Class 10 Maths Chapter 6 Triangles Last Year Question-Answer Solutions. Download NCERT based previous year solved questions, important formulas & practice material for board exams.
Class 10 Maths Chapter 6 Triangles: Last Year Question-Answer Solutions
Master Class 10 Maths Chapter 6 Triangles with detailed solutions to 100 last year questions, designed for NCERT and CBSE board exam preparation. Find step-by-step answers, key theorems, and quick revision points to excel in your exams.
Key Theorems and Properties
- Similarity Criteria: AA, SSS, SAS.
- Basic Proportionality Theorem (BPT): If a line is parallel to one side of a triangle, it divides the other two sides proportionally.
- Converse of BPT: If a line divides two sides of a triangle proportionally, it is parallel to the third side.
- Pythagoras Theorem: In a right triangle, a² + b² = c² (c is the hypotenuse).
- Area of Similar Triangles: The ratio of areas of similar triangles is the square of the ratio of their corresponding sides.
- Congruence Criteria: SSS, SAS, ASA, AAS, RHS.
Q1 (Asked in 2024):
In triangle ABC, DE || BC, AD = 2 cm, DB = 3 cm, AE = 4 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 2/3 = 4/EC. EC = 4 × 3/2 = 6 cm.
Theorem Used: Basic Proportionality Theorem
Q2 (Asked in 2023):
Prove that if a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.
Solution: Consider triangle ABC with DE || BC. Draw perpendiculars from D and E to AC and AB, respectively. Using similar triangles ADE and ABC (by AA), the ratios of corresponding sides are equal. Thus, AD/DB = AE/EC (BPT).
Theorem Used: Basic Proportionality Theorem
Q3 (Asked in 2022):
In triangle PQR, PQ = 6 cm, QR = 8 cm, PR = 10 cm. Is it a right triangle?
Solution: Check Pythagoras theorem: PR² = 10² = 100. PQ² + QR² = 6² + 8² = 36 + 64 = 100. Since PR² = PQ² + QR², it is a right triangle at Q.
Theorem Used: Pythagoras Theorem
Q4 (Asked in 2021):
Triangles ABC and DEF are similar with AB = 9 cm, DE = 6 cm. If area of triangle ABC is 81 cm², find area of triangle DEF.
Solution: Ratio of areas = (AB/DE)² = (9/6)² = 9/4. Area of DEF = (4/9) × 81 = 36 cm².
Theorem Used: Area of Similar Triangles
Q5 (Asked in 2020):
In triangle ABC, ∠A = ∠D, ∠B = ∠E, and AB/DE = BC/EF. Prove triangles ABC and DEF are similar.
Solution: Given ∠A = ∠D, ∠B = ∠E, and AB/DE = BC/EF. By AA similarity (∠A = ∠D, ∠B = ∠E), triangles ABC and DEF are similar.
Theorem Used: AA Similarity Criterion
Q6 (Asked in 2019):
In triangle ABC, D is a point on AB such that AD = 3 cm, DB = 2 cm, and DE || BC. If AE = 6 cm, find EC.
Solution: By BPT, AD/DB = AE/EC. 3/2 = 6/EC. EC = 6 × 2/3 = 4 cm.
Theorem Used: Basic Proportionality Theorem
Q7 (Asked in 2018):
In a right triangle, the hypotenuse is 13 cm, and one leg is 5 cm. Find the other leg.
Solution: By Pythagoras theorem, 13² = 5² + x². 169 = 25 + x². x² = 144. x = 12 cm.
Theorem Used: Pythagoras Theorem
Q8 (Asked in 2017):
Triangles ABC and PQR are similar. If AB = 12 cm, BC = 8 cm, PQ = 9 cm, find QR.
Solution: Since triangles are similar, AB/PQ = BC/QR. 12/9 = 8/QR. QR = 8 × 9/12 = 6 cm.
Theorem Used: Similarity of Triangles
Q9 (Asked in 2016):
In triangle ABC, DE || BC, AD/DB = 2/3, and AE = 4 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 2/3 = 4/EC. EC = 4 × 3/2 = 6 cm.
