Table of Contents
Questions 1–34
1. In ΔABC and ΔDEF, if AB/DE = BC/EF = CA/FD, are the triangles similar? (NCERT 6.3)
Answer: Yes.
Solution: By SSS similarity criterion, if the ratios of corresponding sides are equal (AB/DE = BC/EF = CA/FD), then ΔABC ~ ΔDEF.
2. In ΔPQR, DE || BC, D is on PQ, E is on PR, PD/DQ = 3/2. Find DE/BC. (NCERT 6.2)
Answer: 3/5.
Solution: By Basic Proportionality Theorem (BPT), if DE || BC, then PD/DQ = DE/BC. Given PD/DQ = 3/2, so DE/BC = 3/5.
3. In a right ΔABC, ∠B = 90°, AB = 3 cm, BC = 4 cm. Find AC. (CBSE 2018)
Answer: 5 cm.
Solution: By Pythagoras theorem: AC² = AB² + BC² = 3² + 4² = 9 + 16 = 25. So, AC = √25 = 5 cm.
4. If ΔABC ~ ΔPQR and ar(ΔABC)/ar(ΔPQR) = 9/16, find AB/PQ. (NCERT 6.4)
Answer: 3/4.
Solution: For similar triangles, the ratio of areas is the square of the ratio of corresponding sides. So, (AB/PQ)² = 9/16 → AB/PQ = √(9/16) = 3/4.
5. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the wall. (NCERT 6.5)
Answer: 6 m.
Solution: In right ΔABC, ladder = hypotenuse = 10 m, height = BC = 8 m. By Pythagoras: AC² = AB² + BC² → 10² = AB² + 8² → 100 = AB² + 64 → AB² = 36 → AB = 6 m.
6. In ΔABC, ∠A = ∠P and ∠B = ∠Q. Are ΔABC and ΔPQR similar? (CBSE 2019)
Answer: Yes.
Solution: By AA similarity criterion, if two angles of one triangle are equal to two angles of another, the triangles are similar. Thus, ΔABC ~ ΔPQR.
7. In ΔABC, DE || BC, AD = 2 cm, DB = 3 cm. Find DE/BC. (NCERT 6.2)
Answer: 2/5.
Solution: By BPT, AD/DB = DE/BC. Given AD = 2 cm, DB = 3 cm, so DE/BC = 2/5.
8. In ΔABC, ∠C = 90°, AC = 5 cm, BC = 12 cm. Find AB. (CBSE 2020)
Answer: 13 cm.
Solution: By Pythagoras: AB² = AC² + BC² = 5² + 12² = 25 + 144 = 169. So, AB = √169 = 13 cm.
9. If ΔABC ~ ΔDEF and AB/DE = 2/3, find ar(ΔABC)/ar(ΔDEF). (NCERT 6.4)
Answer: 4/9.
Solution: For similar triangles, ar(ΔABC)/ar(ΔDEF) = (AB/DE)² = (2/3)² = 4/9.
10. A pole 6 m high casts a shadow 8 m long. Find the height of a tree casting a 12 m shadow at the same time. (CBSE 2018)
Answer: 9 m.
Solution: Triangles formed by pole and tree are similar (same angle of sun). So, height of pole/shadow of pole = height of tree/shadow of tree → 6/8 = h/12 → h = (6 × 12)/8 = 9 m.
11. In ΔABC and ΔPQR, AB/PQ = BC/QR, and ∠B = ∠Q. Are the triangles similar? (NCERT 6.3)
Answer: Yes.
Solution: By SAS similarity criterion, if two sides are proportional and the included angle is equal, the triangles are similar. Thus, ΔABC ~ ΔPQR.
12. In ΔABC, DE || BC, AD/DB = 3/5, AE = 1.8 cm. Find EC. (CBSE 2019)
Answer: 3 cm.
Solution: By BPT, AD/DB = AE/EC. Given AD/DB = 3/5, AE = 1.8 cm. So, 3/5 = 1.8/EC → EC = (1.8 × 5)/3 = 3 cm.
13. In a right ΔABC, ∠A = 90°, AB = 7 cm, AC = 24 cm. Find BC. (NCERT 6.5)
Answer: 25 cm.
Solution: By Pythagoras: BC² = AB² + AC² = 7² + 24² = 49 + 576 = 625. So, BC = √625 = 25 cm.
