MCQs on Chapter 6: Triangles for Class 10 Maths

Triangles Class 10 MCQs are available in this Maths article. These carefully prepared multiple-choice questions help students revise Chapter 6 of the CBSE Class 10 Maths syllabus in an easy and exam-oriented way. The MCQs with answers and detailed solutions cover similarity of triangles, criteria for similarity, Pythagoras theorem, area-based properties, and important theorems related to triangles. This article is useful for strengthening conceptual understanding, improving problem-solving speed, and preparing effectively for board exams.

MCQS on Chapter 6: Triangles for Class 10 With Answers

Question 1: In △ABC, DE ∥ BC, AD = 4 cm, DB = 6 cm and AE = 3 cm. Find EC.

(a) 4 cm

(b) 4.5 cm

(c) 5 cm

(d) 6 cm

Answer: (b) 4.5 cm

Explanation: By Basic Proportionality Theorem (since DE ∥ BC):

AD/DB = AE/EC

4/6 = 3/EC

EC = (3 × 6)/4 = 18/4 = 4.5 cm

Question 2: D and E are the midpoints of sides AB and AC of △ABC respectively. If BC = 8 cm and DE ∥ BC, the length of DE is:

(a) 4 cm

(b) 7 cm

(c) 6 cm

(d) 2 cm

Answer:  (a) 4 cm

Explanation: By the Midpoint Theorem (a special case of BPT), the line joining midpoints of two sides is parallel to the third side and half its length.

DE = ½ × BC = ½ × 8 = 4 cm

DE = 4 cm.

 Question 3: In the trapezium ABCD, AB ∥ CD. The diagonals AC and BD intersect at O. If OA = 3x − 1 and OC = 5x − 3 and OB = 5x + 1 and OD = 7x − 1, find x.

(a) x = 1

(b) x = 2

(c) x = 3

(d) x = 4

Answer: (a) x = 1 

Explanation: In a trapezium with AB ∥ CD, the diagonals divide each other in the same ratio: OA/OC = OB/OD

(3x−1)/(5x−3) = (5x+1)/(7x−1)

Cross-multiply: (3x−1)(7x−1) = (5x+1)(5x−3)

⇒ 21x² − 10x + 1 = 25x² − 10x − 3

0 = 4x² − 4 ⇒  x² = 1 ⇒  x = 1 (taking positive value)

Question 4: △ABC ∼ △DEF. If AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, find the perimeter of △ABC.

(a) 20 cm

(b) 16 cm

(c) 18 cm

(d) 22 cm

Answer: (c) 18 cm

Explanation: Since △ABC ∼ △DEF, the ratio of corresponding sides is constant:

AB/DE = 4/6 = 2/3

BC/EF = BC/9 ⇒ BC = 9 × (2/3) = 6 cm

AC/FD = AC/12 ⇒ AC = 12 × (2/3) = 8 cm

Perimeter of △ABC = 4 + 6 + 8 = 18 cm.

Question 5: Two triangles have sides 5 cm, 12 cm, 13 cm and 10 cm, 24 cm, 26 cm. The triangles are:

(a) Congruent

(b) Similar but not congruent

(c) Neither similar nor congruent

(d) Congruent and similar

Answer:  (b) Similar but not congruent

Explanation: Check ratios of corresponding sides:

5/10 = 1/2, 12/24 = 1/2, 13/26 = 1/2

All ratios are equal. Therefore, SSS similarity applies. Hence, the triangles are Similar.

But since the sides are different in length, the triangles are not congruent.

Question 6: Corresponding sides of two similar triangles are in the ratio 2:3. If the area of the smaller triangle is 48 cm², the area of the larger triangle is:

(a) 96 cm²

(b) 72 cm²

(c) 108 cm²

(d) 120 cm²

Answer: (c) 108 cm²

Explanation: Area ratio = (2/3)² = 4/9

48 /  Area2 = 4/9

  Area2 = 48 × 9/4 = 48 × 2.25 = 108 cm²

Area of larger triangle = 108 cm².

Question 7: Two isosceles triangles have equal vertical angles. Their areas are in the ratio 16:25. Find the ratio of their corresponding altitudes.