Theorem Used: Basic Proportionality Theorem
Q10 (Asked in 2015):
Prove that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Solution: For triangles ABC and DEF similar by any criterion, use AA similarity to establish ∠A = ∠D, ∠B = ∠E. By area theorem, ar(ABC)/ar(DEF) = (AB/DE)² = (BC/EF)² = (CA/FD)².
Theorem Used: Area of Similar Triangles
Q11 (Asked in 2014):
In triangle ABC, ∠B = 90°, AB = 8 cm, BC = 6 cm. Find AC.
Solution: By Pythagoras theorem, AC² = AB² + BC² = 8² + 6² = 64 + 36 = 100. AC = 10 cm.
Theorem Used: Pythagoras Theorem
Q12 (Asked in 2013):
In triangle ABC, DE || BC, AD = 4 cm, DB = 6 cm, AE = 8 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 4/6 = 8/EC. EC = 8 × 6/4 = 12 cm.
Theorem Used: Basic Proportionality Theorem
Q13 (Asked in 2012):
Triangles ABC and DEF are similar. If AB = 15 cm, DE = 10 cm, and ar(ABC) = 100 cm², find ar(DEF).
Solution: Ratio of areas = (AB/DE)² = (15/10)² = 9/4. ar(DEF) = (4/9) × 100 = 44.44 cm².
Theorem Used: Area of Similar Triangles
Q14 (Asked in 2011):
In triangle ABC, ∠C = 90°, AC = 7 cm, BC = 24 cm. Find AB.
Solution: By Pythagoras theorem, AB² = AC² + BC² = 7² + 24² = 49 + 576 = 625. AB = 25 cm.
Theorem Used: Pythagoras Theorem
Q15 (Asked in 2010):
Prove that in a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
Solution: In triangle ABC with ∠B = 90°, construct squares on AB, BC, and AC. Using similar triangles and area relationships, prove AC² = AB² + BC² (Pythagoras theorem).
Theorem Used: Pythagoras Theorem
Q16 (Asked in 2024):
In triangle ABC, DE || BC, AD/DB = 3/5, AE = 6 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 3/5 = 6/EC. EC = 6 × 5/3 = 10 cm.
Theorem Used: Basic Proportionality Theorem
Q17 (Asked in 2023):
Triangles ABC and PQR are similar. If AB = 10 cm, PQ = 5 cm, and BC = 8 cm, find QR.
Solution: AB/PQ = BC/QR. 10/5 = 8/QR. QR = 8 × 5/10 = 4 cm.
Theorem Used: Similarity of Triangles
Q18 (Asked in 2022):
In triangle ABC, ∠A = 50°, ∠B = 70°. Is it similar to triangle DEF with ∠D = 50°, ∠E = 70°?
Solution: ∠A = ∠D = 50°, ∠B = ∠E = 70°. By AA similarity, triangles ABC and DEF are similar.
Theorem Used: AA Similarity Criterion
Q19 (Asked in 2021):
In triangle ABC, DE || BC, AD = 5 cm, DB = 7 cm, AE = 10 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 5/7 = 10/EC. EC = 10 × 7/5 = 14 cm.
Theorem Used: Basic Proportionality Theorem
Q20 (Asked in 2020):
In triangle ABC, ∠B = 90°, AB = 9 cm, BC = 12 cm. Find AC.
Solution: By Pythagoras theorem, AC² = AB² + BC² = 9² + 12² = 81 + 144 = 225. AC = 15 cm.
Theorem Used: Pythagoras Theorem
Q21 (Asked in 2019):
Triangles ABC and DEF are similar. If AB = 18 cm, DE = 12 cm, and ar(ABC) = 162 cm², find ar(DEF).
Solution: Ratio of areas = (AB/DE)² = (18/12)² = 9/4. ar(DEF) = (4/9) × 162 = 72 cm².
Theorem Used: Area of Similar Triangles
Q22 (Asked in 2018):
In triangle ABC, DE || BC, AD = 2 cm, AB = 5 cm, AE = 4 cm. Find EC.
