Practice Questions →

Chapter 6: Triangles

NCERT Solutions for CBSE Class 10 Mathematics — 30 solved questions with detailed explanations.

30

Questions

5

Topics

Important Formulas

BPT: AD/DB = AE/EC (when DE ∥ BC)
Area ratio = (side ratio)²
Pythagoras: hyp² = base² + height²
If altitude from right angle vertex to hypotenuse in △ABC (∠B=90°): BD² = AD×DC

Solved Questions

Q1. Which criterion requires two equal angles for similarity?

Difficulty: Easy · Topic: Criteria for Similarity

Solution

AA (Angle-Angle) criterion: two equal angles ⟹ similarity.

Q2. In △ABC, DE ∥ BC, AD = 1.5 cm, DB = 3 cm, AE = 1 cm. Find EC.

Difficulty: Easy-Medium · Topic: Basic Proportionality Theorem

Solution

AD/DB = AE/EC → 1.5/3 = 1/EC → EC = 2 cm.

Q3. If △ABC ~ △DEF and AB/DE = 3/5, find ar(△ABC)/ar(△DEF).

Difficulty: Easy-Medium · Topic: Areas of Similar Triangles

Solution

Area ratio = (side ratio)² = (3/5)² = 9/25.

Q4. A ladder 10 m long reaches a window 8 m high. How far is the foot of the ladder from the wall?

Difficulty: Easy-Medium · Topic: Pythagoras Theorem

Solution

10² = 8² + x² → 100 = 64 + x² → x² = 36 → x = 6 m.

Q5. Two poles 6 m and 11 m tall stand 12 m apart. Find the distance between their tops.

Difficulty: Easy-Medium · Topic: Pythagoras Theorem

Solution

Height difference = 5 m, horizontal = 12 m. Distance = √(5²+12²) = √169 = 13 m.

Q6. Sides of two similar triangles are in ratio 4:9. Their areas are in ratio:

Difficulty: Easy-Medium · Topic: Areas of Similar Triangles

Solution

Area ratio = (side ratio)² = (4/9)² = 16/81.

Q7. In △ABC, DE ∥ BC, AD = 2, BD = 3, AE = 4. Then AC =

Difficulty: Easy-Medium · Topic: Basic Proportionality Theorem

Solution

AD/AB = AE/AC → 2/5 = 4/AC → AC = 10.

Q8. Triangles ABC and PQR are similar. If AB = 3 cm, BC = 4 cm, CA = 5 cm, and PQ = 6 cm, find QR.

A) 7
B) 4.5
C) 10.0
D) 8.0

Difficulty: Easy-Medium · Topic: Similar triangles - finding missing side

Solution

Since triangles are similar: \(\frac{AB}{PQ} = \frac{BC}{QR}\)

\(QR = \frac{BC \times PQ}{AB} = \frac{4 \times 6}{3} = 8.0\) cm

Q9. Triangles ABC and PQR are similar. If AB = 5 cm, BC = 12 cm, CA = 13 cm, and PQ = 10 cm, find QR.

A) 24.0
B) 4.17
C) 17
D) 26.0

Difficulty: Easy-Medium · Topic: Similar triangles - finding missing side

Solution

Since triangles are similar: \(\frac{AB}{PQ} = \frac{BC}{QR}\)

\(QR = \frac{BC \times PQ}{AB} = \frac{12 \times 10}{5} = 24.0\) cm

Q10. Triangles ABC and PQR are similar. If AB = 3 cm, BC = 6 cm, CA = 7 cm, and PQ = 9 cm, find QR.

A) 21.0
B) 18.0
C) 4.5
D) 12

Difficulty: Easy-Medium · Topic: Similar triangles - finding missing side

Solution

Since triangles are similar: \(\frac{AB}{PQ} = \frac{BC}{QR}\)

\(QR = \frac{BC \times PQ}{AB} = \frac{6 \times 9}{3} = 18.0\) cm

Q11. Triangles ABC and PQR are similar. If AB = 9 cm, BC = 12 cm, CA = 15 cm, and PQ = 12 cm, find QR.

A) 9.0
B) 20.0
C) 15
D) 16.0

Difficulty: Easy-Medium · Topic: Similar triangles - finding missing side

Solution

Since triangles are similar: \(\frac{AB}{PQ} = \frac{BC}{QR}\)

\(QR = \frac{BC \times PQ}{AB} = \frac{12 \times 12}{9} = 16.0\) cm

Q12. Triangles ABC and PQR are similar. If AB = 4 cm, BC = 5 cm, CA = 6 cm, and PQ = 8 cm, find QR.