(a) 4:5

(b) 5:4

(c) 16:25

(d) 2:5

Answer: (a) 4:5 

Explanation: Equal vertical angles and isosceles ⇒ the base angles are also equal ⇒ triangles are similar by AA.

Area ratio = (altitude ratio)²

h₁²/h₂² = 16/25 ⇒ h₁/h₂ = 4/5

Altitude ratio = 4:5.

Question 8: The area of an equilateral triangle with side length a is:

(a) (√3/2)a²

(b) (√3/4)a²

(c) (√3/4)a

(d) (1/2)a²

Answer: (b) (√3/4)a²

Explanation: Height of equilateral triangle with side a is (√3/2)a.

Area = ½ × base × height = ½ × a × (√3/2)a = (√3/4)a²

Question 9: In △ABC, ∠BAC = 90° and AD ⊥ BC. If BD = 4 cm and DC = 9 cm, find AD.

(a) 5 cm

(b) 7 cm

(c) 6 cm

(d) 8 cm

Answer:  (c) 6 cm

Explanation: When the altitude is drawn from the right angle to the hypotenuse in a right triangle:

AD² = BD × DC

AD² = 4 × 9 = 36

AD = 6 cm

Question 10: In △ABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B equals:

(a) 30°

(b) 45°

(c) 60°

(d) 90°

Answer: (d) 90°

Explanation: Check: BC² + AB² = 6² + (6√3)² = 36 + 108 = 144 = 12² = AC²

So ∠B = 90°. 

Question 11: In △ABC, medians AD and BE intersect at G. The ratio AG:GD is:

(a) 2:1

(b) 1:2

(c) 1:1

(d) 3:1

Answer: (a) 2:1

Explanation: The centroid G divides each median in the ratio 2:1 from the vertex to the midpoint of the opposite side.

So AG: GD = 2:1 and BG: GE = 2:1.

Question 12: In △ABC, AB = 3 cm and AC = 4 cm. AD is the bisector of ∠A. Then BD:DC equals:

(a) 3:4

(b) 4:3

(c) 9:16

(d) 1:1

Answer: (a) 3:4 

Explanation: By the Angle Bisector Theorem, the bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides.

BD/DC = AB/AC = 3/4

BD:DC = 3:4.

Question 13: △ABC and △BDE are two equilateral triangles such that D is the midpoint of BC. Find the ratio of the areas of triangles ABC and BDE.

(a) 1:2

(b) 2:1

(c) 4:1

(d) 1:4

Answer: (c) 4:1

Explanation: Both triangles are equilateral; therefore, they are similar.

D is the midpoint of BC, so BD = BC/2.

In equilateral △BDE, the side is BD = BC/2.

Ratio of sides = BC/(BC/2) = 2/1.

Area(△ABC)/Area(△BDE) = (BC/BD)² = (2/1)² = 4/1

Area(△ABC):Area(△BDE) = 4:1.

Question 14: The perimeters of two similar triangles △ABC and △PQR are 60 cm and 36 cm respectively. If PQ = 9 cm, find AB.

(a) 12 cm

(b) 15 cm

(c) 18 cm

(d) 20 cm

Answer: (b) 15 cm

Explanation: For similar triangles, the ratio of perimeters equals the ratio of corresponding sides.

Perimeter(△ABC)/Perimeter(△PQR) = AB/PQ

60/36 = AB/9

AB = 9 × 60/36 = 9 × 5/3 = 15 cm

AB = 15 cm.

Question 15: The hypotenuse of a right triangle is 6 m more than twice the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle.

(a) 10, 22, 24

(b) 10, 24, 26

(c) 8, 20, 22

(d) 6, 18, 20

Answer: (b) 10, 24, 26

Explanation: Let the shortest side = x. Hypotenuse = 2x + 6. Third side = (2x + 6) − 2 = 2x + 4.

By Pythagoras: x² + (2x + 4)² = (2x + 6)²

x² + 4x² + 16x + 16 = 4x² + 24x + 36

x² − 8x − 20 = 0 ⇒ (x − 10)(x + 2) = 0 ⇒ x = 10

Sides: 10, 24, 26.

 

Click here to download the free PDF of MCQs worksheet on Chapter 6: Triangles for Class 10 Maths based on the updated NCERT & CBSE pattern with important multiple-choice questions and answers.

MCQs Worksheet on Chapter 6: Triangles for Class 10