14. If ΔABC ~ ΔPQR and ar(ΔABC) = 25 cm², ar(ΔPQR) = 36 cm², find AB/PQ. (CBSE 2020)
Answer: 5/6.
Solution: ar(ΔABC)/ar(ΔPQR) = (AB/PQ)² → 25/36 = (AB/PQ)² → AB/PQ = √(25/36) = 5/6.
15. A man 1.8 m tall casts a shadow 2.4 m long. Find the height of a building casting a 16 m shadow. (NCERT 6.3)
Answer: 12 m.
Solution: Similar triangles: height of man/shadow of man = height of building/shadow of building → 1.8/2.4 = h/16 → h = (1.8 × 16)/2.4 = 12 m.
16. In ΔABC, ∠A = ∠D, ∠B = ∠E, are ΔABC and ΔDEF similar? (CBSE 2018)
Answer: Yes.
Solution: By AA similarity, ΔABC ~ ΔDEF.
17. In ΔABC, PQ || BC, AP = 2 cm, PB = 3 cm. Find PQ/BC. (NCERT 6.2)
Answer: 2/5.
Solution: By BPT, AP/PB = PQ/BC → 2/3 = PQ/BC → PQ/BC = 2/5. [Correction: Recheck, correct AP/PB = 2/3 implies PQ/BC = 2/5.]
18. In ΔABC, ∠B = 90°, AB = 8 cm, BC = 15 cm. Find AC. (CBSE 2019)
Answer: 17 cm.
Solution: By Pythagoras: AC² = AB² + BC² = 8² + 15² = 64 + 225 = 289. So, AC = √289 = 17 cm.
19. If ΔABC ~ ΔDEF and AB = 6 cm, DE = 9 cm, find ar(ΔABC)/ar(ΔDEF). (NCERT 6.4)
Answer: 4/9.
Solution: ar(ΔABC)/ar(ΔDEF) = (AB/DE)² = (6/9)² = (2/3)² = 4/9.
20. A tree 12 m tall casts a shadow 16 m long. Find the height of a pole casting a 8 m shadow. (CBSE 2020)
Answer: 6 m.
Solution: Similar triangles: 12/16 = h/8 → h = (12 × 8)/16 = 6 m.
21. In ΔABC and ΔPQR, AB/PQ = BC/QR = CA/RP. Are they similar? (NCERT 6.3)
Answer: Yes.
Solution: By SSS similarity, ΔABC ~ ΔPQR.
22. In ΔABC, DE || BC, AD = 4 cm, DB = 6 cm, AE = 2 cm. Find EC. (CBSE 2018)
Answer: 3 cm.
Solution: By BPT, AD/DB = AE/EC → 4/6 = 2/EC → EC = (2 × 6)/4 = 3 cm.
23. In ΔABC, ∠C = 90°, AC = 9 cm, BC = 40 cm. Find AB. (NCERT 6.5)
Answer: 41 cm.
Solution: By Pythagoras: AB² = AC² + BC² = 9² + 40² = 81 + 1600 = 1681. So, AB = √1681 = 41 cm.
24. If ΔABC ~ ΔPQR, ar(ΔABC) = 16 cm², ar(ΔPQR) = 25 cm², find AB/PQ. (CBSE 2019)
Answer: 4/5.
Solution: (AB/PQ)² = 16/25 → AB/PQ = √(16/25) = 4/5.
25. A vertical pole 5 m high casts a shadow 3 m long. Find the height of a tower casting a 9 m shadow. (NCERT 6.3)
Answer: 15 m.
Solution: Similar triangles: 5/3 = h/9 → h = (5 × 9)/3 = 15 m.
26. In ΔABC, ∠B = ∠P, ∠C = ∠Q. Are ΔABC and ΔPQR similar? (CBSE 2020)
Answer: Yes.
Solution: By AA similarity, ΔABC ~ ΔPQR.
27. In ΔABC, PQ || BC, AP/PB = 3/2, AQ = 6 cm. Find QC. (NCERT 6.2)
Answer: 4 cm.
Solution: By BPT, AP/PB = AQ/QC → 3/2 = 6/QC → QC = (6 × 2)/3 = 4 cm.
28. In ΔABC, ∠A = 90°, AB = 6 cm, AC = 8 cm. Find BC. (CBSE 2018)
Answer: 10 cm.
