Ex 6.4, 5 - Chapter 6 Class 8 Squares and Square Roots - Part 7

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Ex 6.4, 5 - Chapter 6 Class 8 Squares and Square Roots - Part 8

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Ex 6.4, 5 - Chapter 6 Class 8 Squares and Square Roots - Part 9

  1. Chapter 6 Class 8 Squares and Square Roots
  2. Serial order wise

Transcript

Ex 6.4, 5 Find the least number which must be added to each of the following numbers so as to get a perfect square. Also find the square root of the perfect square so obtained. (iii) 252Rough 24 × 4 = 96 25 × 5 = 125 26 × 6 = 156 Here, Remainder = 27 Since remainder is not 0, So, 252 is not a perfect square We need to find the least number that must be added to 252 so as to get a perfect square Now, Thus, we add 162 – 252 to the number ∴ Number to added = 162 − 252 = 256 – 252 = 4 Thus, Perfect square = 252 + 4 Let’s check Thus, we add 4 to 252 to get a perfect square. Perfect square = 256 & Square root of 256 = 16 Rough 24 × 4 = 96 25 × 5 = 125 26 × 6 = 156

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.