CBSE Class 10 Mathematics Triangles MCQs Set 20

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 06 Triangles

Review structured MCQ sets for Class 10 Mathematics Chapter 06 Triangles. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Chapter 06 Triangles MCQs for Class 10 Mathematics

View or download the dedicated Chapter 06 Triangles MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: In triangle ABC, DE is parallel to BC with D on AB and E on AC. If AD = 4x - 3, AE = 8x - 7, BD = 3x - 1 and CE = 5x - 3, then x equals:
(a) 1
(b) -1/2
(c) 2
(d) 3
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. AD/BD = AE/CE gives (4x - 3)(5x - 3) = (8x - 7)(3x - 1), i.e. 2x^2 - x - 1 = 0, so x = 1 or x = -1/2. For x = -1/2 the length AD = -5 is negative, so x = 1.

Teacher's Note:
1. Basic proportionality (Thales) theorem: a line parallel to one side of a triangle that meets the other two sides divides them in the same ratio. Its converse also holds: if a line divides two sides in the same ratio it is parallel to the third side.

Question 2: D and E lie on sides AB and AC of triangle ABC with AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm. Then:
(a) DE is not parallel to BC
(b) DE is perpendicular to BC
(c) AD is the bisector of angle A
(d) DE is parallel to BC
Show Answer & Explanation

Answer: (d) DE is parallel to BC

Explanation:
1. BD = 4 cm and CE = 6 cm. AD/BD = 8/4 = 2 and AE/CE = 12/6 = 2. The ratios are equal, so by the converse of the basic proportionality theorem DE is parallel to BC.

Teacher's Note:
1. Basic proportionality (Thales) theorem: a line parallel to one side of a triangle that meets the other two sides divides them in the same ratio. Its converse also holds: if a line divides two sides in the same ratio it is parallel to the third side.

Question 3: The ratio of the area of an equilateral triangle on a side of a square to that of an equilateral triangle on its diagonal is:
(a) 1 : 4
(b) 1 : sqrt2
(c) 1 : 2
(d) 2 : 1
Show Answer & Explanation

Answer: (c) 1 : 2

Explanation:
1. The triangles are similar and the diagonal is sqrt2 times the side, so the areas are in the ratio 1 : 2.

Teacher's Note:
1. Areas of similar triangles are in the ratio of the squares of corresponding sides (also of corresponding altitudes or medians).

Question 4: In an equilateral triangle of side 2a, the altitude and the area are:
(a) \( \sqrt3 a\ \text{and}\ \frac{\sqrt3}{2}a^2 \)
(b) \( \sqrt3 a\ \text{and}\ \sqrt3 a^2 \)
(c) \( a\ \text{and}\ a^2 \)
(d) \( 2a\ \text{and}\ 2a^2 \)
Show Answer & Explanation

Answer: (b) \( \sqrt3 a\ \text{and}\ \sqrt3 a^2 \)

Explanation:
1. The altitude bisects the base, so AD^2 = (2a)^2 - a^2 = 3a^2 and AD = sqrt3 a. Area = (1/2)(2a)(sqrt3 a) = sqrt3 a^2.

Teacher's Note:
1. Pythagoras theorem: in a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides; the converse also holds. For a median m to side a: b^2 + c^2 = 2m^2 + a^2/2.

Question 5: BL and CM are medians of a triangle right-angled at A with AB = 6 cm and AC = 8 cm (L on AC, M on AB). The value of 4(BL^2 + CM^2) is:
(a) 400
(b) 500
(c) 600
(d) 250
Show Answer & Explanation

Answer: (b) 500

Explanation:
1. BL^2 = AB^2 + AL^2 = 36 + 16 = 52 and CM^2 = AC^2 + AM^2 = 64 + 9 = 73. So 4(52 + 73) = 500, which equals 5 BC^2 = 5(100).

Teacher's Note:
1. Pythagoras theorem: in a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides; the converse also holds. For a median m to side a: b^2 + c^2 = 2m^2 + a^2/2.

