Find the square of the following numbers.
(i) 32 (ii) 35 (iii) 86
(iv) 93 (v) 71 (vi) 46
(i) (32)2 = (30 + 2)2
= 302 + 2(30)(2)+(2)2
= 900 + 120 + 4
= 1024
(i) (35)2 = (30 + 5)2
= 302 + 2(30)(5)+(5)2
= 900 + 300 + 25
= 1225
Second method
(35)2 = 3 x (3 + 1) x 100 + 25
= 3 x 4 x 100 + 25
= 1200 + 25
= 1225\
(iii) (86)2 = (80 + 6)2
= (80)2 + 2(80)(6)+(6)2
= 6400 + 960 + 9
= 8649
(iv) (93)2 = (90 + 3)2
= (90)2 + 2(90)(3)+(3)2
= 8100 + 540 + 9
= 8649
(v) (71)2 = (70 + 1)2
= (70)2 + 2(70)(1)+(1)2
= 4900 + 140 + 1
= 5041
(vi) (46)2 = (40 + 6)2
= (40)2 + 2(40)(6)+(6)2
= 1600 + 480 + 36
= 2116
(i) 112 = 121. What is the square root of 121?
(ii) 142 = 196. What is the square root of 196?
(i) The square root of 121 is 11.
(ii) The square root of 196 is 14.
Find the square of the following numbers containing 5 in unit’s place.
(i) 15 (ii) 95 (iii) 105 (iv) 205
(i) (15)2 = 1 x (1 + 1) x 100 + 25
= 1 x 2 x 100 + 25
= 200 + 25 = 225
(ii) (95)2 = 9 x (9 + 1) x 100 + 25
= 9 x 10 x 100 + 25
= 9000 + 25 = 9025
(iii) (105)2 = 10 x (10 + 1) x 100 + 25
= 10 x 11 x 100 + 25
= 11000 + 25 = 11025
(iv) (205)2 = 20 x (20 + 1) x 100 + 25
= 20 x 21 x 100 + 25
= 42000 + 25 = 42025
(–1) = 1. Is –1, a square root of 1?
(–2)2 = 4. Is –2, a square root of 4?
(–9)2 = 81. Is –9, a square root of 81?
(i) Since (–1) × (–1) = 1
i.e. (–1)2 = 1
∴ Square root of 1 can also be –1.
Similarly,
(ii) Yes (–2) is a square root of 4.
(iii) Yes (–9) is a square root of 81.
Since we have to consider the positive square roots only and the symbol for a positive square root is 
.
∴ 
= 4 (and not (-4)
Similarly, 
means, the positive square root of 25 i.e. 5.
We can also find the square root by subtracting successive odd numbers starting from 1.
Write a Pytagorean triplet whose one member is.
(i) 6 (ii) 14 (iii) 16 (iv) 18
(i) Let 2n = 6 ∴ n = 3
Now, n2 -1 = 32 - 1 = 8
and n2 +1 = 32 + 1 = 10
Thus the required Pythagorean triplet is 6,8,10.
(ii) Let 2n = 14 ∴ n = 7
Now, n2 -1 = 72 - 1 = 48
and n2 +1 = 72 + 1 = 50
Thus the required Pythagorean triplet is 14,48,50.
(iii) Let 2n = 16 ∴ n = 8
Now, n2 -1 = 82 - 1 = 63
and n2 +1 = 82 + 1 = 65
∴ The required Pythagorean triplet is 16,63,65.
(iv) Let 2n = 18 ∴ n = 9
Now, n2 -1 = 92 - 1 = 80
and n2 +1 = 92 + 1 = 82
∴ The required Pythagorean triplet is 18,80,82.
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