CBSE Class 10 Mathematics Triangles MCQs Set 18

Mathematics Objective Questions and Answers: Chapter 06 Triangles

Access targeted multiple-choice questions for Chapter 06 Triangles designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Download Chapter 06 Triangles MCQs with Answers

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 ∥ BC with D on AB and E on AC. If AD = x, DB = x − 2, AE = x + 2 and EC = x − 1, then x equals:
(a) 2
(b) 4
(c) 3
(d) 5
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. x/(x − 2) = (x + 2)/(x − 1).
2. x(x − 1) = (x + 2)(x − 2), so x² − x = x² − 4.
3. x = 4.

Teacher's Note:
1. Use AD/DB = AE/EC.
2. The x² terms cancel.

Question 2: In triangle ABC, AD bisects angle A and meets BC at D. If AC = 4.2 cm, DC = 6 cm and BC = 10 cm, then AB equals:
(a) 2.8 cm
(b) 3.5 cm
(c) 4.8 cm
(d) 7 cm
Show Answer & Explanation

Answer: (a) 2.8 cm

Explanation:
1. BD = 10 − 6 = 4 cm. By the angle bisector theorem AB/AC = BD/DC, so AB = 4.2 × 4/6 = 2.8 cm.

Teacher's Note:
1. Find BD first.
2. AB/AC = BD/DC.

Question 3: In triangle ABC the bisector AD of angle A meets BC at D. If AB = 6 cm and AC = 9 cm, then Area(ABD) : Area(ACD) is:
(a) 3 : 2
(b) 4 : 9
(c) 1 : 1
(d) 2 : 3
Show Answer & Explanation

Answer: (d) 2 : 3

Explanation:
1. Triangles ABD and ACD have the same height from A, so their areas are in the ratio of the bases BD : DC. By the angle bisector theorem BD : DC = AB : AC = 6 : 9 = 2 : 3.

Teacher's Note:
1. Same height means areas are in the ratio of the bases.
2. Apply the angle bisector theorem.

Question 4: In triangle ABC, D is the midpoint of BC, DE bisects angle ADB (E on AB) and EF ∥ BC meets AC at F. The angle EDF is:
(a) 60°
(b) 45°
(c) 90°
(d) 120°
Show Answer & Explanation

Answer: (c) 90°

Explanation:
1. DE bisects ∠ADB and, by the angle bisector theorem in triangle ADC with DF bisecting ∠ADC (since EF ∥ BC gives AE/EB = AF/FC and AD/DB = AD/DC), DF bisects ∠ADC. The two bisectors of the supplementary angles ADB and ADC are perpendicular, so ∠EDF = 90°.

Teacher's Note:
1. The angles ADB and ADC are supplementary.
2. The bisectors of adjacent supplementary angles are perpendicular.

Question 5: In triangle ABC, AD is the internal bisector of angle A, AB = 10 cm, AC = 14 cm and BC = 6 cm. Then BD and DC are:
(a) 3 cm and 3 cm
(b) 2.5 cm and 3.5 cm
(c) 2 cm and 4 cm
(d) 3.5 cm and 2.5 cm
Show Answer & Explanation

Answer: (b) 2.5 cm and 3.5 cm

Explanation:
1. BD : DC = AB : AC = 10 : 14 = 5 : 7. With BC = 6, BD = 6 × 5/12 = 2.5 cm and DC = 3.5 cm.

Teacher's Note:
1. Divide BC in the ratio AB : AC.
2. Check BD + DC = 6.

Question 6: In trapezium ABCD, AB ∥ DC, DC = 2AB and a line EF parallel to AB meets AD at F and BC at E with BE/EC = 3/4. If AB = 14 cm, then EF is:
(a) 20 cm
(b) 21 cm
(c) 17.5 cm
(d) 24 cm
Show Answer & Explanation

Answer: (a) 20 cm

Explanation:
1. EF is a weighted mean: EF = AB + (BE/BC)(DC − AB) = 14 + (3/7)(14).
2. EF = 14 + 6 = 20 cm.

Teacher's Note:
1. Use the weighted-average formula for a line parallel to the bases.
2. BE : EC = 3 : 4, so BC is 7 parts.

Question 7: In triangles ABC and DBC on the same base BC, AD meets BC at O. If AO = 6 cm and DO = 4 cm, then area(ABC) : area(DBC) is:
(a) 2 : 3
(b) 9 : 4
(c) 6 : 4 × 2
(d) 3 : 2
Show Answer & Explanation

Answer: (d) 3 : 2

Explanation:
1. The triangles have the same base BC, so their areas are in the ratio of the heights from A and D, which equals AO : DO = 6 : 4 = 3 : 2.

Teacher's Note:
1. Same base means area ratio = ratio of heights.
2. The heights are proportional to AO and DO.

Question 8: In triangle ABC with AB = AC = 5 cm, BD ⟂ AC with AD = 3 cm. The value of BD² − CD² is:
(a) 16 cm²
(b) 9 cm²
(c) 12 cm²
(d) 25 cm²
Show Answer & Explanation

Answer: (c) 12 cm²

Explanation:
1. CD = 5 − 3 = 2 and BD² = AB² − AD² = 25 − 9 = 16. So BD² − CD² = 16 − 4 = 12 = 2 × AD × CD.