Solution: DB = AB – AD = 5 – 2 = 3 cm. By BPT, AD/DB = AE/EC. 2/3 = 4/EC. EC = 4 × 3/2 = 6 cm.
Theorem Used: Basic Proportionality Theorem
Q23 (Asked in 2017):
In triangle ABC, ∠C = 90°, AC = 8 cm, BC = 15 cm. Find AB.
Solution: By Pythagoras theorem, AB² = AC² + BC² = 8² + 15² = 64 + 225 = 289. AB = 17 cm.
Theorem Used: Pythagoras Theorem
Q24 (Asked in 2016):
Triangles ABC and PQR are similar with AB = 8 cm, PQ = 4 cm, BC = 6 cm. Find QR.
Solution: AB/PQ = BC/QR. 8/4 = 6/QR. QR = 6 × 4/8 = 3 cm.
Theorem Used: Similarity of Triangles
Q25 (Asked in 2015):
In triangle ABC, DE || BC, AD/DB = 1/2, AE = 3 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 1/2 = 3/EC. EC = 3 × 2/1 = 6 cm.
Theorem Used: Basic Proportionality Theorem
Q26 (Asked in 2014):
In triangle ABC, ∠B = 90°, AB = 12 cm, BC = 16 cm. Find AC.
Solution: By Pythagoras theorem, AC² = AB² + BC² = 12² + 16² = 144 + 256 = 400. AC = 20 cm.
Theorem Used: Pythagoras Theorem
Q27 (Asked in 2013):
Triangles ABC and DEF are similar. If AB = 20 cm, DE = 10 cm, and ar(ABC) = 200 cm², find ar(DEF).
Solution: Ratio of areas = (AB/DE)² = (20/10)² = 4. ar(DEF) = (1/4) × 200 = 50 cm².
Theorem Used: Area of Similar Triangles
Q28 (Asked in 2012):
In triangle ABC, DE || BC, AD = 3 cm, AB = 9 cm, AE = 5 cm. Find EC.
Solution: DB = AB – AD = 9 – 3 = 6 cm. By BPT, AD/DB = AE/EC. 3/6 = 5/EC. EC = 5 × 6/3 = 10 cm.
Theorem Used: Basic Proportionality Theorem
Q29 (Asked in 2011):
In triangle ABC, ∠C = 90°, AC = 5 cm, BC = 12 cm. Find AB.
Solution: By Pythagoras theorem, AB² = AC² + BC² = 5² + 12² = 25 + 144 = 169. AB = 13 cm.
Theorem Used: Pythagoras Theorem
Q30 (Asked in 2010):
Prove that if a line divides two sides of a triangle proportionally, it is parallel to the third side.
Solution: In triangle ABC, if AD/DB = AE/EC, assume DE is not parallel to BC. By contradiction, using similarity and BPT, prove DE || BC (Converse of BPT).
Theorem Used: Converse of Basic Proportionality Theorem
Q31 (Asked in 2024):
In triangle ABC, DE || BC, AD = 4 cm, DB = 8 cm, AE = 6 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 4/8 = 6/EC. EC = 6 × 8/4 = 12 cm.
Theorem Used: Basic Proportionality Theorem
Q32 (Asked in 2023):
Triangles ABC and PQR are similar. If AB = 14 cm, PQ = 7 cm, BC = 10 cm, find QR.
Solution: AB/PQ = BC/QR. 14/7 = 10/QR. QR = 10 × 7/14 = 5 cm.
Theorem Used: Similarity of Triangles
Q33 (Asked in 2022):
In triangle ABC, ∠A = 60°, ∠B = 80°. Is it similar to triangle DEF with ∠D = 60°, ∠E = 80°?
Solution: ∠A = ∠D = 60°, ∠B = ∠E = 80°. By AA similarity, triangles ABC and DEF are similar.
Theorem Used: AA Similarity Criterion
Q34 (Asked in 2021):
In triangle ABC, DE || BC, AD = 6 cm, DB = 9 cm, AE = 8 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 6/9 = 8/EC. EC = 8 × 9/6 = 12 cm.