Question 6: In a right triangle with legs 6 cm and 8 cm, the length of the altitude drawn to the hypotenuse is:
(a) 4.8 cm
(b) 4 cm
(c) 5 cm
(d) 6 cm
Show Answer & Explanation

Answer: (a) 4.8 cm

Explanation:
1. The hypotenuse is 10 cm. Area = (1/2)(6)(8) = 24 = (1/2)(10)p, so p = 4.8 cm (that is, cp = ab).

Teacher's Note:
1. Pythagoras theorem: in a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides; the converse also holds. For a median m to side a: b^2 + c^2 = 2m^2 + a^2/2.

Question 7: In an equilateral triangle ABC of side 6 cm the point D trisects BC (BD = 2 cm). Then AD^2 equals:
(a) 27 cm^2
(b) 36 cm^2
(c) 30 cm^2
(d) 28 cm^2
Show Answer & Explanation

Answer: (d) 28 cm^2

Explanation:
1. The foot E of the altitude is the mid-point of BC, so BE = 3 cm and AE^2 = 36 - 9 = 27. ED = 3 - 2 = 1, so AD^2 = 27 + 1 = 28. This agrees with 9AD^2 = 7AB^2.

Teacher's Note:
1. Pythagoras theorem: in a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides; the converse also holds. For a median m to side a: b^2 + c^2 = 2m^2 + a^2/2.

Question 8: The perimeters of two similar triangles are 25 cm and 15 cm. If a side of the first triangle is 9 cm, the corresponding side of the other is:
(a) 6.2 cm
(b) 3.4 cm
(c) 5.4 cm
(d) 8.4 cm
Show Answer & Explanation

Answer: (c) 5.4 cm

Explanation:
1. Corresponding sides are in the ratio of the perimeters, 25 : 15 = 5 : 3. So the side is 9 x 3/5 = 5.4 cm.

Teacher's Note:
1. Areas of similar triangles are in the ratio of the squares of corresponding sides (also of corresponding altitudes or medians).

Question 9: In triangle ABC, angle B = 90 degrees and D is a point on BC with BD = 2 DC. Then:
(a) AC^2 = AD^2 + 3 CD^2
(b) AC^2 = AD^2 + 7 CD^2
(c) AC^2 = AD^2 + 5 CD^2
(d) AC^2 = AB^2 + 5 BD^2
Show Answer & Explanation

Answer: (c) AC^2 = AD^2 + 5 CD^2

Explanation:
1. Let DC = x so BD = 2x and BC = 3x. AC^2 = AB^2 + 9x^2 and AD^2 = AB^2 + 4x^2. Subtracting, AC^2 = AD^2 + 5x^2.

Teacher's Note:
1. Pythagoras theorem: in a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides; the converse also holds. For a median m to side a: b^2 + c^2 = 2m^2 + a^2/2.

Question 10: P and Q are the mid-points of AB and BC of a triangle ABC right-angled at B. Then AQ^2 + CP^2 equals:
(a) \( AC^2 \)
(b) \( \frac54 AC^2 \)
(c) \( \frac34 AC^2 \)
(d) \( \frac54 AB^2 \)
Show Answer & Explanation

Answer: (b) \( \frac54 AC^2 \)

Explanation:
1. Let AB = 2p and BC = 2q. AQ^2 = 4p^2 + q^2 and CP^2 = p^2 + 4q^2, so AQ^2 + CP^2 = 5(p^2 + q^2) = (5/4)(4p^2 + 4q^2) = (5/4)AC^2.

Teacher's Note:
1. Pythagoras theorem: in a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides; the converse also holds. For a median m to side a: b^2 + c^2 = 2m^2 + a^2/2.

Question 11: In triangle ABC, AD bisects angle A and meets BC at D. If AB = 10 cm, AC = 6 cm and BC = 12 cm, BD equals:
(a) 7.5 cm
(b) 3.3 cm
(c) 18 cm
(d) 4.5 cm
Show Answer & Explanation

Answer: (a) 7.5 cm

Explanation:
1. BD/DC = AB/AC = 10/6 = 5/3. So BD = 12 x 5/8 = 7.5 cm.

