CBSE Class 10 Mathematics Triangles MCQs Set 16

Multiple Choice Questions (MCQs) for Class 10 Mathematics: 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.

Practice Chapter 06 Triangles MCQs for Class 10 Mathematics

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Multiple Choice Questions

Question 1: In △ABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC. If AD = 6 cm, DB = 9 cm and AE = 8 cm, then AC is:
(a) 16 cm
(b) 20 cm
(c) 12 cm
(d) 15 cm
Show Answer & Explanation

Answer: (b) 20 cm

Explanation:
1. By the basic proportionality theorem, \( \frac{AD}{DB}=\frac{AE}{EC}\Rightarrow\frac69=\frac8{EC}\Rightarrow EC=12\ \text{cm} \).
2. AC = AE + EC = 8 + 12 = 20 cm.

Teacher's Note:
1. DE ∥ BC gives AD/DB = AE/EC.
2. AC is the whole side, not just EC.

Question 2: In △ABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC. If AD/DB = 3/4 and AC = 15 cm, then AE is:
(a) 6.43 cm
(b) 9 cm
(c) 7.5 cm
(d) 11.25 cm
Show Answer & Explanation

Answer: (a) 6.43 cm

Explanation:
1. AD/DB = AE/EC = 3/4, so AE : AC = 3 : 7.
2. \( AE=\frac37\times15=\frac{45}7\approx6.43\ \text{cm} \).

Teacher's Note:
1. Convert the ratio AE : EC into AE : AC.
2. AC = AE + EC.

Question 3: In △ABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC. If AD/DB = 2/3 and AC = 18 cm, then AE is:
(a) 10.8 cm
(b) 6 cm
(c) 12 cm
(d) 7.2 cm
Show Answer & Explanation

Answer: (d) 7.2 cm

Explanation:
1. AE : EC = 2 : 3, so AE : AC = 2 : 5.
2. \( AE=\frac25\times18=7.2\ \text{cm} \).

Teacher's Note:
1. Use AE/EC = AD/DB.
2. AE is two out of five equal parts of AC.

Question 4: In △ABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC. If AD = 4 cm, AE = 8 cm, DB = x − 4 and EC = 3x − 19, then x is:
(a) 11
(b) 9
(c) 8
(d) 12
Show Answer & Explanation

Answer: (a) 11

Explanation:
1. \( \frac{AD}{DB}=\frac{AE}{EC}\Rightarrow\frac4{x-4}=\frac8{3x-19} \).
2. Cross-multiplying: 12x − 76 = 8x − 32, so 4x = 44 and x = 11.
3. Check: DB = 7 and EC = 14, both positive.

Teacher's Note:
1. Form the equation from the proportionality theorem.
2. Check that the lengths come out positive.

Question 5: In △ABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC. If AD = 8 cm, AB = 12 cm and AE = 12 cm, then CE is:
(a) 18 cm
(b) 4 cm
(c) 6 cm
(d) 8 cm
Show Answer & Explanation

Answer: (c) 6 cm

Explanation:
1. DB = 12 − 8 = 4 cm. \( \frac{AD}{DB}=\frac{AE}{EC}\Rightarrow\frac84=\frac{12}{EC}\Rightarrow EC=6\ \text{cm} \).

Teacher's Note:
1. First find DB = AB − AD.
2. Apply AD/DB = AE/EC.

Question 6: In △ABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC. If AD = 4 cm, DB = 4.5 cm and AE = 8 cm, then AC is:
(a) 12.5 cm
(b) 17 cm
(c) 9 cm
(d) 16 cm
Show Answer & Explanation

Answer: (b) 17 cm

Explanation:
1. \( \frac{AD}{DB}=\frac{AE}{EC}\Rightarrow EC=\frac{8\times4.5}{4}=9\ \text{cm} \).
2. AC = AE + EC = 8 + 9 = 17 cm.

Teacher's Note:
1. EC comes from the proportionality theorem.
2. Add AE to get AC.

