NCERT Solutions for Class 8 Maths Chapter 5 Squares and Square Roots Exercise 5.2
1. Find the square of the following numbers.
(i) 32
Ans: `32^2` = `(30+2)^2` = (30+2)(30+2)
= `30^2` + 30×2 + 2×30 + `2^2`
= 900 + 120 + 4 = 1024
(ii) 35
Ans: `35^2` = `(30+5)^2` = (30+5)(30+5)
= `30^2` + 30×5 + 5×30 + `5^2`
= 900 + 300 + 25 = 1225
Alternate method:
`35^2` = (3 × 4) hundreds + 25
= 12×100 + 25 = 1225
(iii) 86
Ans: `86^2` = `(80+6)^2` = (80+6)(80+6)
= `80^2` + 80×6 + 6×80 + `6^2`
= 6400 + 960 + 36 = 7396
(iv) 93
Ans: `93^2` = `(90+3)^2` = (90+3)(90+3)
= `90^2` + 90×3 + 3×90 + `3^2`
= 8100 + 540 + 9 = 8649
(v) 71
Ans: `71^2` = `(70+1)^2` = (70+1)(70+1)
= `70^2` + 70×1 + 1×70 + `1^2`
= 4900 + 140 + 1 = 5041
(vi) 46
Ans: `46^2` = `(40+6)^2` = (40+6)(40+6)
= `40^2` + 40×6 + 6×40 + `6^2`
= 1600 + 480 + 36 = 2116
2. Write a Pythagorean triplet whose one member is.
(i) 6
Ans: Let 2m = 6
then, m = 3
thus, `m^2`-1 = `3^2`-1 = 9-1 = 8
and `m^2`+1 = `3^2`+1 = 9+1 = 10
Therefore, the required triplet is 6,8,10
(ii) 14
Ans: Let 2m = 14
then, m = 7
thus, `m^2`-1 = `7^2`-1 = 49-1 = 48
and `m^2`+1 = `7^2`+1 = 49+1 = 50
Therefore, the required triplet is 14,48,50
(iii) 16
Ans: Let 2m = 16
then, m = 8
thus, `m^2`-1 = `8^2`-1 = 64-1 = 63
and `m^2`+1 = `8^2`+1 = 64+1 = 65
Therefore, the required triplet is 16,63,65
(iv) 18
Ans: Let 2m = 18
then, m = 9
thus, `m^2`-1 = `9^2`-1 = 81-1 = 80
and `m^2`+1 = `9^2`+1 = 81+1 = 82
Therefore, the required triplet is 16,80,82
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