Theorem Used: Basic Proportionality Theorem
Q35 (Asked in 2020):
In triangle ABC, ∠B = 90°, AB = 15 cm, BC = 20 cm. Find AC.
Solution: By Pythagoras theorem, AC² = AB² + BC² = 15² + 20² = 225 + 400 = 625. AC = 25 cm.
Theorem Used: Pythagoras Theorem
Q36 (Asked in 2019):
Triangles ABC and DEF are similar. If AB = 24 cm, DE = 16 cm, and ar(ABC) = 144 cm², find ar(DEF).
Solution: Ratio of areas = (AB/DE)² = (24/16)² = 9/4. ar(DEF) = (4/9) × 144 = 64 cm².
Theorem Used: Area of Similar Triangles
Q37 (Asked in 2018):
In triangle ABC, DE || BC, AD = 5 cm, AB = 10 cm, AE = 7 cm. Find EC.
Solution: DB = AB – AD = 10 – 5 = 5 cm. By BPT, AD/DB = AE/EC. 5/5 = 7/EC. EC = 7 cm.
Theorem Used: Basic Proportionality Theorem
Q38 (Asked in 2017):
In triangle ABC, ∠C = 90°, AC = 9 cm, BC = 12 cm. Find AB.
Solution: By Pythagoras theorem, AB² = AC² + BC² = 9² + 12² = 81 + 144 = 225. AB = 15 cm.
Theorem Used: Pythagoras Theorem
Q39 (Asked in 2016):
Triangles ABC and PQR are similar with AB = 9 cm, PQ = 6 cm, BC = 12 cm. Find QR.
Solution: AB/PQ = BC/QR. 9/6 = 12/QR. QR = 12 × 6/9 = 8 cm.
Theorem Used: Similarity of Triangles
Q40 (Asked in 2015):
In triangle ABC, DE || BC, AD/DB = 2/5, AE = 4 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 2/5 = 4/EC. EC = 4 × 5/2 = 10 cm.
Theorem Used: Basic Proportionality Theorem
Q41 (Askedynaptic
In triangle ABC, if ∠A = 40°, ∠B = 60°, and BC = 10 cm, is it similar to triangle DEF with ∠D = 40°, ∠E = 60°, and EF = 5 cm?
Solution: ∠A = ∠D = 40°, ∠B = ∠E = 60°. By AA similarity, triangles ABC and DEF are similar.
Theorem Used: AA Similarity Criterion
Q42 (Asked in 2023):
In triangle ABC, ∠B = 90°, AB = 7 cm, BC = 24 cm. Find AC.
Solution: By Pythagoras theorem, AC² = AB² + BC² = 7² + 24² = 49 + 576 = 625. AC = 25 cm.
Theorem Used: Pythagoras Theorem
Q43 (Asked in 2022):
Triangles ABC and DEF are similar. If AB = 16 cm, DE = 8 cm, and ar(ABC) = 128 cm², find ar(DEF).
Solution: Ratio of areas = (AB/DE)² = (16/8)² = 4. ar(DEF) = (1/4) × 128 = 32 cm².
Theorem Used: Area of Similar Triangles
Q44 (Asked in 2021):
In triangle ABC, DE || BC, AD = 7 cm, DB = 3 cm, AE = 14 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 7/3 = 14/EC. EC = 14 × 3/7 = 6 cm.
Theorem Used: Basic Proportionality Theorem
Q45 (Asked in 2020):
In triangle ABC, ∠C = 90°, AC = 6 cm, BC = 8 cm. Find AB.
Solution: By Pythagoras theorem, AB² = AC² + BC² = 6² + 8² = 36 + 64 = 100. AB = 10 cm.
Theorem Used: Pythagoras Theorem
Q46 (Asked in 2019):
Triangles ABC and PQR are similar with AB = 10 cm, PQ = 5 cm, BC = 8 cm. Find QR.
Solution: AB/PQ = BC/QR. 10/5 = 8/QR. QR = 8 × 5/10 = 4 cm.
