Ex 5.2, 2 (i) - Chapter 5 Class 8 Squares and Square Roots

Transcript

Ex 5.2, 2 Write a Pythagorean triplet whose one member is. (i) 6We know 2m, 𝑚^2−1 and 𝑚^2+1 form a Pythagorean triplet. Given, One member of the triplet = 6. Let 2m = 6 2m = 6 m = 6/2 m = 3 Let 𝒎^𝟐−𝟏" = 6" 𝑚^2 = 6 + 1 𝑚^2 = 7 Since, 7 is not a square number, ∴ 𝑚^2−1 ≠ 6 It is not possible. Let 𝒎^𝟐+𝟏 = 6 𝑚^2 = 6 − 1 𝑚^2 = 5 Since, 5 is not a square number, ∴ 𝑚^2+1 ≠ 6 It is not possible. Therefore, m = 3 Finding Triplets for m = 3 1st number = 2m 2nd number = 𝑚^2−1 3rd number = 𝑚^2+1 ∴ The required triplet is 6, 8, 10