Teacher's Note:
1. Angle bisector theorem: the internal bisector of an angle of a triangle divides the opposite side in the ratio of the other two sides; the external bisector divides it externally in that ratio.

Question 12: A line parallel to BC meets AB and AC at D and E. If the area of triangle ADE is one-fifth of the area of triangle ABC and BC = 10 cm, then DE equals:
(a) \( 2\ \text{cm} \)
(b) \( 4\ \text{cm} \)
(c) \( 5\sqrt2\ \text{cm} \)
(d) \( 2\sqrt5\ \text{cm} \)
Show Answer & Explanation

Answer: (d) \( 2\sqrt5\ \text{cm} \)

Explanation:
1. The triangles are similar, so (DE/BC)^2 = 1/5 and DE = 10/sqrt5 = 2 sqrt5 cm.

Teacher's Note:
1. Areas of similar triangles are in the ratio of the squares of corresponding sides (also of corresponding altitudes or medians).

Question 13: The two sides containing the right angle of a right triangle are 6 cm and 8 cm. The radius of the inscribed circle is:
(a) 3 cm
(b) 4 cm
(c) 1.5 cm
(d) 2 cm
Show Answer & Explanation

Answer: (d) 2 cm

Explanation:
1. The hypotenuse is 10 cm. For a right triangle r = (a + b - c)/2 = (6 + 8 - 10)/2 = 2 cm (also area/semi-perimeter = 24/12 = 2).

Teacher's Note:
1. Pythagoras theorem: in a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides; the converse also holds. For a median m to side a: b^2 + c^2 = 2m^2 + a^2/2.

Question 14: In triangle ABC, DE is parallel to BC and AD : DB = 5 : 3. The ratio of the area of triangle ADE to the area of triangle ABC is:
(a) 5 : 8
(b) 25 : 9
(c) 25 : 64
(d) 5 : 3
Show Answer & Explanation

Answer: (c) 25 : 64

Explanation:
1. AD : AB = 5 : 8, so the triangles are similar with sides in the ratio 5 : 8 and areas in the ratio 25 : 64.

Teacher's Note:
1. Areas of similar triangles are in the ratio of the squares of corresponding sides (also of corresponding altitudes or medians).

Question 15: In a triangle ABC, angle C = 90 degrees, AB = 17 cm and BC = 8 cm. The external bisector of angle B meets AC produced at D. Then AC and BD are:
(a) \( 15\ \text{cm and}\ \frac{40}{3}\ \text{cm} \)
(b) \( 15\ \text{cm and}\ \frac{8\sqrt{34}}{3}\ \text{cm} \)
(c) \( 15\ \text{cm and}\ 17\ \text{cm} \)
(d) \( 16\ \text{cm and}\ \frac{8\sqrt{34}}{3}\ \text{cm} \)
Show Answer & Explanation

Answer: (b) \( 15\ \text{cm and}\ \frac{8\sqrt{34}}{3}\ \text{cm} \)

Explanation:
1. AC = sqrt(289 - 64) = 15 cm. The external bisector divides AC externally: AD/CD = AB/BC = 17/8, so with CD = c, (15 + c)/c = 17/8 and c = 40/3. Then BD^2 = 64 + 1600/9 = 2176/9, so BD = 8 sqrt34 / 3 cm.

Teacher's Note:
1. Angle bisector theorem: the internal bisector of an angle of a triangle divides the opposite side in the ratio of the other two sides; the external bisector divides it externally in that ratio.

Question 16: PQRS is a parallelogram with PQ = 16 cm and QR = 10 cm. L is a point on PR with RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N. Then PN and RM are:
(a) \( 15\ \text{cm and}\ \frac{32}{3}\ \text{cm} \)
(b) \( 10\ \text{cm and}\ \frac{32}{3}\ \text{cm} \)
(c) \( 15\ \text{cm and}\ 8\ \text{cm} \)
(d) \( \frac{20}{3}\ \text{cm and}\ \frac{32}{3}\ \text{cm} \)
Show Answer & Explanation

Answer: (a) \( 15\ \text{cm and}\ \frac{32}{3}\ \text{cm} \)

Explanation:
1. Triangles PLN and RLQ are similar (PN is parallel to QR): PN/QR = LP/RL = 3/2, so PN = 15 cm. Triangles RLM and PLQ are similar (RM is parallel to PQ): RM/PQ = RL/LP = 2/3, so RM = 32/3 = 10.67 cm.