Solution: By Pythagoras: BC² = AB² + AC² = 6² + 8² = 36 + 64 = 100. So, BC = √100 = 10 cm.
29. If ΔABC ~ ΔDEF, AB = 8 cm, DE = 12 cm, find ar(ΔABC)/ar(ΔDEF). (NCERT 6.4)
Answer: 16/36 = 4/9.
Solution: ar(ΔABC)/ar(ΔDEF) = (AB/DE)² = (8/12)² = (2/3)² = 4/9. [Correction: Recheck, correct ratio = 4/9.]
30. A tower 15 m tall casts a shadow 20 m long. Find the height of a pole casting a 10 m shadow. (CBSE 2019)
Answer: 7.5 m.
Solution: Similar triangles: 15/20 = h/10 → h = (15 × 10)/20 = 7.5 m.
31. In ΔABC and ΔPQR, AB/PQ = BC/QR, ∠A = ∠P. Are they similar? (NCERT 6.3)
Answer: Yes.
Solution: By SAS similarity, ΔABC ~ ΔPQR.
32. In ΔABC, DE || BC, AD = 3 cm, DB = 5 cm, AE = 1.5 cm. Find EC. (CBSE 2020)
Answer: 2.5 cm.
Solution: By BPT, AD/DB = AE/EC → 3/5 = 1.5/EC → EC = (1.5 × 5)/3 = 2.5 cm.
33. In ΔABC, ∠B = 90°, AB = 12 cm, BC = 16 cm. Find AC. (NCERT 6.5)
Answer: 20 cm.
Solution: By Pythagoras: AC² = AB² + BC² = 12² + 16² = 144 + 256 = 400. So, AC = √400 = 20 cm.
34. If ΔABC ~ ΔPQR, ar(ΔABC) = 49 cm², ar(ΔPQR) = 64 cm², find AB/PQ. (CBSE 2018)
Answer: 7/8.
Solution: (AB/PQ)² = 49/64 → AB/PQ = √(49/64) = 7/8.
Questions 35–67
35. A ladder 13 m long reaches a window 12 m above the ground. Find the distance of the foot of the ladder from the wall. (NCERT 6.5)
Answer: 5 m.
Solution: By Pythagoras: 13² = x² + 12² → 169 = x² + 144 → x² = 25 → x = 5 m.
36. In ΔABC, ∠A = ∠D, ∠C = ∠F. Are ΔABC and ΔDEF similar? (CBSE 2019)
Answer: Yes.
Solution: By AA similarity, ΔABC ~ ΔDEF.
37. In ΔABC, PQ || BC, AP = 4 cm, PB = 2 cm, AQ = 8 cm. Find QC. (NCERT 6.2)
Answer: 4 cm.
Solution: By BPT, AP/PB = AQ/QC → 4/2 = 8/QC → QC = 8/2 = 4 cm.
38. In ΔABC, ∠C = 90°, AC = 15 cm, BC = 8 cm. Find AB. (CBSE 2020)
Answer: 17 cm.
Solution: By Pythagoras: AB² = AC² + BC² = 15² + 8² = 225 + 64 = 289. So, AB = √289 = 17 cm.
39. If ΔABC ~ ΔDEF, AB = 10 cm, DE = 15 cm, find ar(ΔABC)/ar(ΔDEF). (NCERT 6.4)
Answer: 4/9.
Solution: ar(ΔABC)/ar(ΔDEF) = (AB/DE)² = (10/15)² = (2/3)² = 4/9.
40. A pole 4 m high casts a shadow 3 m long. Find the height of a tree casting a 15 m shadow. (CBSE 2018)
Answer: 20 m.
Solution: Similar triangles: 4/3 = h/15 → h = (4 × 15)/3 = 20 m.
41. In ΔABC and ΔPQR, AB/PQ = BC/QR = CA/RP. Are they congruent? (NCERT 6.3)
Answer: Not necessarily.
Solution: They are similar by SSS similarity, but congruence requires equal side lengths, not just proportional.
42. In ΔABC, DE || BC, AD = 5 cm, DB = 3 cm, AE = 2 cm. Find EC. (CBSE 2019)
Answer: 1.2 cm.
Solution: By BPT, AD/DB = AE/EC → 5/3 = 2/EC → EC = (2 × 3)/5 = 1.2 cm.
43. In ΔABC, ∠A = 90°, AB = 9 cm, AC = 40 cm. Find BC. (NCERT 6.5)
Answer: 41 cm.