Teacher's Note:
1. Find CD and BD² separately.
2. Compare with 2·AD·CD.

Question 9: PB and QA are perpendicular to segment AB, and PQ meets AB at O. If PO = 5 cm, QO = 7 cm and area of triangle POB is 150 cm², then the area of triangle QOA is:
(a) 210 cm²
(b) 294 cm²
(c) 150 cm²
(d) 250 cm²
Show Answer & Explanation

Answer: (b) 294 cm²

Explanation:
1. ∠B = ∠A = 90° and ∠POB = ∠QOA, so the triangles are similar.
2. Area POB : area QOA = PO² : QO² = 25 : 49.
3. Area QOA = 150 × 49/25 = 294 cm².

Teacher's Note:
1. Prove the two triangles similar by AA.
2. Areas are in the ratio of the squares of corresponding sides.

Question 10: Two poles of heights 6 m and 3 m stand p metres apart. The height of the point where the lines joining the top of each pole to the foot of the other cross is:
(a) 2 m
(b) 4.5 m
(c) 1.5 m
(d) 3 m
Show Answer & Explanation

Answer: (a) 2 m

Explanation:
1. The height is ab/(a + b) = 6 × 3/9 = 2 m, independent of the distance between the poles.

Teacher's Note:
1. Use similar triangles: the height h satisfies 1/h = 1/a + 1/b.

Question 11: The areas of two similar triangles are 81 cm² and 49 cm². If the altitude of the larger triangle is 4.5 cm, the corresponding altitude of the smaller triangle is:
(a) 2.5 cm
(b) 4 cm
(c) 3 cm
(d) 3.5 cm
Show Answer & Explanation

Answer: (d) 3.5 cm

Explanation:
1. Altitudes are in the ratio of the square roots of the areas: 9 : 7. So the smaller altitude = 4.5 × 7/9 = 3.5 cm.

Teacher's Note:
1. Ratio of sides = square root of ratio of areas.
2. Altitudes are proportional to corresponding sides.

Question 12: The area of the equilateral triangle described on the side of a square compared with that on its diagonal is:
(a) 1 : 4
(b) 1 : √2
(c) 1 : 2
(d) 2 : 1
Show Answer & Explanation

Answer: (c) 1 : 2

Explanation:
1. The diagonal is √2 times the side, so the equilateral triangles have areas in the ratio 1² : (√2)² = 1 : 2.

Teacher's Note:
1. Areas of similar figures are in the ratio of the squares of the sides.

Question 13: In an equilateral triangle ABC, AD ⟂ BC with D on BC. Then AD² is equal to:
(a) 2BD²
(b) 3BD²
(c) 4BD²
(d) BD²
Show Answer & Explanation

Answer: (b) 3BD²

Explanation:
1. BD = s/2 and AD² = s² − s²/4 = 3s²/4 = 3(s/2)² = 3BD².

Teacher's Note:
1. BD is half the side.
2. Apply Pythagoras in triangle ABD.

Question 14: In a triangle ABC right-angled at C, P and Q are the midpoints of CA and CB. If AC = 6 cm and BC = 8 cm, the value of AQ² + BP² is:
(a) 125 cm²
(b) 100 cm²
(c) 150 cm²
(d) 200 cm²
Show Answer & Explanation

Answer: (a) 125 cm²

Explanation:
1. AQ² = AC² + CQ² = 36 + 16 = 52. BP² = BC² + CP² = 64 + 9 = 73. Sum = 125, which equals 5AB²/4 = 5 × 100/4.

Teacher's Note:
1. Use Pythagoras in triangles ACQ and BCP.
2. AB = 10 cm.

Question 15: Similar triangles ACD and ABE are drawn on the hypotenuse AC and on side AB of an isosceles right triangle ABC (right angle at B), with AC and AB as corresponding sides. The ratio of area(ABE) to area(ACD) is:
(a) 1 : 1
(b) 1 : 4
(c) 2 : 1
(d) 1 : 2
Show Answer & Explanation

Answer: (d) 1 : 2

Explanation:
1. AC = √2·AB, so the ratio of the areas of the similar triangles is AB² : AC² = 1 : 2.

Teacher's Note:
1. Find AC in terms of AB.
2. Areas are in the ratio of the squares of corresponding sides.

Question 16: A right triangle has area 24 cm² and one leg b = 6 cm. The altitude on the hypotenuse is:
(a) 5 cm
(b) 4 cm
(c) 4.8 cm
(d) 6 cm
Show Answer & Explanation

Answer: (c) 4.8 cm

Explanation:
1. The other leg is 2 × 24/6 = 8 and the hypotenuse is 10. Altitude = (6 × 8)/10 = 4.8 cm, which agrees with 2Ab/√(b⁴ + 4A²).

Teacher's Note:
1. Find the other leg from the area.
2. Altitude × hypotenuse = product of legs.