Teacher's Note:
1. Areas of similar triangles are in the ratio of the squares of corresponding sides (also of corresponding altitudes or medians).

Question 17: In triangle ABC, DE is parallel to BC with AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm. Then BD and CE are:
(a) 3.6 cm and 4.8 cm
(b) 2.4 cm and 3.2 cm
(c) 3 cm and 4 cm
(d) 3.6 cm and 3.2 cm
Show Answer & Explanation

Answer: (a) 3.6 cm and 4.8 cm

Explanation:
1. AD/AB = DE/BC = 2/5, so AB = 6 cm and BD = 3.6 cm. Similarly AE/AC = 2/5 gives AC = 8 cm and CE = 4.8 cm.

Teacher's Note:
1. Areas of similar triangles are in the ratio of the squares of corresponding sides (also of corresponding altitudes or medians).

Question 18: In a right triangle with the right angle at C, CD is perpendicular to the hypotenuse AB, AD = 9 cm and BD = 4 cm. CD equals:
(a) 5 cm
(b) 6.5 cm
(c) 36 cm
(d) 6 cm
Show Answer & Explanation

Answer: (d) 6 cm

Explanation:
1. The triangles ACD and CBD are similar, so CD^2 = AD x BD = 36 and CD = 6 cm.

Teacher's Note:
1. Pythagoras theorem: in a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides; the converse also holds. For a median m to side a: b^2 + c^2 = 2m^2 + a^2/2.

Question 19: In a triangle PQR right-angled at Q, S is the mid-point of QR. Then PR^2 equals:
(a) 4 PS^2 - PQ^2
(b) 3 PS^2 - 4 PQ^2
(c) 4 PS^2 - 3 PQ^2
(d) 4 PS^2 + 3 PQ^2
Show Answer & Explanation

Answer: (c) 4 PS^2 - 3 PQ^2

Explanation:
1. Let QS = SR = s. PS^2 = PQ^2 + s^2 and PR^2 = PQ^2 + 4s^2. So 4PS^2 - 3PQ^2 = 4PQ^2 + 4s^2 - 3PQ^2 = PQ^2 + 4s^2 = PR^2. Check: PQ = 6, QR = 8 gives 4(52) - 108 = 100.

Teacher's Note:
1. Pythagoras theorem: in a right triangle the square of the hypotenuse equals the sum of the squares of the other two sides; the converse also holds. For a median m to side a: b^2 + c^2 = 2m^2 + a^2/2.

Question 20: In triangle ABC, angle B = 135 degrees, AB = 2 sqrt2 cm and BC = 3 cm. Then AC^2 equals:
(a) 17 cm^2
(b) 29 cm^2
(c) 23 cm^2
(d) 41 cm^2
Show Answer & Explanation

Answer: (b) 29 cm^2

Explanation:
1. By the cosine-type result for an obtuse angle, AC^2 = AB^2 + BC^2 + 4 ar(ABC). The area is (1/2)(2 sqrt2)(3) sin 135 = 3, so AC^2 = 8 + 9 + 12 = 29.

Teacher's Note:
1. Vector or coordinate check: place B at the origin, A at (2 sqrt2 cos 135, 2 sqrt2 sin 135) = (-2, 2) and C at (3, 0); AC^2 = 25 + 4 = 29.

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Chapter 06 Triangles Objective Questions & Solutions for Class 10 Mathematics

About Chapter 06 Triangles MCQs for Class 10 Mathematics

Test your conceptual understanding of Chapter 06 Triangles with these targeted multiple-choice questions. Designed in alignment with the latest CBSE curriculum for Class 10 Mathematics, these problem sets build accuracy and prepare students for objective exams.

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FAQs

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