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  1. Chapter 6 Class 8 Squares and Square Roots
  2. Serial order wise
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Ex 6.4, 7 In a right triangle ABC, ∠B = 90°. (a) If AB = 6 cm, BC = 8 cm, find ACHypotenuse is the side opposite to right angle. Given, ∠B = 90° AB = 6 cm BC = 8 cm In right angled triangle ∆ABC, Using Pythagoras theorem, (Hypotenuse)2 = (Base)2 + (Perpendicular)2 AC2 = BC2 + AB2 AC2 = 82 + 62 AC2 = 64 + 36 AC2 = 100 AC = √100 Finding Square root of 100 by Prime factorization :- We see that, 100 = 2 × 2 × 5 × 5 ∴ √100 = 2 × 5 = 10 Therefore, AC = 10 cm Ex 6.4, 7 In a right triangle ABC, ∠B = 90°. (b) If AC = 13 cm, BC = 5 cm, find ABHypotenuse is the side of the triangle opposite to the right angle. Given, ∠B = 90° AC = 13 cm BC = 5 cm In ∆ABC right − angled at B, Using Pythagoras theorem, Ex 6.4, 7 In a right triangle ABC, ∠B = 90°. (b) If AC = 13 cm, BC = 5 cm, find ABHypotenuse is the side opposite to right angle. Given, ∠B = 90° AC = 13 cm BC = 5 cm In right angled triangle ∆ABC, Using Pythagoras theorem, (Hypotenuse)2 = (Base)2 + (Perpendicular)2 AC2 = BC2 + AB2 132 = 52 + AB2 AB2 = 132 – 52 AB2 = 169 – 25 AB2 = 144 AB = √144 Finding Square root of 144 by Prime factorization :- We see that, 144 = 2 × 2 × 2 × 2 × 3 × 3 ∴ √144 = 2 × 2 × 3 = 12 Therefore, AB = 12 cm

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.