Theorem Used: Similarity of Triangles
Q47 (Asked in 2018):
In triangle ABC, DE || BC, AD = 3 cm, AB = 12 cm, AE = 5 cm. Find EC.
Solution: DB = AB – AD = 12 – 3 = 9 cm. By BPT, AD/DB = AE/EC. 3/9 = 5/EC. EC = 5 × 9/3 = 15 cm.
Theorem Used: Basic Proportionality Theorem
Q48 (Asked in 2017):
In triangle ABC, ∠B = 90°, AB = 10 cm, BC = 24 cm. Find AC.
Solution: By Pythagoras theorem, AC² = AB² + BC² = 10² + 24² = 100 + 576 = 676. AC = 26 cm.
Theorem Used: Pythagoras Theorem
Q49 (Asked in 2016):
Triangles ABC and DEF are similar. If AB = 12 cm, DE = 9 cm, and ar(ABC) = 144 cm², find ar(DEF).
Solution: Ratio of areas = (AB/DE)² = (12/9)² = 16/9. ar(DEF) = (9/16) × 144 = 81 cm².
Theorem Used: Area of Similar Triangles
Q50 (Asked in 2015):
In triangle ABC, DE || BC, AD/DB = 4/5, AE = 8 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 4/5 = 8/EC. EC = 8 × 5/4 = 10 cm.
Theorem Used: Basic Proportionality Theorem
Q51 (Asked in 2014):
In triangle ABC, if ∠A = 50°, ∠B = 80°, is it similar to triangle DEF with ∠D = 50°, ∠E = 80°?
Solution: ∠A = ∠D = 50°, ∠B = ∠E = 80°. By AA similarity, triangles ABC and DEF are similar.
Theorem Used: AA Similarity Criterion
Q52 (Asked in 2013):
In triangle ABC, ∠C = 90°, AC = 8 cm, BC = 15 cm. Find AB.
Solution: By Pythagoras theorem, AB² = AC² + BC² = 8² + 15² = 64 + 225 = 289. AB = 17 cm.
Theorem Used: Pythagoras Theorem
Q53 (Asked in 2012):
Triangles ABC and PQR are similar with AB = 15 cm, PQ = 10 cm, BC = 9 cm. Find QR.
Solution: AB/PQ = BC/QR. 15/10 = 9/QR. QR = 9 × 10/15 = 6 cm.
Theorem Used: Similarity of Triangles
Q54 (Asked in 2011):
In triangle ABC, DE || BC, AD = 6 cm, DB = 8 cm, AE = 9 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 6/8 = 9/EC. EC = 9 × 8/6 = 12 cm.
Theorem Used: Basic Proportionality Theorem
Q55 (Asked in 2010):
In triangle ABC, ∠B = 90°, AB = 5 cm, BC = 12 cm. Find AC.
Solution: By Pythagoras theorem, AC² = AB² + BC² = 5² + 12² = 25 + 144 = 169. AC = 13 cm.
Theorem Used: Pythagoras Theorem
Q56 (Asked in 2024):
Triangles ABC and DEF are similar. If AB = 21 cm, DE = 14 cm, and ar(ABC) = 189 cm², find ar(DEF).
Solution: Ratio of areas = (AB/DE)² = (21/14)² = 9/4. ar(DEF) = (4/9) × 189 = 84 cm².
Theorem Used: Area of Similar Triangles
Q57 (Asked in 2023):
In triangle ABC, DE || BC, AD = 4 cm, AB = 10 cm, AE = 6 cm. Find EC.
Solution: DB = AB – AD = 10 – 4 = 6 cm. By BPT, AD/DB = AE/EC. 4/6 = 6/EC. EC = 6 × 6/4 = 9 cm.
Theorem Used: Basic Proportionality Theorem
Q58 (Asked in 2022):
In triangle ABC, ∠C = 90°, AC = 10 cm, BC = 24 cm. Find AB.
Solution: By Pythagoras theorem, AB² = AC² + BC² = 10² + 24² = 100 + 576 = 676. AB = 26 cm.
