Ex 5.2, 2 (i) - Write a Pythagorean triplet whose one member is 6 - Ex 5.2

part 2 - Ex 5.2, 2 (i) - Ex 5.2 - Serial order wise - Chapter 5 Class 8 Squares and Square Roots

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Ex 5.2, 2 Write a Pythagorean triplet whose one member is. (i) 6We know 2m, ๐‘š^2โˆ’1 and ๐‘š^2+1 form a Pythagorean triplet. Given, One member of the triplet = 6. Let 2m = 6 2m = 6 m = 6/2 m = 3 Let ๐’Ž^๐Ÿโˆ’๐Ÿ" = 6" ๐‘š^2 = 6 + 1 ๐‘š^2 = 7 Since, 7 is not a square number, โˆด ๐‘š^2โˆ’1 โ‰  6 It is not possible. Let ๐’Ž^๐Ÿ+๐Ÿ = 6 ๐‘š^2 = 6 โˆ’ 1 ๐‘š^2 = 5 Since, 5 is not a square number, โˆด ๐‘š^2+1 โ‰  6 It is not possible. Therefore, m = 3 Finding Triplets for m = 3 1st number = 2m 2nd number = ๐‘š^2โˆ’1 3rd number = ๐‘š^2+1 โˆด The required triplet is 6, 8, 10

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