Question 7: In △ABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC. If AD = 2 cm, AB = 6 cm and AC = 9 cm, then AE is:
(a) 3 cm
(b) 4.5 cm
(c) 2 cm
(d) 6 cm
Show Answer & Explanation

Answer: (a) 3 cm

Explanation:
1. \( \frac{AD}{AB}=\frac{AE}{AC}\Rightarrow AE=\frac{2\times9}6=3\ \text{cm} \).

Teacher's Note:
1. DE ∥ BC also gives AD/AB = AE/AC.
2. Multiply and divide carefully.

Question 8: In △ABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC. If AD/BD = 4/5 and EC = 2.5 cm, then AE is:
(a) 3.125 cm
(b) 4 cm
(c) 1.5 cm
(d) 2 cm
Show Answer & Explanation

Answer: (d) 2 cm

Explanation:
1. \( \frac{AD}{BD}=\frac{AE}{EC}\Rightarrow AE=\frac45\times2.5=2\ \text{cm} \).

Teacher's Note:
1. AD/BD = AE/EC by the theorem.
2. Multiply 4/5 by 2.5.

Question 9: In △ABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC. If AD = x, DB = x − 2, AE = x + 2 and EC = x − 1, then x is:
(a) 5
(b) 6
(c) 4
(d) 3
Show Answer & Explanation

Answer: (c) 4

Explanation:
1. \( \frac x{x-2}=\frac{x+2}{x-1}\Rightarrow x(x-1)=(x+2)(x-2) \).
2. x² − x = x² − 4, so x = 4.

Teacher's Note:
1. Cross-multiply and cancel x².
2. Check the lengths: DB = 2, EC = 3.

Question 10: In △ABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC. If AD = 8x − 7, DB = 5x − 3, AE = 4x − 3 and EC = 3x − 1, then x is:
(a) 1
(b) 2
(c) −1/2
(d) 3
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. \( \frac{8x-7}{5x-3}=\frac{4x-3}{3x-1}\Rightarrow(8x-7)(3x-1)=(5x-3)(4x-3) \).
2. 24x² − 29x + 7 = 20x² − 27x + 9, so 2x² − x − 1 = 0, giving x = 1 or x = −1/2.
3. x = −1/2 makes the lengths negative, so x = 1.

Teacher's Note:
1. Cross-multiply and solve the quadratic.
2. Reject any root giving negative lengths.

Question 11: In △ABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC. If AD = 4x − 3, AE = 8x − 7, BD = 3x − 1 and CE = 5x − 3, then x is:
(a) 5/2
(b) 1
(c) 2
(d) 3/4
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. \( \frac{4x-3}{3x-1}=\frac{8x-7}{5x-3}\Rightarrow(4x-3)(5x-3)=(8x-7)(3x-1) \).
2. 20x² − 27x + 9 = 24x² − 29x + 7, so 2x² − x − 1 = 0, giving x = 1 or −1/2.
3. Only x = 1 gives positive lengths.

Teacher's Note:
1. Same method as the previous item.
2. A negative root is rejected.

Question 12: In △ABC, AD is the bisector of ∠A meeting side BC at D. If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, then DC is:
(a) 2.1 cm
(b) 1.5 cm
(c) 3 cm
(d) 4.2 cm
Show Answer & Explanation

Answer: (a) 2.1 cm

Explanation:
1. By the angle bisector theorem, \( \frac{BD}{DC}=\frac{AB}{AC}\Rightarrow DC=\frac{2.5\times4.2}{5}=2.1\ \text{cm} \).

Teacher's Note:
1. BD/DC = AB/AC.
2. Substitute the values.

Question 13: In △ABC, AD is the bisector of ∠A meeting side BC at D. If BD = 2 cm, AB = 5 cm and DC = 3 cm, then AC is:
(a) 6 cm
(b) 3.33 cm
(c) 4 cm
(d) 7.5 cm
Show Answer & Explanation

Answer: (d) 7.5 cm

Explanation:
1. \( \frac{BD}{DC}=\frac{AB}{AC}\Rightarrow AC=\frac{5\times3}2=7.5\ \text{cm} \).

Teacher's Note:
1. Use BD/DC = AB/AC.
2. Solve for AC.

