# HBSE 8th Class Maths Solutions Chapter 6 Square and Square Roots Ex 6.2

Haryana State Board HBSE 8th Class Maths Solutions Chapter 6 Square and Square Roots Ex 6.2 Textbook Exercise Questions and Answers.

## Haryana Board 8th Class Maths Solutions Chapter 6 Square and Square Roots Exercise 6.2

Question 1.
Find the square of the following numbers :
(i) 32
(ii) 35
(iii) 86
(iv) 93
(v) 71
(vi) 46.
Solution:
(i) 322 = (30 + 2)2
= (30 + 2) (30 + 2)
= 30(30 + 2) + 2(30 + 2)
= 900 + 60 + 60 + 4 = 1024.

(ii) 352 = (3 × 4) hundreds + 25
= 12 × 100 + 25 = 1225

(iii) 862 = (80 + 6)2
= (80+ 6) (80+ 6)
= 80(80 + 6) + 6(80 + 6)
= 6400 + 480 + 480 + 36
= 7396

(iv) 932 = (90 + 3)2
= (90 + 3) (90 + 3)
= (90)2 + 90 × 3 + 3 × 90 + (3)2
= 8100 + 270 + 270 + 9
= 8649

(v) 712 = (70 + 1)2
= (70)2 + 2 × 70 × 1 + (1)2
[Using (a + b)2 = a2 + 2ab + b2]
= 4900 + 140 + 1 = 5041

(vi) 462 = (40 + 6)2
= (40)2 + 2 × 40 × 6 + (6)2
= 1600 + 480 + 36 = 2116.

Question 2.
Write a Pythagorean triplet whose one member is
(i) 6
(ii) 14
(iii) 16
(iv) 18.
Solution:
We have Pythagorean triplet (2m, m2 – 1, m2 + 1); where m > 1 (natural number).
Let us try one by one. .
Now, we take . .
m2 – 1 = 6 ⇒ m2 = 6 + 1 = 7
⇒ m = $$\sqrt {7}$$ (not an integer)
m2 + 1 = 6 ⇒ m2 = 5
⇒ m = $$\sqrt {5}$$ (not an integer)
2m = 6
⇒ m = $$\frac{6}{2}$$ = 3 (is an integer)
Thus, m2 – 1 = 32 – 1 = 9 – 1 = 8
and m2 + 1 = 32 + 1 = 9 + 1 = 10
Therefore, the required triplet is (6, 8, 10).

(ii) m2 – 1 = 14
or m2 = 15
Then the value of m will not be an integer So, we try to take
m2 + 1 = 14
or m2 = 14 – 1 = 13
or m = $$\sqrt {13}$$
which is also not an integer.
So, take 2m = 14
⇒ m = 7.
Thus, m2 – 1 = 72 – 1 = 49 – 1 = 48
and m2 + 1 = 72 + 1 = 49 + 1 = 50
∴ The required triplet is (14, 48, 50).

(iii) Let m2 – 1 = 16
⇒ m2 = 17
or m = $$\sqrt {17}$$
which is not an integer
m2 + 1 = 16
⇒ m2 = 15
⇒ m = $$\sqrt {15}$$ (not an integer)
and 2m = 16
⇒ m = 8.
Here, the value of m is an integer.
∴ m2 – 1 = 82 – 1 = 64 – 1 = 63
and m2+ 1 = 82 + 1 = 64 + 1 = 65
Therefore, the required triplet is (16, 63, 65)

(iv) Let m2 – 1 = 18
⇒ m2 = 19
or m = $$\sqrt {19}$$
which is not an integer
Now, take m2 + 1 = 18
⇒ m2 = 17
or m = $$\sqrt {17}$$
which is also not an integer
So, let us take 2m = 18
⇒ m = 9 which is an integer.
∴ The required triplet is (18, 92 – 1, 92 + 1)
i.e., (18, 80, 81)