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  1. Chapter 6 Class 8 Squares and Square Roots
  2. Serial order wise

Transcript

Ex 6.2, 2 Write a Pythagorean triplet whose one member is. (ii) 14We know 2m, ๐‘š^2โˆ’1 and ๐‘š^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’Ž^๐Ÿโˆ’๐Ÿ" = 14" ๐‘š^2 = 14 + 1 ๐‘š^2 = 15 Since, 15 is not a square number, โˆด ๐‘š^2โˆ’1 โ‰  14 It is not possible. Let ๐’Ž^๐Ÿ+๐Ÿ = 14 ๐‘š^2 = 14 โˆ’ 1 ๐‘š^2 = 13 Since, 13 is not a square number, โˆด ๐‘š^2+1 โ‰  14 It is not possible. Therefore, m = 7 Finding Triplets for m = 7 1st number = 2m 2nd number = ๐‘š^2โˆ’1 3rd number = ๐‘š^2+1 โˆด The required triplet is 14, 48, 50

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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.