A) 10.0
B) 9
C) 12.0
D) 6.4

Difficulty: Easy-Medium · Topic: Similar triangles - finding missing side

Solution

Since triangles are similar: \(\frac{AB}{PQ} = \frac{BC}{QR}\)

\(QR = \frac{BC \times PQ}{AB} = \frac{5 \times 8}{4} = 10.0\) cm

Q13. Triangles ABC and PQR are similar. If AB = 7 cm, BC = 24 cm, CA = 25 cm, and PQ = 14 cm, find QR.

A) 50.0
B) 4.08
C) 48.0
D) 31

Difficulty: Easy-Medium · Topic: Similar triangles - finding missing side

Solution

Since triangles are similar: \(\frac{AB}{PQ} = \frac{BC}{QR}\)

\(QR = \frac{BC \times PQ}{AB} = \frac{24 \times 14}{7} = 48.0\) cm

Q14. Triangles ABC and PQR are similar. If AB = 5 cm, BC = 7 cm, CA = 9 cm, and PQ = 10 cm, find QR.

A) 12
B) 7.14
C) 14.0
D) 18.0

Difficulty: Easy-Medium · Topic: Similar triangles - finding missing side

Solution

Since triangles are similar: \(\frac{AB}{PQ} = \frac{BC}{QR}\)

\(QR = \frac{BC \times PQ}{AB} = \frac{7 \times 10}{5} = 14.0\) cm

Q15. Triangles ABC and PQR are similar. If AB = 8 cm, BC = 15 cm, CA = 17 cm, and PQ = 16 cm, find QR.

A) 23
B) 8.53
C) 30.0
D) 34.0

Difficulty: Easy-Medium · Topic: Similar triangles - finding missing side

Solution

Since triangles are similar: \(\frac{AB}{PQ} = \frac{BC}{QR}\)

\(QR = \frac{BC \times PQ}{AB} = \frac{15 \times 16}{8} = 30.0\) cm

Q16. Triangles ABC and PQR are similar. If AB = 10 cm, BC = 15 cm, CA = 20 cm, and PQ = 6 cm, find QR.

A) 12.0
B) 11
C) 4.0
D) 9.0

Difficulty: Easy-Medium · Topic: Similar triangles - finding missing side

Solution

Since triangles are similar: \(\frac{AB}{PQ} = \frac{BC}{QR}\)

\(QR = \frac{BC \times PQ}{AB} = \frac{15 \times 6}{10} = 9.0\) cm

Q17. Triangles ABC and PQR are similar. If AB = 6 cm, BC = 8 cm, CA = 10 cm, and PQ = 9 cm, find QR.

A) 15.0
B) 11
C) 12.0
D) 6.75

Difficulty: Easy-Medium · Topic: Similar triangles - finding missing side

Solution

Since triangles are similar: \(\frac{AB}{PQ} = \frac{BC}{QR}\)

\(QR = \frac{BC \times PQ}{AB} = \frac{8 \times 9}{6} = 12.0\) cm

Q18. In △ABC, DE ∥ BC, AD/DB = 2/3. If AC = 15 cm, find AE.

Difficulty: Medium · Topic: Basic Proportionality Theorem

Solution

AD/DB = AE/EC = 2/3. AE/(AC−AE) = 2/3 → 3AE = 2(15−AE) = 30−2AE → 5AE = 30 → AE = 6.

Q19. In △ABC, ∠A = ∠P = 60°, AB/PQ = AC/PR = 1/2. Are the triangles similar? If yes, by which criterion?

Difficulty: Medium · Topic: Criteria for Similarity

Solution

One angle equal (60°) and the including sides proportional (ratio 1:2). By SAS similarity, △ABC ~ △PQR.

Q20. Prove that in an equilateral triangle with side a, the altitude h = a√3/2.

Difficulty: Medium · Topic: Pythagoras Theorem

Solution

The altitude bisects the base. In the right triangle: a² = h² + (a/2)² → h² = a² − a²/4 = 3a²/4 → h = a√3/2.

Q21. The areas of two similar triangles are 81 cm² and 49 cm². If the altitude of the larger triangle is 63 cm, find the altitude of the smaller.

Difficulty: Medium · Topic: Areas of Similar Triangles

Solution

Altitude ratio = √(area ratio) = √(81/49) = 9/7. Alt₂ = 63 × 7/9 = 49 cm.

Wait: ratio of altitudes = ratio of sides = 9/7 (larger to smaller). Smaller altitude = 63 × 7/9 = 49 cm.

