² – AD² = BD.DC

(b) AB² – AD² = BD² – DC²

(c) AB² + AD² = BD.DC

(d) AB² + AD² = BD² – DC²

**Solution:**

**(a)** If ∆ABC, AB = AC

D is a point on BC such that

AD ⊥ BC

AD bisects BC at D

In right ∆ABD,

AB² = AD² + BD²

AB² – AD² = BD² = BD x BD = BD x DC (BD = DC)

**Question 46.**

∆ABC is a right triangle right-angled at A and AD ⊥ BC. Then , \(\frac { BD }{ DC }\) =

(a) \(\left( \frac { AB }{ AC } \right) ^{ 2 }\)

(b) \(\frac { AB }{ AC }\)

(c) \(\left( \frac { AB }{ AD } \right) ^{ 2 }\)

(d) \(\frac { AB }{ AD }\)

**Solution:**

**(a)** In right angled ∆ABC, ∠A = 90°

AD ⊥ BC

**Question 47.**

If E is a point on side CA of an equilateral triangle ABC such that BE ⊥ CA, then AB² + BC² + CA² =

(a) 2 BE²

(b) 3 BE²

(c) 4 BE²

(d) 6 BE²

**Solution:**

**(c)** ∆ABC is an equilateral triangle

BE ⊥ AC

**Question 48.**

In a right triangle ABC right-angled at B, if P and Q are points on the sides AB and AC respectively, then

(a) AQ² + CP² = 2 (AC² + PQ²)

(b) 2 (AQ² + CP²) = AC² + PQ²

(c) AQ² + CP² = AC² + PQ²

(d) AQ + CP = \(\frac { 1 }{ 2 }\) (AC + PQ)

**Solution:**

**(c)** In right ∆ABC, ∠B = 90°

P and Q are points on AB and BC respectively

AQ, CP and PQ are joined

In right ∆ABC,

AC² = AB² + BC² ….(i)

(Pythagoras Theorem)

Similarly in right ∆PBQ,

PQ² = PB² + BQ² ………(ii)

In right ∆ABQ

AQ² = AB² + BQ² ….(iii)

and in right ∆CPB,

CP² = PB² + BC² ….(iv)

Adding (iii) and (iv)

AQ² + CP² = AB² + BQ² + PB² + BC²

= AB² + BC² + BQ² + PB²

= AC² + PQ2 {From (i) and (ii)}

**Question 49.**

If ∆ABC ~ ∆DEF such that DE = 3 cm, EF = 2 cm, DF = 2.5 cm, BC = 4 cm, then perimeter of ∆ABC is

(a) 18 cm

(b) 20 cm

(c) 12 cm

(d) 15 cm

**Solution:**

**(d)** ∆ABC ~ ∆DEF

DE = 3 cm, EF = 2 cm, DF = 2.5 cm, BC = 4 cm

**Question 50.**

If ∆ABC ~ ∆DEF such that AB = 9.1 cm and DE = 6.5 cm. If the perimeter of ∆DEF is 25 cm, then the perimeter of ∆ABC is

(a) 36 cm

(b) 30 cm

(c) 34 cm

(d) 35 cm

**Solution:**

**(d)** ∆ABC ~ ∆DEF

AB = 9.1 cm and DE = 6.5 cm

Perimeter of ∆DEF = 25 cm

**Question 51.**

In an isosceles triangle ABC, if AB = AC = 25 cm and BC = 14 cm, then the measure of altitude from A on BC is

(a) 20 cm

(b) 22 cm

(c) 18 cm

(d) 24 cm

**Solution:**

**(d)** ∆ABC is an isosceles triangle in which AB = AC = 25 cm, BC = 14 cm

From A, draw AD ⊥ BC

D is mid-point of BC

BD = \(\frac { 1 }{ 2 }\) BC = \(\frac { 1 }{ 2 }\) x 14 = 7 cm

Now in right ∆ABD

AD² = AB² – BD²

= (25)² – (7)² = 625 – 49 = 576 = (24)²

AD = 24 cm

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles MCQS are helpful to complete your math homework.

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