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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 5.2, 2 Write a Pythagorean triplet whose one member is. (iv) 18We know 2m, 𝑚^2−1 and 𝑚^2+1 form a Pythagorean triplet. Given, One member of the triplet = 18. Let 2m = 18 2m = 18 m = 18/2 m = 9 Let 𝒎^𝟐−𝟏" = 18" 𝑚^2 = 18 + 1 𝑚^2 = 19 Since, 19 is not a square number, ∴ 𝑚^2−1 ≠ 18 It is not possible. Let 𝒎^𝟐+𝟏 = 18 𝑚^2 = 18 − 1 𝑚^2 = 17 Since, 17 is not a square number, ∴ 𝑚^2+1 ≠ 18 It is not possible. Therefore, m = 9 Finding Triplets for m = 9 1st number = 2m 2nd number = 𝑚^2−1 3rd number = 𝑚^2+1 ∴ The required triplet is 18, 80, 82

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.