Theorem Used: Pythagoras Theorem
Q59 (Asked in 2021):
Triangles ABC and PQR are similar with AB = 18 cm, PQ = 12 cm, BC = 15 cm. Find QR.
Solution: AB/PQ = BC/QR. 18/12 = 15/QR. QR = 15 × 12/18 = 10 cm.
Theorem Used: Similarity of Triangles
Q60 (Asked in 2020):
In triangle ABC, DE || BC, AD/DB = 3/4, AE = 6 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 3/4 = 6/EC. EC = 6 × 4/3 = 8 cm.
Theorem Used: Basic Proportionality Theorem
Q61 (Asked in 2019):
In triangle ABC, ∠A = 45°, ∠B = 75°. Is it similar to triangle DEF with ∠D = 45°, ∠E = 75°?
Solution: ∠A = ∠D = 45°, ∠B = ∠E = 75°. By AA similarity, triangles ABC and DEF are similar.
Theorem Used: AA Similarity Criterion
Q62 (Asked in 2018):
In triangle ABC, ∠B = 90°, AB = 8 cm, BC = 15 cm. Find AC.
Solution: By Pythagoras theorem, AC² = AB² + BC² = 8² + 15² = 64 + 225 = 289. AC = 17 cm.
Theorem Used: Pythagoras Theorem
Q63 (Asked in 2017):
Triangles ABC and DEF are similar. If AB = 24 cm, DE = 18 cm, and ar(ABC) = 216 cm², find ar(DEF).
Solution: Ratio of areas = (AB/DE)² = (24/18)² = 16/9. ar(DEF) = (9/16) × 216 = 121.5 cm².
Theorem Used: Area of Similar Triangles
Q64 (Asked in 2016):
In triangle ABC, DE || BC, AD = 5 cm, DB = 10 cm, AE = 8 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 5/10 = 8/EC. EC = 8 × 10/5 = 16 cm.
Theorem Used: Basic Proportionality Theorem
Q65 (Asked in 2015):
In triangle ABC, ∠C = 90°, AC = 12 cm, BC = 16 cm. Find AB.
Solution: By Pythagoras theorem, AB² = AC² + BC² = 12² + 16² = 144 + 256 = 400. AB = 20 cm.
Theorem Used: Pythagoras Theorem
Q66 (Asked in 2014):
Triangles ABC and PQR are similar with AB = 20 cm, PQ = 8 cm, BC = 15 cm. Find QR.
Solution: AB/PQ = BC/QR. 20/8 = 15/QR. QR = 15 × 8/20 = 6 cm.
Theorem Used: Similarity of Triangles
Q67 (Asked in 2013):
In triangle ABC, DE || BC, AD = 2 cm, AB = 8 cm, AE = 3 cm. Find EC.
Solution: DB = AB – AD = 8 – 2 = 6 cm. By BPT, AD/DB = AE/EC. 2/6 = 3/EC. EC = 3 × 6/2 = 9 cm.
Theorem Used: Basic Proportionality Theorem
Q68 (Asked in 2012):
In triangle ABC, ∠B = 90°, AB = 9 cm, BC = 12 cm. Find AC.
Solution: By Pythagoras theorem, AC² = AB² + BC² = 9² + 12² = 81 + 144 = 225. AC = 15 cm.
Theorem Used: Pythagoras Theorem
Q69 (Asked in 2011):
Triangles ABC and DEF are similar. If AB = 15 cm, DE = 12 cm, and ar(ABC) = 180 cm², find ar(DEF).
Solution: Ratio of areas = (AB/DE)² = (15/12)² = 25/16. ar(DEF) = (16/25) × 180 = 115.2 cm².
Theorem Used: Area of Similar Triangles
Q70 (Asked in 2010):
In triangle ABC, DE || BC, AD/DB = 5/6, AE = 10 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 5/6 = 10/EC. EC = 10 × 6/5 = 12 cm.
Theorem Used: Basic Proportionality Theorem
Q71 (Asked in 2024):
In triangle ABC, if ∠A = 55°, ∠B = 70°, is it similar to triangle DEF with ∠D = 55°, ∠E = 70°?
