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

Last updated at Sept. 17, 2018 by Teachoo

Last updated at Sept. 17, 2018 by Teachoo

Transcript

Ex 6.2, 2 Write a Pythagorean triplet whose one member is. (ii) 14We know 2m, ๐^2โ1 and ๐^2+1 form a Pythagorean triplet. Given, One member of the triplet = 14. Let 2m = 14 2m = 14 m = 14/2 m = 7 Let ๐^๐โ๐" = 14" ๐^2 = 14 + 1 ๐^2 = 15 Since, 15 is not a square number, โด ๐^2โ1 โ 14 It is not possible. Let ๐^๐+๐ = 14 ๐^2 = 14 โ 1 ๐^2 = 13 Since, 13 is not a square number, โด ๐^2+1 โ 14 It is not possible. Therefore, m = 7 Finding Triplets for m = 7 1st number = 2m 2nd number = ๐^2โ1 3rd number = ๐^2+1 โด The required triplet is 14, 48, 50

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