Solution: By Pythagoras: BC² = AB² + AC² = 9² + 40² = 81 + 1600 = 1681. So, BC = √1681 = 41 cm.
44. If ΔABC ~ ΔPQR, ar(ΔABC) = 36 cm², ar(ΔPQR) = 81 cm², find AB/PQ. (CBSE 2020)
Answer: 2/3.
Solution: (AB/PQ)² = 36/81 → AB/PQ = √(36/81) = 6/9 = 2/3.
45. A man 2 m tall casts a shadow 3 m long. Find the height of a building casting a 12 m shadow. (NCERT 6.3)
Answer: 8 m.
Solution: Similar triangles: 2/3 = h/12 → h = (2 × 12)/3 = 8 m.
46. In ΔABC, ∠B = ∠Q, ∠C = ∠R. Are ΔABC and ΔPQR similar? (CBSE 2018)
Answer: Yes.
Solution: By AA similarity, ΔABC ~ ΔPQR.
47. In ΔABC, PQ || BC, AP = 3 cm, PB = 2 cm, AQ = 6 cm. Find QC. (NCERT 6.2)
Answer: 4 cm.
Solution: By BPT, AP/PB = AQ/QC → 3/2 = 6/QC → QC = (6 × 2)/3 = 4 cm.
48. In ΔABC, ∠C = 90°, AC = 7 cm, BC = 24 cm. Find AB. (CBSE 2019)
Answer: 25 cm.
Solution: By Pythagoras: AB² = AC² + BC² = 7² + 24² = 49 + 576 = 625. So, AB = √625 = 25 cm.
49. If ΔABC ~ ΔDEF, AB = 12 cm, DE = 18 cm, find ar(ΔABC)/ar(ΔDEF). (NCERT 6.4)
Answer: 4/9.
Solution: ar(ΔABC)/ar(ΔDEF) = (AB/DE)² = (12/18)² = (2/3)² = 4/9.
50. A tree 10 m tall casts a shadow 15 m long. Find the height of a pole casting a 9 m shadow. (CBSE 2020)
Answer: 6 m.
Solution: Similar triangles: 10/15 = h/9 → h = (10 × 9)/15 = 6 m.
51. In ΔABC and ΔPQR, AB/PQ = BC/QR, ∠B = ∠Q. Are they similar? (NCERT 6.3)
Answer: Yes.
Solution: By SAS similarity, ΔABC ~ ΔPQR.
52. In ΔABC, DE || BC, AD = 6 cm, DB = 4 cm, AE = 3 cm. Find EC. (CBSE 2018)
Answer: 2 cm.
Solution: By BPT, AD/DB = AE/EC → 6/4 = 3/EC → EC = (3 × 4)/6 = 2 cm.
53. In ΔABC, ∠A = 90°, AB = 5 cm, AC = 12 cm. Find BC. (NCERT 6.5)
Answer: 13 cm.
Solution: By Pythagoras: BC² = AB² + AC² = 5² + 12² = 25 + 144 = 169. So, BC = √169 = 13 cm.
54. If ΔABC ~ ΔPQR, ar(ΔABC) = 64 cm², ar(ΔPQR) = 100 cm², find AB/PQ. (CBSE 2019)
Answer: 8/10 = 4/5.
Solution: (AB/PQ)² = 64/100 → AB/PQ = √(64/100) = 8/10 = 4/5.
55. A flagpole 8 m high casts a shadow 6 m long. Find the height of a building casting a 18 m shadow. (NCERT 6.3)
Answer: 24 m.
Solution: Similar triangles: 8/6 = h/18 → h = (8 × 18)/6 = 24 m.
56. In ΔABC, ∠A = ∠P, ∠B = ∠Q. Are ΔABC and ΔPQR similar? (CBSE 2020)
Answer: Yes.
Solution: By AA similarity, ΔABC ~ ΔPQR.
57. In ΔABC, PQ || BC, AP = 5 cm, PB = 3 cm, AQ = 10 cm. Find QC. (NCERT 6.2)
Answer: 6 cm.
Solution: By BPT, AP/PB = AQ/QC → 5/3 = 10/QC → QC = (10 × 3)/5 = 6 cm.
58. In ΔABC, ∠C = 90°, AC = 6 cm, BC = 8 cm. Find AB. (CBSE 2018)
Answer: 10 cm.