Question 14: In △ABC, AD is the bisector of ∠A meeting side BC at D. If AB = 3.5 cm, AC = 4.2 cm and DC = 2.8 cm, then BD is:
(a) 1.5 cm
(b) 2.8 cm
(c) 2.33 cm
(d) 3.36 cm
Show Answer & Explanation

Answer: (c) 2.33 cm

Explanation:
1. \( BD=\frac{AB\times DC}{AC}=\frac{3.5\times2.8}{4.2}=\frac73\approx2.33\ \text{cm} \).

Teacher's Note:
1. BD/DC = AB/AC.
2. 7/3 cm is about 2.33 cm.

Question 15: In △ABC, AD is the bisector of ∠A meeting side BC at D. If AB = 10 cm, AC = 14 cm and BC = 6 cm, then BD and DC are respectively:
(a) 3.5 cm and 2.5 cm
(b) 2.5 cm and 3.5 cm
(c) 3 cm and 3 cm
(d) 2 cm and 4 cm
Show Answer & Explanation

Answer: (b) 2.5 cm and 3.5 cm

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

Teacher's Note:
1. Divide BC in the ratio AB : AC.
2. BD is the part next to AB, the shorter side.

Question 16: In △ABC, AD is the bisector of ∠A meeting side BC at D. If AC = 4.2 cm, DC = 6 cm and BC = 10 cm, then AB is:
(a) 2.8 cm
(b) 4 cm
(c) 6.3 cm
(d) 5.6 cm
Show Answer & Explanation

Answer: (a) 2.8 cm

Explanation:
1. BD = BC − DC = 4 cm. \( \frac{AB}{AC}=\frac{BD}{DC}\Rightarrow AB=\frac{4\times4.2}6=2.8\ \text{cm} \).

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

Question 17: In △ABC, AD is the bisector of ∠A meeting side BC at D. If AB = 5.6 cm, AC = 6 cm and DC = 3 cm, then BC is:
(a) 6 cm
(b) 2.8 cm
(c) 8.8 cm
(d) 5.8 cm
Show Answer & Explanation

Answer: (d) 5.8 cm

Explanation:
1. \( BD=\frac{AB\times DC}{AC}=\frac{5.6\times3}6=2.8\ \text{cm} \).
2. BC = BD + DC = 2.8 + 3 = 5.8 cm.

Teacher's Note:
1. Find BD from the bisector theorem.
2. BC = BD + DC.

Question 18: In △ABC, AD is the bisector of ∠A meeting side BC at D. If AB = 5.6 cm, BC = 6 cm and BD = 3.2 cm, then AC is:
(a) 3.2 cm
(b) 6.4 cm
(c) 4.9 cm
(d) 5.6 cm
Show Answer & Explanation

Answer: (c) 4.9 cm

Explanation:
1. DC = 6 − 3.2 = 2.8 cm. \( \frac{AB}{AC}=\frac{BD}{DC}\Rightarrow AC=\frac{5.6\times2.8}{3.2}=4.9\ \text{cm} \).

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

Question 19: In △ABC, AD is the bisector of ∠A meeting side BC at D. If AB = 10 cm, AC = 6 cm and BC = 12 cm, then BD and DC are respectively:
(a) 8 cm and 4 cm
(b) 7.5 cm and 4.5 cm
(c) 4.5 cm and 7.5 cm
(d) 6 cm and 6 cm
Show Answer & Explanation

Answer: (b) 7.5 cm and 4.5 cm

Explanation:
1. BD : DC = AB : AC = 10 : 6 = 5 : 3.
2. BD = 12 × 5/8 = 7.5 cm and DC = 12 × 3/8 = 4.5 cm.

Teacher's Note:
1. Divide BC in the ratio AB : AC.
2. The longer side AB gets the longer segment.

Question 20: In △ABC, AE is the bisector of the exterior ∠CAD (D lies on BA produced) meeting BC produced at E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, then CE is:
(a) 18 cm
(b) 12 cm
(c) 15 cm
(d) 9 cm
Show Answer & Explanation

Answer: (a) 18 cm

Explanation:
1. The external bisector divides BC externally in the ratio of the sides: \( \frac{BE}{CE}=\frac{AB}{AC}=\frac{10}6=\frac53 \).
2. BE = 12 + CE, so \( 3(12+CE)=5\,CE\Rightarrow CE=18\ \text{cm} \).