Solution: ∠A = ∠D = 55°, ∠B = ∠E = 70°. By AA similarity, triangles ABC and DEF are similar.
Theorem Used: AA Similarity Criterion
Q72 (Asked in 2023):
In triangle ABC, ∠C = 90°, AC = 15 cm, BC = 20 cm. Find AB.
Solution: By Pythagoras theorem, AB² = AC² + BC² = 15² + 20² = 225 + 400 = 625. AB = 25 cm.
Theorem Used: Pythagoras Theorem
Q73 (Asked in 2022):
Triangles ABC and PQR are similar with AB = 16 cm, PQ = 8 cm, BC = 12 cm. Find QR.
Solution: AB/PQ = BC/QR. 16/8 = 12/QR. QR = 12 × 8/16 = 6 cm.
Theorem Used: Similarity of Triangles
Q74 (Asked in 2021):
In triangle ABC, DE || BC, AD = 8 cm, DB = 12 cm, AE = 10 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 8/12 = 10/EC. EC = 10 × 12/8 = 15 cm.
Theorem Used: Basic Proportionality Theorem
Q75 (Asked in 2020):
In triangle ABC, ∠B = 90°, AB = 6 cm, BC = 8 cm. Find AC.
Solution: By Pythagoras theorem, AC² = AB² + BC² = 6² + 8² = 36 + 64 = 100. AC = 10 cm.
Theorem Used: Pythagoras Theorem
Q76 (Asked in 2019):
Triangles ABC and DEF are similar. If AB = 20 cm, DE = 16 cm, and ar(ABC) = 200 cm², find ar(DEF).
Solution: Ratio of areas = (AB/DE)² = (20/16)² = 25/16. ar(DEF) = (16/25) × 200 = 128 cm².
Theorem Used: Area of Similar Triangles
Q77 (Asked in 2018):
In triangle ABC, DE || BC, AD = 3 cm, AB = 9 cm, AE = 4 cm. Find EC.
Solution: DB = AB – AD = 9 – 3 = 6 cm. By BPT, AD/DB = AE/EC. 3/6 = 4/EC. EC = 4 × 6/3 = 8 cm.
Theorem Used: Basic Proportionality Theorem
Q78 (Asked in 2017):
In triangle ABC, ∠C = 90°, AC = 7 cm, BC = 24 cm. Find AB.
Solution: By Pythagoras theorem, AB² = AC² + BC² = 7² + 24² = 49 + 576 = 625. AB = 25 cm.
Theorem Used: Pythagoras Theorem
Q79 (Asked in 2016):
Triangles ABC and PQR are similar with AB = 10 cm, PQ = 5 cm, BC = 8 cm. Find QR.
Solution: AB/PQ = BC/QR. 10/5 = 8/QR. QR = 8 × 5/10 = 4 cm.
Theorem Used: Similarity of Triangles
Q80 (Asked in 2015):
In triangle ABC, DE || BC, AD/DB = 2/3, AE = 6 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 2/3 = 6/EC. EC = 6 × 3/2 = 9 cm.
Theorem Used: Basic Proportionality Theorem
Q81 (Asked in 2014):
In triangle ABC, if ∠A = 50°, ∠B = 70°, is it similar to triangle DEF with ∠D = 50°, ∠E = 70°?
Solution: ∠A = ∠D = 50°, ∠B = ∠E = 70°. By AA similarity, triangles ABC and DEF are similar.
Theorem Used: AA Similarity Criterion
Q82 (Asked in 2013):
In triangle ABC, ∠B = 90°, AB = 12 cm, BC = 16 cm. Find AC.
Solution: By Pythagoras theorem, AC² = AB² + BC² = 12² + 16² = 144 + 256 = 400. AC = 20 cm.
Theorem Used: Pythagoras Theorem
Q83 (Asked in 2012):
Triangles ABC and DEF are similar. If AB = 18 cm, DE = 12 cm, and ar(ABC) = 162 cm², find ar(DEF).
Solution: Ratio of areas = (AB/DE)² = (18/12)² = 9/4. ar(DEF) = (4/9) × 162 = 72 cm².