Question 17: In triangle ABC, ∠A = 90° and AD is its bisector (D on BC). If DE ⟂ AC with E on AC, AB = 6 cm and AC = 12 cm, then DE is:
(a) 3 cm
(b) 4 cm
(c) 2 cm
(d) 6 cm
Show Answer & Explanation

Answer: (b) 4 cm

Explanation:
1. BD : DC = AB : AC = 1 : 2, so DC = (2/3) BC.
2. Triangles CDE and CBA are similar, so DE/AB = DC/BC = 2/3.
3. DE = 4 cm; check 4 × 18 = 72 = 6 × 12.

Teacher's Note:
1. DE is parallel to AB.
2. Use the bisector theorem for BD : DC.

Question 18: In triangle ABC, right-angled at B, AC = 5 cm and the median AD to BC has length 3√5/2 cm. The length of the median CE to AB is:
(a) \( 2\sqrt{5}\ \text{cm} \)
(b) \( \sqrt{5}\ \text{cm} \)
(c) \( 3\sqrt{5}\ \text{cm} \)
(d) \( 4\ \text{cm} \)
Show Answer & Explanation

Answer: (a) \( 2\sqrt{5}\ \text{cm} \)

Explanation:
1. AD² = AB² + (BC/2)², CE² = BC² + (AB/2)².
2. AD² + CE² = (5/4)(AB² + BC²) = (5/4)(25) = 125/4.
3. CE² = 125/4 − 45/4 = 20, so CE = 2√5.

Teacher's Note:
1. Write AD² and CE² using Pythagoras in the triangles ABD and CBE.
2. Add them and use AC = 5.

Question 19: In triangle ABC, AB = AC = 10 cm and D is a point on BC with BD = 4 cm and BC = 12 cm. The length AD is:
(a) \( 8\ \text{cm} \)
(b) \( 2\sqrt{13}\ \text{cm} \)
(c) \( \sqrt{68}-2\ \text{cm} \)
(d) \( 2\sqrt{17}\ \text{cm} \)
Show Answer & Explanation

Answer: (d) \( 2\sqrt{17}\ \text{cm} \)

Explanation:
1. CD = 8 cm and AB² − AD² = BD × CD = 32, so AD² = 100 − 32 = 68 and AD = 2√17 cm.

Teacher's Note:
1. Use AB² − AD² = BD·CD for an isosceles triangle.
2. CD = BC − BD.

Question 20: In triangle ABC, ∠A = 80° and ∠B = 60°. The bisectors of ∠B and ∠C meet at D inside the triangle. If ∠C = 2x° and ∠BDC = y°, then x and y are respectively:
(a) 15° and 70°
(b) 10° and 160°
(c) 20° and 130°
(d) 20° and 125°
Show Answer & Explanation

Answer: (c) 20° and 130°

Explanation:
1. ∠C = 180° − 80° − 60° = 40°, so x = 20. In triangle BDC, ∠DBC = 30° and ∠DCB = 20°, so y = 180° − 50° = 130°.

Teacher's Note:
1. Find ∠C first.
2. ∠BDC = 90° + ∠A/2.

Question 21: If a + b + c = 2s, then \( (s-a)^2 + (s-b)^2 + (s-c)^2 \) equals:
(a) \( s^2 + a^2 + b^2 + c^2 \)
(b) \( a^2 + b^2 + c^2 - s^2 \)
(c) \( s^2 - a^2 - b^2 - c^2 \)
(d) \( 4s^2 - a^2 - b^2 - c^2 \)
Show Answer & Explanation

Answer: (b) \( a^2 + b^2 + c^2 - s^2 \)

Explanation:
1. Expanding: 3s² − 2s(a + b + c) + a² + b² + c² = 3s² − 4s² + a² + b² + c² = a² + b² + c² − s².

Teacher's Note:
1. Expand each square.
2. Use a + b + c = 2s.

Question 22: D lies on side BC = 12 cm of triangle ABC with BD = 9 cm and ∠ADC = ∠BAC. The length of AC is:
(a) 6 cm
(b) \( 6\sqrt{3}\ \text{cm} \)
(c) 9 cm
(d) 3 cm
Show Answer & Explanation

Answer: (a) 6 cm

Explanation:
1. Triangles CDA and CAB are similar (common angle C and ∠CDA = ∠CAB). So AC/CB = CD/CA and AC² = CD × CB = 3 × 12 = 36, giving AC = 6 cm.

Teacher's Note:
1. DC = 12 − 9 = 3.
2. Use the AA similarity of triangles CDA and CAB.

Question 23: In triangle ABC the medians BE and CF meet at G, and the line AG meets BC at D. If GD = 1.5 cm, then AD is:
(a) 2.5 cm
(b) 3.0 cm
(c) 4.0 cm
(d) 4.5 cm
Show Answer & Explanation

Answer: (d) 4.5 cm

Explanation:
1. G is the centroid and AD is a median, so AG = 2 × GD = 3 cm and AD = AG + GD = 4.5 cm.

Teacher's Note:
1. The centroid divides a median 2 : 1 from the vertex.

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

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