Teacher's Note:
1. For an external bisector, E lies on BC produced.
2. BE = BC + CE.

Question 21: In △ABC, D, E and F are points on the sides BC, CA and AB such that AD bisects ∠A, BE bisects ∠B and CF bisects ∠C. If AB = 5 cm, BC = 8 cm and CA = 4 cm, then AF is:
(a) 32/13 cm
(b) 40/9 cm
(c) 2 cm
(d) 5/3 cm
Show Answer & Explanation

Answer: (d) 5/3 cm

Explanation:
1. CF bisects ∠C, so \( \frac{AF}{FB}=\frac{CA}{CB}=\frac48=\frac12 \).
2. AF = (1/3) × AB = 5/3 cm.

Teacher's Note:
1. The bisector from C divides AB in the ratio CA : CB.
2. AF is one of the three equal parts of AB.

Question 22: In △ABC with AB = 5 cm, BC = 8 cm and CA = 4 cm, BE bisects ∠B with E on CA. Then CE is:
(a) 40/9 cm
(b) 20/13 cm
(c) 32/13 cm
(d) 5/3 cm
Show Answer & Explanation

Answer: (c) 32/13 cm

Explanation:
1. \( \frac{AE}{EC}=\frac{BA}{BC}=\frac58 \).
2. CE = CA × 8/13 = 32/13 cm.

Teacher's Note:
1. The bisector of ∠B divides AC in the ratio BA : BC.
2. CE is 8 of the 13 parts.

Question 23: In △ABC with AB = 5 cm, BC = 8 cm and CA = 4 cm, AD bisects ∠A with D on BC. Then BD is:
(a) 4 cm
(b) 40/9 cm
(c) 5/3 cm
(d) 32/13 cm
Show Answer & Explanation

Answer: (b) 40/9 cm

Explanation:
1. \( \frac{BD}{DC}=\frac{AB}{AC}=\frac54 \).
2. BD = BC × 5/9 = 40/9 cm.

Teacher's Note:
1. The bisector of ∠A divides BC in the ratio AB : AC.
2. BD is 5 of the 9 parts.

Question 24: For a triangle ABC, the true statement is:
(a) AC < AB + BC
(b) AC² = AB² + BC²
(c) AC = AB + BC
(d) AC > AB + BC
Show Answer & Explanation

Answer: (a) AC < AB + BC

Explanation:
1. In any triangle the sum of two sides exceeds the third side (triangle inequality).
2. So AC < AB + BC; the other relations hold only in special cases (right angle) or are impossible.

Teacher's Note:
1. The straight path between two points is the shortest.
2. Equality would mean a degenerate triangle.

Question 25: If AD, BE and CF are the medians of a triangle ABC, then the true statement is:
(a) \( 3(AB^2+BC^2+AC^2)=4(AD^2+BE^2+CF^2) \)
(b) \( AB^2+BC^2+AC^2=AD^2+BE^2+CF^2 \)
(c) \( 2(AB^2+BC^2+AC^2)=3(AD^2+BE^2+CF^2) \)
(d) \( AB^2+BC^2+AC^2=3(AD^2+BE^2+CF^2) \)
Show Answer & Explanation

Answer: (a) \( 3(AB^2+BC^2+AC^2)=4(AD^2+BE^2+CF^2) \)

Explanation:
1. Apollonius' theorem: \( AB^2+AC^2=2AD^2+\frac12BC^2 \) and similarly for the other two medians.
2. Adding the three: \( 2(AB^2+BC^2+CA^2)=2(AD^2+BE^2+CF^2)+\frac12(AB^2+BC^2+CA^2) \), so \( \frac32(AB^2+BC^2+CA^2)=2(AD^2+BE^2+CF^2) \), i.e. \( 3(AB^2+BC^2+CA^2)=4(AD^2+BE^2+CF^2) \).

Teacher's Note:
1. Apply Apollonius' theorem to each median and add.
2. Collect the terms in the sum of squares of the sides.

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