Class 10 Maths MCQs on Chapter 6- Triangles are prepared to help students recognise the important topics and concepts for the objective type questions. In CBSE Class 10 Maths Exam 2020, there will be increased number of objective type questions as compared to the previous years. So, students must practice with the important questions provided here to prepare well for the exam and increase their chances of scoring high marks. All the MCQ questions provided here are thoroughly solved so that students can easily understand the appropriate logic behind each correct answer.

**Check below the solved MCQs from Class 10 Maths Chapter 6Triangles:**

**1.** O is the point of intersection of two equal chords ABand CD such that OB = OD, then triangles OAC and ODB are

(A) Equilateral but not similar

(B) Isosceles but not similar

(C) Equilateral and similar

(D) Isosceles and similar

**Answer:**** (D)**

**Explanation:**

Since O is the point of intersection of two equal chords AB and CD such that OB = OD,

As chords are equal and OB = OD, so AO will also be equal to OC

Also ∠AOC = ∠DOB = 45

Now in triangles OAC and ODB

AO/OB = CO/OD

And ∠AOC = ∠DOB = 45

So triangles are isosceles and similar

**2. **D and E are respectively the midpoints on the sides AB and AC of a triangle ABC and BC = 6 cm. If DE || BC, then the length of DE (in cm) is

(A) 2.5

(B) 3

(C) 5

(D) 6

**Answer:**** B**

**Explanation:**

By midpoint theorem,

If D and E are respectively the midpoints on the sides AB and AC of a triangle ABC, DE||BC and BC = 6 cm

So, DE will be half of BC i.e. 3cm

**Also Check: ****CBSE Class 10 Science Chapter-wise Important MCQs with Answers**

**3. **In triangle PQR, if PQ = 6 cm, PR = 8 cm, QS = 3 cm, and PS is the bisector of angle QPR, what is the length of SR?

(A) 2

(B) 4

(C) 6

(D) 8

**Answer:**** (B)**

**Explanation:**

Since, PS is the angle bisector of angle QPR

So, by angle bisector theorem,

QS/SR = PQ/PR

⇒ 3/SR = 6/8

⇒ SR = (3 X 8)/6 cm = 4 cm

**4.** The lengths of the diagonals of a rhombus are 16 cm and 12cm. Then, thelength of the side of the rhombus is

(A) 9 cm

(B) 10 cm

(C) 8 cm

(D) 20 cm

**Answer:****(B)**

**Explanation:**

The diagonals of rhombus bisect each other at right angle, so side of rhombus is the hypotenuse for the triangles formed.

Therefore,

By Pythagoras theorem

(16/2)^{2} + (12/2)^{2} = Side^{2}

⇒ 8^{2} + 6^{2} = Side^{2}

⇒ 64 + 36 = Side^{2}

⇒ Side = 10 cm

**5. **A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.

(A) 25.6

(B) 20.4

(C) 23.7

(D) 32.5

**Answer:**(B)

**Explanation:**

According to given question

The far end of shadow is represented by point A,

Therefore we need to Find AC

By Pythagoras theorem,

(18)^{2} + (9.6)^{2} = (AC)^{2}

⇒ AC^{2} = 416.16

⇒ AC = 20.4 m (approx)

**6. **Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3 RS, Then the ratio of areas of triangles POQ and ROS is:

(A) 1:9

(B) 9:1

(C) 3:1

(D) 1:3

**Answer:(B)**

**Explanation:**According to given Question

Since

SR || PQ,

So, ∠OSR= ∠OQP (alternate interior angles)

Also ∠SOR= ∠POQ (vertically opposite angles)

So triangles SOR and POQ are similar,

Therefore,

ar(POQ)/ar(SOR) = (PQ/SR)^{2}

ar(POQ)/ar(SOR) = (3 SR/SR)^{2}

ar(POQ)/ar(SOR) = 9/1

**7.** ABCD is a trapezium in which AB|| DC and P, Q are points on ADand BC respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm andQC = 15 cm, find AD.

