1. Chapter 5 Class 8 Squares and Square Roots
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  3. Ex 5.4
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Ex 5.4, 7 - Chapter 5 Class 8 Squares and Square Roots

Last updated at Dec. 16, 2024 by Teachoo

 

 

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Ex 5.4, 7 In a right triangle ABC, ∠B = 90°. (a) If AB = 6 cm, BC = 8 cm, find AC Given, ∠B = 90° AB = 6 cm BC = 8 cm In right angled triangle ∆ABC, Using Pythagoras theorem, Hypotenuse is the side opposite to right angle. (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 5.4, 7 In a right triangle ABC, ∠B = 90°. (b) If AC = 13 cm, BC = 5 cm, find AB 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
  1. Chapter 5 Class 8 Squares and Square Roots
  2. Serial order wise

    3. Ex 5.4

    Examples →
Examples →
Chapter 5 Class 8 Squares and Square Roots
Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo