# Ex 6.2, 2 (iii) - 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. (iii) 16We know 2m, ๐^2โ1 and ๐^2+1 form a Pythagorean triplet. Given, One member of the triplet = 16. Let 2m = 16 2m = 16 m = 16/2 m = 8 Let ๐^๐โ๐" = 16" ๐^2 = 16 + 1 ๐^2 = 17 Since, 17 is not a square number, โด ๐^2โ1 โ 16 It is not possible. Let ๐^๐โ๐" = 16" ๐^2 = 16 + 1 ๐^2 = 17 Since, 17 is not a square number, โด ๐^2โ1 โ 16 It is not possible. Therefore, m = 8 Finding Triplets for m = 8 1st number = 2m 2nd number = ๐^2โ1 3rd number = ๐^2+1 โด The required triplet is 16, 63, 65

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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.