(A) 55cm

(B) 57cm

(C) 60cm

(D) 62cm

**Answer:(C)**

**Explanation:**

According to question

ABCD is a trapezium in which AB || DC and P and Q are points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm,

In triangle ABD

DP/AP = OD/OB

In triangle BDC

BQ/QC = OB/OD

This implies

DP/AP = QC/BQ

18/AP = 15/35

AP = (18 x 35)/15

AP = 42

Therefore, AD = AP + DP = 42 + 18 = 60cm

**8.** Areas of two similar triangles are 36 cm^{2 }and 100 cm^{2}. If the length of a side of the larger triangle is 20 cm, then the length of the corresponding side of the smaller triangle is:

(A) 12cm

(B) 13cm

(C) 14cm

(D) 15cm

**Answer:****(A)**

**Explanation:**

Let the side of smaller triangle be x cm.

ar(Larger Triangle)/ar(Smaller Triangle) = (side of larger triangle/side of smaller triangle)^{2}

100/36 = (20/x)^{2}

x = √144

X = 12 cm

**9.** In the figure if ∠ACB = ∠CDA, AC = 8 cm and AD = 3 cm, find BD.

(A) 53/3 cm

(B) 55/3 cm

(C) 64/3 cm

(D) 35/7 cm

**Answer:(B)**

**Explanation:**

In triangle ACB and ADC

∠A=∠A

∠ACB = ∠CDA

Therefore triangle ACB and ADC are similar,

Hence

AC/AD = AB/AC

AC^{2} = AD X AB

8^{2} = 3 x AB

⇒ AB = 64/3

This implies,

BD = 64/3 – AD

⇒ BD = 55/3

**10.** If ABCD is parallelogram, P is a point on side BC and DP when produced meets AB produced at L, then select the correct option

(A) DP/BL = DC/PL

(B) DP/PL = DC/BL

(C) DP/PL = BL/DC

(D) DP/PL = AB/DC

**Answer: (B)**

**Explanation:**

In ΔALD, we have

BP || AD

∴ LB/BA = LP/PD

⇒ BL/AB = PL/DP

⇒ BL/DC = PL/DP [∵ AB = DC

⇒ DP/PL = DC/BL

**11**. In the figure given below DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, the value of x is:

(A) 4

(B) 8

(C) 16

(D) 32

**Answer: (A)**

**Explanation:**

In triangle ABC, we have DE || BC

∴ AD/DB = AE/EC (By Thale’s Theorem)

⇒ x/x – 2 = (x + 2)/(x – 1)

⇒ x (x – 1) = (x – 2)(x + 2)

⇒ x^{2} – x = x^{2} – 4

⇒ x = 4

**12.** The length of altitude of an equilateral triangle of side 8cm is

(A) √3 cm

(B) 2√3 cm

(C) 3√3 cm

(D) 4√3 cm

**Answer:(D)**

**Explanation:**

The altitude divides the opposite side into two equal parts,

Therefore, BD = DC = 4 cm

In triangle ABD

AB^{2} = AD^{2} + BD^{2}

8^{2} = AD^{2} + 4^{2}

AD^{2} = 64 – 16

AD^{2} = 48

AD = 4√3 cm

**13.** If ΔABC ~ ΔDEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, find the perimeter of ABC.

(A) 18 cm

(B) 20 cm

(C) 21 cm

(D) 22 cm

**Answer:(A)**

**Explanation:**

According to question,

ΔABC ~ ΔDEF,

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

Therefore,

AB/DE = BC/EF = AC/DF

4/6 = BC/9 = AC/12

⇒ 4/6 = BC/9

⇒ BC = 6 cm

And

4/6 = AC/12

⇒ AC = 8 cm

Perimeter = AB + BC + CA

= 4 + 6 + 8

= 18 cm

**14.** A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then the distance by which the top of the ladder would slide upwards on the wall is:

(A) 2 m

(B) 1.2 m

(C) 0.8 m

(D) 0.5 m

**Answer:(C)**

**Explanation:**

Let AC be the ladder of length 5m and BC = 4m be the height of the wall where ladder is placed. If the foot of the ladder is moved 1.6m towards the wall i.e. AD = 1.6 m, then the ladder is slided upward to position E i.e. CE = x m.

In right triangle ABC

AC^{2} = AB^{2} + BC^{2}

⇒5^{2} = AB^{2} + 4^{2}

⇒ AB = 3m

⇒ DB = AB – AD = 3 – 1.6 = 1.4m

In right angled ΔEBD

ED^{2} = EB^{2} + BD^{2}

⇒ 5^{2} = EB^{2} + (1.4)^{2}

⇒ EB = 4.8m

EC = EB – BC = 4.8 – 4 = 0.8m

Hence the top of the ladder would slide upwards on the wall at distance 0.8 m.

**15.** Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm^{2}, then the area of the larger triangle is:

(A) 108 m^{2}

(B) 107 m^{2}

(C) 106 m^{2}

(D) 230 m^{2}

**Answer:(A)**

**Explanation:**

According to given Question

ar(Larger Triangle)/ar(Smaller Triangle) = (side of larger triangle/side of larger triangle)^{2}

ar(Larger Triangle)/48 = (3/2)^{2}

ar(Larger Triangle) = (9 x 48 )/4

ar(Larger Triangle) = 108 cm^{2}

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