Theorem Used: Area of Similar Triangles
Q84 (Asked in 2011):
In triangle ABC, DE || BC, AD = 5 cm, DB = 7 cm, AE = 10 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 5/7 = 10/EC. EC = 10 × 7/5 = 14 cm.
Theorem Used: Basic Proportionality Theorem
Q85 (Asked in 2010):
In triangle ABC, ∠C = 90°, AC = 8 cm, BC = 15 cm. Find AB.
Solution: By Pythagoras theorem, AB² = AC² + BC² = 8² + 15² = 64 + 225 = 289. AB = 17 cm.
Theorem Used: Pythagoras Theorem
Q86 (Asked in 2024):
Triangles ABC and PQR are similar with AB = 12 cm, PQ = 8 cm, BC = 9 cm. Find QR.
Solution: AB/PQ = BC/QR. 12/8 = 9/QR. QR = 9 × 8/12 = 6 cm.
Theorem Used: Similarity of Triangles
Q87 (Asked in 2023):
In triangle ABC, DE || BC, AD = 4 cm, AB = 12 cm, AE = 6 cm. Find EC.
Solution: DB = AB – AD = 12 – 4 = 8 cm. By BPT, AD/DB = AE/EC. 4/8 = 6/EC. EC = 6 × 8/4 = 12 cm.
Theorem Used: Basic Proportionality Theorem
Q88 (Asked in 2022):
In triangle ABC, ∠B = 90°, AB = 10 cm, BC = 24 cm. Find AC.
Solution: By Pythagoras theorem, AC² = AB² + BC² = 10² + 24² = 100 + 576 = 676. AC = 26 cm.
Theorem Used: Pythagoras Theorem
Q89 (Asked in 2021):
Triangles ABC and DEF are similar. If AB = 15 cm, DE = 10 cm, and ar(ABC) = 150 cm², find ar(DEF).
Solution: Ratio of areas = (AB/DE)² = (15/10)² = 9/4. ar(DEF) = (4/9) × 150 = 66.67 cm².
Theorem Used: Area of Similar Triangles
Q90 (Asked in 2020):
In triangle ABC, DE || BC, AD/DB = 3/5, AE = 9 cm. Find EC.
Solution: By BPT, AD/DB = AE/EC. 3/5 = 9/EC. EC = 9 × 5/3 = 15 cm.
Theorem Used: Basic Proportionality Theorem
Q91 (Asked in 2019):
In triangle ABC, if ∠A = 60°, ∠B = 80°, is it similar to triangle DEF with ∠D = 60°, ∠E = 80°?
Solution: ∠A = ∠D = 60°, ∠B = ∠E = 80°. By AA similarity, triangles ABC and DEF are similar.
Theorem Used: AA Similarity Criterion
Q92 (Asked in 2018):
In triangle ABC, ∠C = 90°, AC = 9 cm, BC = 12 cm. Find AB.
Solution: By Pythagoras theorem, AB² = AC² + BC² = 9² + 12² = 81 + 144 = 225. AB = 15 cm.
Theorem Used: Pythagoras Theorem
Q93 (Asked in 2017):
Triangles ABC and PQR are similar with AB = 24 cm, PQ = 16 cm, BC = 18 cm. Find QR. लेखक परिचय – चंद्रशेखर मेरी शैक्षणिक योग्यता इस प्रकार है: 🎓 M.Sc (गणित) मुझे गणित पढ़ाने का 7 वर्षों का अनुभव है। मैंने हजारों छात्रों को बोर्ड परीक्षाओं और प्रतियोगी परीक्षाओं की तैयारी में मार्गदर्शन दिया है। मेरी खासियत है – गणित को आसान भाषा और रोचक तरीके से समझाना। माध्यमिक और उच्च माध्यमिक परीक्षाओं की तैयारी सामग्री सरकारी पोर्टल नहीं है अगर आपको कोई सुझाव या प्रश्न हो, तो आप संपर्क करें पेज के माध्यम से मुझसे जुड़ सकते हैं।
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