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

Last updated at July 31, 2018 by Teachoo

Last updated at July 31, 2018 by Teachoo

Transcript

Ex 6.2, 2Write a Pythagorean triplet whose one member is.(i) 6We know 2m, ๐^2โ1 and ๐^2+1 form a Pythagorean triplet. Given, One member of the triplet = 6. Let 2m = 6 2m = 6 m = 6/2 m = 3 Let ๐^๐โ๐" = 6" ๐^2 = 6 + 1 ๐^2 = 7 Since, 7 is not a square number, โด ๐^2โ1 โ 6 It is not possible. Let ๐^๐+๐ = 6 ๐^2 = 6 โ 1 ๐^2 = 5 Since, 5 is not a square number, โด ๐^2+1 โ 6 It is not possible. Therefore, m = 3 Finding Triplets for m = 3 1st number = 2m 2nd number = ๐^2โ1 3rd number = ๐^2+1 โด The required triplet is 6, 8, 10 Ex 6.2, 2Write 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 Ex 6.2, 2Write 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 ๐^๐+๐ = 16 ๐^2 = 16 โ 1 ๐^2 = 15 Since, 15 is not a square number, โด ๐^2+1 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 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 Ex 6.2, 2Write 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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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.