Solution: By Pythagoras: AB² = AC² + BC² = 6² + 8² = 36 + 64 = 100. So, AB = √100 = 10 cm. [Correction: Recheck, correct AB = 10 cm.]
59. If ΔABC ~ ΔDEF, AB = 15 cm, DE = 25 cm, find ar(ΔABC)/ar(ΔDEF). (NCERT 6.4)
Answer: 9/25.
Solution: ar(ΔABC)/ar(ΔDEF) = (AB/DE)² = (15/25)² = (3/5)² = 9/25.
60. A tower 20 m tall casts a shadow 30 m long. Find the height of a pole casting a 15 m shadow. (CBSE 2019)
Answer: 10 m.
Solution: Similar triangles: 20/30 = h/15 → h = (20 × 15)/30 = 10 m.
61. In ΔABC and ΔPQR, AB/PQ = BC/QR = CA/RP. Are they similar? (NCERT 6.3)
Answer: Yes.
Solution: By SSS similarity, ΔABC ~ ΔPQR.
62. In ΔABC, DE || BC, AD = 2 cm, DB = 2 cm, AE = 3 cm. Find EC. (CBSE 2020)
Answer: 3 cm.
Solution: By BPT, AD/DB = AE/EC → 2/2 = 3/EC → EC = 3/1 = 3 cm.
63. In ΔABC, ∠A = 90°, AB = 8 cm, AC = 15 cm. Find BC. (NCERT 6.5)
Answer: 17 cm.
Solution: By Pythagoras: BC² = AB² + AC² = 8² + 15² = 64 + 225 = 289. So, BC = √289 = 17 cm.
64. If ΔABC ~ ΔPQR, ar(ΔABC) = 81 cm², ar(ΔPQR) = 144 cm², find AB/PQ. (CBSE 2018)
Answer: 3/4.
Solution: (AB/PQ)² = 81/144 → AB/PQ = √(81/144) = 9/12 = 3/4.
65. A man 1.5 m tall casts a shadow 2 m long. Find the height of a tree casting a 10 m shadow. (NCERT 6.3)
Answer: 7.5 m.
Solution: Similar triangles: 1.5/2 = h/10 → h = (1.5 × 10)/2 = 7.5 m.
66. In ΔABC, ∠B = ∠P, ∠C = ∠Q. Are ΔABC and ΔPQR similar? (CBSE 2019)
Answer: Yes.
Solution: By AA similarity, ΔABC ~ ΔPQR.
67. In ΔABC, PQ || BC, AP = 6 cm, PB = 4 cm, AQ = 9 cm. Find QC. (NCERT 6.2)
Answer: 6 cm.
Solution: By BPT, AP/PB = AQ/QC → 6/4 = 9/QC → QC = (9 × 4)/6 = 6 cm.
Questions 68–100
68. In ΔABC, ∠C = 90°, AC = 8 cm, BC = 6 cm. Find AB. (CBSE 2020)
Answer: 10 cm.
Solution: By Pythagoras: AB² = AC² + BC² = 8² + 6² = 64 + 36 = 100. So, AB = √100 = 10 cm.
69. If ΔABC ~ ΔDEF, AB = 9 cm, DE = 15 cm, find ar(ΔABC)/ar(ΔDEF). (NCERT 6.4)
Answer: 9/25.
Solution: ar(ΔABC)/ar(ΔDEF) = (AB/DE)² = (9/15)² = (3/5)² = 9/25.
70. A pole 6 m high casts a shadow 4 m long. Find the height of a tree casting a 20 m shadow. (CBSE 2018)
Answer: 30 m.
Solution: Similar triangles: 6/4 = h/20 → h = (6 × 20)/4 = 30 m.
71. In ΔABC and ΔPQR, AB/PQ = BC/QR = CA/RP. Are they similar? (NCERT 6.3)
Answer: Yes.
Solution: By SSS similarity, ΔABC ~ ΔPQR.
72. In ΔABC, DE || BC, AD = 4 cm, DB = 4 cm, AE = 2 cm. Find EC. (CBSE 2019)
Answer: 2 cm.
Solution: By BPT, AD/DB = AE/EC → 4/4 = 2/EC → EC = 2/1 = 2 cm.
73. In ΔABC, ∠A = 90°, AB = 10 cm, AC = 24 cm. Find BC. (NCERT 6.5)
Answer: 26 cm.
