# Ex 6.2, 2 (iv) - Chapter 6 Class 8 Squares and Square Roots

Last updated at Sept. 11, 2018 by Teachoo

Last updated at Sept. 11, 2018 by Teachoo

Transcript

Ex 6.2, 2 Write a Pythagorean triplet whose one member is. (iv) 18We know 2m, ๐^2โ1 and ๐^2+1 form a Pythagorean triplet. Given, One member of the triplet = 18. Let 2m = 18 2m = 18 m = 18/2 m = 9 Let ๐^๐โ๐" = 18" ๐^2 = 18 + 1 ๐^2 = 19 Since, 19 is not a square number, โด ๐^2โ1 โ 18 It is not possible. Let ๐^๐+๐ = 18 ๐^2 = 18 โ 1 ๐^2 = 17 Since, 17 is not a square number, โด ๐^2+1 โ 18 It is not possible. Therefore, m = 9 Finding Triplets for m = 9 1st number = 2m 2nd number = ๐^2โ1 3rd number = ๐^2+1 โด The required triplet is 18, 80, 82

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