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  1. Chapter 6 Class 8 Squares and Square Roots
  2. Serial order wise
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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. (i) 525Rough 41 × 1 = 41 42 × 2 = 84 43 × 3 = 129 Here, Remainder = 41 Since remainder is not 0, So, 525 is not a perfect square We need to find the least number that must be added to 525 so as to get a perfect square Now, Thus, we add 232 – 525 to the number ∴ Number to added = 232 − 525 = 529 – 525 = 4 Thus, Perfect square = 525 + 4 Let’s check Thus, we add 4 to 525 to get a perfect square. Perfect square = 529 & Square root of 529 = 23 Rough 41 × 1 = 41 42 × 2 = 84 43 × 3 = 129 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. (ii) 1750Rough 81 × 1 = 81 82 × 2 = 164 Here, Remainder = 69 Since remainder is not 0, So, 1750 is not a perfect square We need to find the least number that must be added to 1750 so as to get a perfect square Now, Thus, we add 422 – 1750 to the number ∴ Number to added = 422 − 1750 = 1764 – 1750 = 14 Thus, Perfect square = 1750 + 14 Let’s check Thus, we add 14 to 1750 to get a perfect square. Perfect square = 1764 & Square root of 1764 = 42 Rough 81 × 1 = 81 82 × 2 = 164 Thus, we add 14 to 1750 to get a perfect square. Perfect square = 1764 & Square root of 1764 = 42 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 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. (iv) 1825Rough 81 × 1 = 81 82 × 2 = 164 83 × 3 = 249 Here, Remainder = 61 Since remainder is not 0, So, 1825 is not a perfect square We need to find the least number that must be added to 1825 so as to get a perfect square Now, Thus, we add 432 – 1825 to the number ∴ Number to added = 432 − 1825 = 1849 – 1825 = 24 Thus, Perfect square = 1825 + 24 Let’s check Thus, we add 24 to 1825 to get a perfect square. Perfect square = 1849 & Square root of 1849 = 43 Rough 81 × 1 = 81 82 × 2 = 164 83 × 3 = 249 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. (v) 6412Rough 160 × 0 = 325 161 × 1 = 161 Here, Remainder = 12 Since remainder is not 0, So, 6412 is not a perfect square We need to find the least number that must be added to 6412 so as to get a perfect square Now, Thus, we add 812 – 6412 to the number ∴ Number to added = 812 − 6412 = 6561 – 6412 = 149 Thus, Perfect square = 6412 + 149 Let’s check Thus, we add 149 to 6412 to get a perfect square. Perfect square = 6561 & Square root of 6561 = 81 Rough 160 × 0 = 325 161 × 1 = 161

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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.