Solution: By Pythagoras: BC² = AB² + AC² = 10² + 24² = 100 + 576 = 676. So, BC = √676 = 26 cm.
74. If ΔABC ~ ΔPQR, ar(ΔABC) = 100 cm², ar(ΔPQR) = 169 cm², find AB/PQ. (CBSE 2020)
Answer: 10/13.
Solution: (AB/PQ)² = 100/169 → AB/PQ = √(100/169) = 10/13.
75. A man 1.6 m tall casts a shadow 2 m long. Find the height of a building casting a 15 m shadow. (NCERT 6.3)
Answer: 12 m.
Solution: Similar triangles: 1.6/2 = h/15 → h = (1.6 × 15)/2 = 12 m.
76. In ΔABC, ∠A = ∠P, ∠C = ∠R. Are ΔABC and ΔPQR similar? (CBSE 2018)
Answer: Yes.
Solution: By AA similarity, ΔABC ~ ΔPQR.
77. In ΔABC, PQ || BC, AP = 8 cm, PB = 2 cm, AQ = 12 cm. Find QC. (NCERT 6.2)
Answer: 3 cm.
Solution: By BPT, AP/PB = AQ/QC → 8/2 = 12/QC → QC = 12/4 = 3 cm.
78. In ΔABC, ∠C = 90°, AC = 12 cm, BC = 9 cm. Find AB. (CBSE 2019)
Answer: 15 cm.
Solution: By Pythagoras: AB² = AC² + BC² = 12² + 9² = 144 + 81 = 225. So, AB = √225 = 15 cm.
79. If ΔABC ~ ΔDEF, AB = 6 cm, DE = 10 cm, find ar(ΔABC)/ar(ΔDEF). (NCERT 6.4)
Answer: 9/25.
Solution: ar(ΔABC)/ar(ΔDEF) = (AB/DE)² = (6/10)² = (3/5)² = 9/25.
80. A tree 12 m tall casts a shadow 18 m long. Find the height of a pole casting a 6 m shadow. (CBSE 2020)
Answer: 4 m.
Solution: Similar triangles: 12/18 = h/6 → h = (12 × 6)/18 = 4 m.
81. In ΔABC and ΔPQR, AB/PQ = BC/QR = CA/RP. Are they similar? (NCERT 6.3)
Answer: Yes.
Solution: By SSS similarity, ΔABC ~ ΔPQR.
82. In ΔABC, DE || BC, AD = 3 cm, DB = 3 cm, AE = 2 cm. Find EC. (CBSE 2018)
Answer: 2 cm.
Solution: By BPT, AD/DB = AE/EC → 3/3 = 2/EC → EC = 2/1 = 2 cm.
83. In ΔABC, ∠A = 90°, AB = 7 cm, AC = 24 cm. Find BC. (NCERT 6.5)
Answer: 25 cm.
Solution: By Pythagoras: BC² = AB² + AC² = 7² + 24² = 49 + 576 = 625. So, BC = √625 = 25 cm.
84. If ΔABC ~ ΔPQR, ar(ΔABC) = 121 cm², ar(ΔPQR) = 169 cm², find AB/PQ. (CBSE 2019)
Answer: 11/13.
Solution: (AB/PQ)² = 121/169 → AB/PQ = √(121/169) = 11/13.
85. A pole 5 m high casts a shadow 4 m long. Find the height of a building casting a 16 m shadow. (NCERT 6.3)
Answer: 20 m.
Solution: Similar triangles: 5/4 = h/16 → h = (5 × 16)/4 = 20 m.
86. In ΔABC, ∠B = ∠Q, ∠C = ∠R. Are ΔABC and ΔPQR similar? (CBSE 2020)
Answer: Yes.
Solution: By AA similarity, ΔABC ~ ΔPQR.
87. In ΔABC, PQ || BC, AP = 7 cm, PB = 3 cm, AQ = 14 cm. Find QC. (NCERT 6.2)
Answer: 6 cm.
Solution: By BPT, AP/PB = AQ/QC → 7/3 = 14/QC → QC = (14 × 3)/7 = 6 cm.
88. In ΔABC, ∠C = 90°, AC = 5 cm, BC = 12 cm. Find AB. (CBSE 2018)
Answer: 13 cm.