Q22. In △ABC, ∠C = 90°, AC = 5 cm, BC = 12 cm. D is the mid-point of AB. Find CD.

Difficulty: Medium · Topic: Pythagoras Theorem

Solution

AB = √(25+144) = 13. D is midpoint of hypotenuse. In a right triangle, median to hypotenuse = half the hypotenuse. CD = 13/2 = 6.5 cm.

Q23. In a trapezium ABCD, AB ∥ DC. Diagonals AC and BD intersect at O. If AB = 2CD, find ar(△AOB)/ar(△COD).

Difficulty: Medium · Topic: Criteria for Similarity

Solution

△AOB ~ △COD (AA: alternate angles + vertical angles). AB/CD = 2. Area ratio = 2² = 4.

Q24. PQR is a right triangle, right-angled at Q. QS ⊥ PR. If PQ = 6 cm, QR = 8 cm, find PS and QS.

Difficulty: Medium · Topic: Pythagoras Theorem

Solution

PR = √(36+64) = 10. △PQS ~ △PQR (AA). PQ²=PS×PR → 36=10PS → PS=3.6.

QS² = PS×SR = 3.6×6.4 = 23.04 → QS = 4.8.

Q25. △ABC and △PQR are similar. AB = 6, BC = 8, CA = 10, PR = 5. Find PQ.

Difficulty: Medium · Topic: Criteria for Similarity

Solution

CA/PR = 10/5 = 2. PQ = AB/2 = 6/2 = 3.

Q26. In △ABC, AD ⊥ BC. Prove that AB² + CD² = AC² + BD².

Difficulty: Medium-Hard · Topic: Pythagoras Theorem

Solution

In △ABD: AB² = AD² + BD² → AD² = AB² − BD².

In △ACD: AC² = AD² + CD² → AD² = AC² − CD².

Equating: AB² − BD² = AC² − CD² → AB² + CD² = AC² + BD².

Q27. In △ABC, D and E are on AB and AC such that DE ∥ BC. If AD = 4x−3, AE = 8x−7, BD = 3x−1, CE = 5x−3, find x.

Difficulty: Medium-Hard · Topic: Basic Proportionality Theorem

Solution

By BPT: (4x−3)/(3x−1) = (8x−7)/(5x−3).

Cross-multiply: (4x−3)(5x−3) = (8x−7)(3x−1).

20x²−12x−15x+9 = 24x²−8x−21x+7.

20x²−27x+9 = 24x²−29x+7.

4x²−2x−2 = 0 → 2x²−x−1 = 0 → (2x+1)(x−1) = 0. x = 1 (positive).

Q28. In an equilateral triangle, prove that three times the square of one side equals four times the square of one altitude.

Difficulty: Medium-Hard · Topic: Pythagoras Theorem

Solution

Side = a, altitude = a√3/2. 4h² = 4 × 3a²/4 = 3a². Hence 3a² = 4h².

Q29. The areas of two similar triangles are 100 cm² and 64 cm². If the median of the larger triangle is 12.5 cm, find the corresponding median of the smaller.

Difficulty: Medium-Hard · Topic: Areas of Similar Triangles

Solution

Ratio of medians = ratio of sides = √(area ratio) = √(100/64) = 10/8 = 5/4.

Smaller median = 12.5 × 4/5 = 10 cm.

Q30. ABC is a triangle with AD perpendicular to BC and AD² = BD × DC. Prove that ∠BAC = 90°.

Difficulty: Hard · Topic: Pythagoras Theorem

Solution

AD² = BD×DC. In △ABD: AB² = AD²+BD² = BD×DC+BD² = BD(BD+DC) = BD×BC.

In △ACD: AC² = AD²+DC² = BD×DC+DC² = DC(BD+DC) = DC×BC.

AB²+AC² = BD×BC+DC×BC = BC(BD+DC) = BC² = BC².

By converse of Pythagoras, ∠BAC = 90°.

Other Chapters in Mathematics

Ch 1: Real Numbers Ch 2: Polynomials Ch 3: Pair of Linear Equations in Two Variables Ch 4: Quadratic Equations Ch 5: Arithmetic Progressions Ch 7: Coordinate Geometry Ch 8: Introduction to Trigonometry Ch 9: Some Applications of Trigonometry Ch 10: Circles Ch 11: Areas Related to Circles Ch 12: Surface Areas and Volumes Ch 13: Statistics Ch 14: Probability

Want to practice these questions interactively?

Get instant feedback, track progress, and improve with adaptive practice.

Start Practicing Free →