Slide7.JPG

Slide8.JPG
Slide9.JPG Slide10.JPG Slide11.JPG Slide12.JPG Slide13.JPG Slide14.JPG

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

Transcript

Ex 6.2, 2 Write a Pythagorean triplet whose one member is. (i) 6We know 2m, ๐‘š^2โˆ’1 and ๐‘š^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’Ž^๐Ÿโˆ’๐Ÿ" = 6" ๐‘š^2 = 6 + 1 ๐‘š^2 = 7 Since, 7 is not a square number, โˆด ๐‘š^2โˆ’1 โ‰  6 It is not possible. Let ๐’Ž^๐Ÿ+๐Ÿ = 6 ๐‘š^2 = 6 โˆ’ 1 ๐‘š^2 = 5 Since, 5 is not a square number, โˆด ๐‘š^2+1 โ‰  6 It is not possible. Therefore, m = 3 Finding Triplets for m = 3 1st number = 2m 2nd number = ๐‘š^2โˆ’1 3rd number = ๐‘š^2+1 โˆด The required triplet is 6, 8, 10 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 Ex 6.2, 2 Write a Pythagorean triplet whose one member is. (iii) 16We know 2m, ๐‘š^2โˆ’1 and ๐‘š^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’Ž^๐Ÿโˆ’๐Ÿ" = 16" ๐‘š^2 = 16 + 1 ๐‘š^2 = 17 Since, 17 is not a square number, โˆด ๐‘š^2โˆ’1 โ‰  16 It is not possible. Let ๐’Ž^๐Ÿ+๐Ÿ = 16 ๐‘š^2 = 16 โˆ’ 1 ๐‘š^2 = 15 Since, 15 is not a square number, โˆด ๐‘š^2+1 โ‰  16 It is not possible. Therefore, m = 8 Finding Triplets for m = 8 1st number = 2m 2nd number = ๐‘š^2โˆ’1 3rd number = ๐‘š^2+1 โˆด The required triplet is 16, 63, 65 Therefore, m = 8 Finding Triplets for m = 8 1st number = 2m 2nd number = ๐‘š^2โˆ’1 3rd number = ๐‘š^2+1 โˆด The required triplet is 16, 63, 65 Ex 6.2, 2 Write a Pythagorean triplet whose one member is. (iv) 18We know 2m, ๐‘š^2โˆ’1 and ๐‘š^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’Ž^๐Ÿโˆ’๐Ÿ" = 18" ๐‘š^2 = 18 + 1 ๐‘š^2 = 19 Since, 19 is not a square number, โˆด ๐‘š^2โˆ’1 โ‰  18 It is not possible. Let ๐’Ž^๐Ÿ+๐Ÿ = 18 ๐‘š^2 = 18 โˆ’ 1 ๐‘š^2 = 17 Since, 17 is not a square number, โˆด ๐‘š^2+1 โ‰  18 It is not possible. Therefore, m = 9 Finding Triplets for m = 9 1st number = 2m 2nd number = ๐‘š^2โˆ’1 3rd number = ๐‘š^2+1 โˆด The required triplet is 18, 80, 82

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.