Solution: By Pythagoras: AB² = AC² + BC² = 5² + 12² = 25 + 144 = 169. So, AB = √169 = 13 cm.
89. If ΔABC ~ ΔDEF, AB = 8 cm, DE = 12 cm, find ar(ΔABC)/ar(ΔDEF). (NCERT 6.4)
Answer: 4/9.
Solution: ar(ΔABC)/ar(ΔDEF) = (AB/DE)² = (8/12)² = (2/3)² = 4/9.
90. A tree 14 m tall casts a shadow 21 m long. Find the height of a pole casting a 9 m shadow. (CBSE 2019)
Answer: 6 m.
Solution: Similar triangles: 14/21 = h/9 → h = (14 × 9)/21 = 6 m.
91. In ΔABC and ΔPQR, AB/PQ = BC/QR = CA/RP. Are they similar? (NCERT 6.3)
Answer: Yes.
Solution: By SSS similarity, ΔABC ~ ΔPQR.
92. In ΔABC, DE || BC, AD = 5 cm, DB = 5 cm, AE = 4 cm. Find EC. (CBSE 2020)
Answer: 4 cm.
Solution: By BPT, AD/DB = AE/EC → 5/5 = 4/EC → EC = 4/1 = 4 cm.
93. In ΔABC, ∠A = 90°, AB = 6 cm, AC = 8 cm. Find BC. (NCERT 6.5)
Answer: 10 cm.
Solution: By Pythagoras: BC² = AB² + AC² = 6² + 8² = 36 + 64 = 100. So, BC = √100 = 10 cm.
94. If ΔABC ~ ΔPQR, ar(ΔABC) = 25 cm², ar(ΔPQR) = 49 cm², find AB/PQ. (CBSE 2018)
Answer: 5/7.
Solution: (AB/PQ)² = 25/49 → AB/PQ = √(25/49) = 5/7.
95. A man 1.8 m tall casts a shadow 3 m long. Find the height of a building casting a 15 m shadow. (NCERT 6.3)
Answer: 9 m.
Solution: Similar triangles: 1.8/3 = h/15 → h = (1.8 × 15)/3 = 9 m.
96. In ΔABC, ∠A = ∠P, ∠B = ∠Q. Are ΔABC and ΔPQR similar? (CBSE 2019)
Answer: Yes.
Solution: By AA similarity, ΔABC ~ ΔPQR.
97. In ΔABC, PQ || BC, AP = 9 cm, PB = 3 cm, AQ = 18 cm. Find QC. (NCERT 6.2)
Answer: 6 cm.
Solution: By BPT, AP/PB = AQ/QC → 9/3 = 18/QC → QC = 18/3 = 6 cm.
98. In ΔABC, ∠C = 90°, AC = 9 cm, BC = 12 cm. Find AB. (CBSE 2020)
Answer: 15 cm.
Solution: By Pythagoras: AB² = AC² + BC² = 9² + 12² = 81 + 144 = 225. So, AB = √225 = 15 cm.
99. If ΔABC ~ ΔDEF, AB = 10 cm, DE = 20 cm, find ar(ΔABC)/ar(ΔDEF). (NCERT 6.4)
Answer: 1/4.
Solution: ar(ΔABC)/ar(ΔDEF) = (AB/DE)² = (10/20)² = (1/2)² = 1/4.
100. A pole 7 m high casts a shadow 5 m long. Find the height of a tree casting a 25 m shadow. (CBSE 2018)
Answer: 35 m.
Solution: Similar triangles: 7/5 = h/25 → h = (7 × 25)/5 = 35 m.
Download the free PDF with 100 Triangles Questions and NCERT book solutions for Class 10 Maths to boost your 2025 exam preparation!
Frequently Asked Questions
What is the Basic Proportionality Theorem (BPT)?
In a triangle, if a line parallel to one side intersects the other two sides, it divides them proportionally. For ΔABC with DE || BC, AD/DB = AE/EC.
What are the criteria for similarity of triangles?
Triangles are similar if they satisfy: (1) AA (two angles equal), (2) SSS (three sides proportional), or (3) SAS (two sides proportional and included angle equal).
How is Pythagoras theorem applied in triangles?
In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides: a² + b² = c², where c is the hypotenuse.
How to find the ratio of areas of similar triangles?
The ratio of areas of similar triangles is the square of the ratio of their corresponding sides: ar(Δ1)/ar(Δ2) = (side1/side2)².
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