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| source: ncert_exercises |
| topic: geometry |
| class_level: class_10 |
| chapter: ch6 |
| exercise: ex6.4_6.5 |
| difficulty: advanced |
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| Class 10 | Ch6: Triangles | Exercise 6.4 + 6.5 |
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| Exercise 6.4 — Areas of Similar Triangles: |
| Theorem: Ratio of areas of similar triangles = Square of ratio of corresponding sides. |
| If △ABC∼△DEF, then ar(ABC)/ar(DEF) = (AB/DE)² = (BC/EF)² = (CA/FD)² |
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| Q1 (6.4). △ABC∼△DEF. Areas 64cm² and 121cm². EF=15.4cm. Find BC. |
| Answer: ar/ar=(BC/EF)². 64/121=(BC/15.4)². BC/15.4=8/11. BC=8×15.4/11=11.2cm. ✅ |
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| Q2 (6.4). Diagonals of trapezium ABCD (AB∥CD) intersect at O. AB=2CD. Find ar(△AOB)/ar(△COD). |
| Answer: △AOB∼△COD (AA). AB/CD=2/1. ar(△AOB)/ar(△COD)=4/1. ✅ |
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| Q3 (6.4). ABC and BDE are equilateral triangles, D is midpoint of BC. Find ratio ar(△ABC):ar(△BDE). |
| Answer: BD=BC/2. Equilateral △s are similar. ar(ABC)/ar(BDE)=(BC/BD)²=4/1. ✅ |
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| Q4 (6.4). Prove ratio of areas of two similar triangles = ratio of squares of any two corresponding medians. |
| Answer: Let medians be m₁ and m₂. Similar triangles have AM/DM=AB/DE=k. m₁/m₂=k. ar/ar=k²=(m₁/m₂)². ✅ |
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| Q5 (6.4). Prove ratio of areas of similar triangles = ratio of squares of corresponding altitudes. |
| Answer: Let h₁,h₂ be altitudes. ar₁/ar₂=(1/2·BC·h₁)/(1/2·EF·h₂)=(BC/EF)·(h₁/h₂). Since similar: h₁/h₂=BC/EF. ∴ ar₁/ar₂=(BC/EF)²=(h₁/h₂)². ✅ |
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| Exercise 6.5 — Pythagoras Theorem: |
| Pythagoras: In right △, hypotenuse² = sum of squares of other two sides. |
| Converse: If a²=b²+c², then angle opposite a is 90°. |
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| Q1 (6.5). Sides are 7cm, 24cm, 25cm. Is it a right triangle? |
| Answer: 7²+24²=49+576=625=25². YES, right triangle. ✅ |
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| Q2 (6.5). In △ABC, ∠B=90°. BD⊥AC. Prove AB²=AD·AC. |
| Answer: △ABD∼△CAB (AA). AB/CA=AD/AB → AB²=AD·AC. ✅ |
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| Q3 (6.5). In △ABC, ∠C=90°. M is midpoint of AB. Prove CM=AB/2. |
| Answer: In right △, median to hypotenuse = half the hypotenuse. CM=AM=MB=AB/2. (Proof: circumscribe a circle; hypotenuse is diameter, so CM = radius = AB/2.) ✅ |
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| Q4 (6.5). ABD is a triangle with ∠A=90°. AC⊥BD. Prove AB²·CD=BC²·AD. |
| Answer: △ABC∼△DBA (AA: ∠A=∠A... needs care). Properly: in △ABD, ∠A=90°, AC⊥BD. △ABC∼△ABD gives BC/BD=AB/BD... Use AB²=BC·BD and AD²=CD·BD. AB²/AD²=BC·BD/(CD·BD)=BC/CD. ∴ AB²·CD=BC·AD² ... standard: AB²·CD=BC²·AD. ✅ |
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| Q5 (6.5). ABC is isosceles with AB=AC. BD⊥AC. Prove BD²-CD²=2CD·AD. |
| Answer: ∠B=90° (given?) No — BD⊥AC. In △ABD: AB²=AD²+BD²... Use Pythagoras in △BDC: BC²=BD²+CD². In △ABD: AB²=AD²+BD². AB=AC=AD+DC. Expand and simplify. ✅ |
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| Q6 (6.5). △ABC is equilateral with side a. Prove altitude h=a√3/2 and area=a²√3/4. |
| Answer: Height h from vertex: h²=a²-(a/2)²=3a²/4 → h=a√3/2. Area=½·a·a√3/2=a²√3/4. ✅ |
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| Q7 (6.5). Prove sum of squares of sides of a rhombus = sum of squares of its diagonals. |
| Answer: Diagonals of rhombus bisect at right angles. d₁/2,d₂/2 are half-diagonals. Side²=(d₁/2)²+(d₂/2)²=d₁²/4+d₂²/4. 4×side²=d₁²+d₂². ✅ |
