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Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.1 solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.2 solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.3 solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.4 solution
SQUARE AND SQUARE ROOTS (Chapter-6) Exercise 6.2 SOLUTION
EXERCISE 6.2
1. Find the square of the following numbers.
(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46
Solution- According to the chapter we know that When we multiply a number by same number. We get the square of a number,
(i) The square of 32 is 322,
322 = 32 X 32 = 1024
or
322 = (30 +2)2
= 302 + 2X30X2 + 22
= 900 + 120 + 4
= 1024
(ii) The square of 35 is 352,
352 = 35 X 35 = 1225
or
352 = (30 +5)2
= 302 + 2X30X5 + 52
= 900 + 300 + 25
= 1225
(iii) The square of 86 is 862,
862 = 86 X 86 = 7396
or
862 = (80 +6)2
= 802 + 2X80X6 + 62
= 6400 + 960 + 36
= 7396
(iv) 93 The square of 93 is 932,
932 = 93 X 93 = 8649
or
932 = (90 +3)2
= 902 + 2X90X3 + 32
= 8100 + 540 + 9
= 8649
(v) 71 The square of 71 is 712,
712 = 71 X 71 = 5041
or
712 = (70 + 1)2
= 702 + 2X70X1 + 12
= 4900 + 140 + 1
= 5041
(vi) 46 The square of 46 is 462,
462 = 46 X 46= 2116
or
462 = (40 +6)2
= 402 + 2X40X6 + 62
= 1600 + 480 + 36
= 2116
2. Write a Pythagorean triplet whose one member is.
(i) 6 (ii) 14 (iii) 16 (iv) 18
Solution- According to the chapter we know that For any natural number m > 1, we have (2m) 2 + (m2 – 1)2 = (m2 + 1)2 . So, 2m, m2 – 1 and m2 + 1 forms a Pythagorean triplet.
To find Pythagorean triplet first we observe the this number if it is the nearest of the perfect square number then we assume m2 – 1 and m2 + 1 according to the conditions.
(i) 6 is the one of the element of Pythagorean triplet. 6 is not a nearest number of the perfect square so we can't assume 6 as m2 – 1 and m2 + 1.
let 2m = 6
m = 3
So, other element of Pythagorean triplet is-
m2 – 1 = 32 – 1 = 9 - 1 = 8
and
m2 + 1 = 32 + 1 = 9 + 1 = 10
Hence, Pythagorean triplet is (6, 8, 10)
(ii) 14 is the one of the element of Pythagorean triplet. 14 is not a nearest number of the perfect square so we can't assume 14 as m2 – 1 and m2 + 1.
let 2m = 14
m = 7
So, other element of Pythagorean triplet is-
m2 – 1 = 72 – 1 = 49 - 1 = 48
and
m2 + 1 = 72 + 1 = 49 + 1 = 50
Hence, Pythagorean triplet is (14, 48, 50)
(iii) 16 is the one of the element of Pythagorean triplet. 16 is not a nearest number of the perfect square so we can't assume 16 as m2 – 1 and m2 + 1.
let 2m = 16
m = 8
So, other element of Pythagorean triplet is-
m2 – 1 = 82 – 1 = 64 - 1 = 63
and
m2 + 1 = 82 + 1 = 64 + 1 = 65
Hence, Pythagorean triplet is (16, 63, 65)
(iv) 18 is the one of the element of Pythagorean triplet. 18 is not a nearest number of the perfect square so we can't assume 18 as m2 – 1 and m2 + 1.
let 2m = 18
m = 9
So, other element of Pythagorean triplet is-
m2 – 1 = 92 – 1 = 81 - 1 = 80
and
m2 + 1 = 92 + 1 = 81 + 1 = 82
Hence, Pythagorean triplet is (9, 80, 82)
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NCERT Solutions For Class 8th MATHEMATICS Chapter 6 SQUARE AND SQUARE ROOTS
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises solution
Class 8th chapter 6 SQUARE AND SQUARE ROOTS all exercises 6.1 solution
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