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

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Ex 6.2, 2 (i) - Write a Pythagorean triplet whose one member is 6

Ex 6.2, 2 (i) - Chapter 6 Class 8 Squares and Square Roots - Part 2
Ex 6.2, 2 (i) - Chapter 6 Class 8 Squares and Square Roots - Part 3 Ex 6.2, 2 (i) - Chapter 6 Class 8 Squares and Square Roots - Part 4 Ex 6.2, 2 (i) - Chapter 6 Class 8 Squares and Square Roots - Part 5 Ex 6.2, 2 (i) - Chapter 6 Class 8 Squares and Square Roots - Part 6 Ex 6.2, 2 (i) - Chapter 6 Class 8 Squares and Square Roots - Part 7 Ex 6.2, 2 (i) - Chapter 6 Class 8 Squares and Square Roots - Part 8

Transcript

Ex 6.2, 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 Ex 6.2, 2 Write a Pythagorean triplet whose one member is. (ii) 14We know 2m, 𝑚^2−1 and 𝑚^2+1 form a Pythagorean triplet. Given, One member of the triplet = 14. Let 2m = 14 2m = 14 m = 14/2 m = 7 Let 𝒎^𝟐−𝟏" = 14" 𝑚^2 = 14 + 1 𝑚^2 = 15 Since, 15 is not a square number, ∴ 𝑚^2−1 ≠ 14 It is not possible. Let 𝒎^𝟐+𝟏 = 14 𝑚^2 = 14 − 1 𝑚^2 = 13 Since, 13 is not a square number, ∴ 𝑚^2+1 ≠ 14 It is not possible. Therefore, m = 7 Finding Triplets for m = 7 1st number = 2m 2nd number = 𝑚^2−1 3rd number = 𝑚^2+1 ∴ The required triplet is 14, 48, 50 Ex 6.2, 2 Write a Pythagorean triplet whose one member is. (iii) 16We know 2m, 𝑚^2−1 and 𝑚^2+1 form a Pythagorean triplet. Given, One member of the triplet = 16. Let 2m = 16 2m = 16 m = 16/2 m = 8 Let 𝒎^𝟐−𝟏" = 16" 𝑚^2 = 16 + 1 𝑚^2 = 17 Since, 17 is not a square number, ∴ 𝑚^2−1 ≠ 16 It is not possible. Let 𝒎^𝟐+𝟏 = 16 𝑚^2 = 16 − 1 𝑚^2 = 15 Since, 15 is not a square number, ∴ 𝑚^2+1 ≠ 16 It is not possible. Therefore, m = 8 Finding Triplets for m = 8 1st number = 2m 2nd number = 𝑚^2−1 3rd number = 𝑚^2+1 ∴ The required triplet is 16, 63, 65 Therefore, m = 8 Finding Triplets for m = 8 1st number = 2m 2nd number = 𝑚^2−1 3rd number = 𝑚^2+1 ∴ The required triplet is 16, 63, 65 Ex 6.2, 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.