# KSEEB Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.2

You can Download KSEEB Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.2 Questions and Answers to help you to revise the complete syllabus.

## KSEEB Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 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
= 1225 (iii) 862 = (80 + 6)2
= (80 + 6)(80 + 6)
= 80(80 + 6) + 6(80 + 6)
= 6400 + 480 + 480 + 36
= 6400 + 960 + 36
= 6400 + 946
= 7396

(iv) 932 = (90 + 3)2
= (90 + 3)(90 + 3)
= 90(90 + 3) + 3(90 + 3)
= 8100 + 270 + 270 + 9
= 8100 + 540 + 9
= 8649

(v) 712 = (70 + 1)2
= (70 + 1) (70 + 1)
= 70(70 + 1) + 1(70 + 1)
= 4900 + 70 + 70 + 1
= 5041

(vi) 462 = (40 + 6)2
= (40 + 6)(40 + 6)
= 40(40 + 6) + 6(40 + 6)
= 1600 + 240 + 240 + 36
= 1600 + 480 + 36
= 2116 Question 2.
Write a Pythagorean triplet whose one member is.
(i) 6
(ii) 14
(iii) 16
(iv) 18
Solution:
For any natural number m > 1
We have
(2m)2 + (m2 – 1)2 = (m2 – 1)2
So, 2m, m2 – 1, m2 + 1 forms a Pythagorean triplet.

(i) We take, m2 – 1 = 6
m2 = 1
m = √7
Then, the value of m is not an integer.
Now, take m2 + 1 = 6
m2 = 5
m = √5
Again value of m is not an integer
2m = 6
m = 3
Thus, m2 – 1 = 32 – 1 = 9 – 1 = 8
m2 + 1 = 32 + 1 = 9 + 1 = 10
2m = 2 × 3 = 6
∴ Required triplet = (6, 8, 10) (ii) Take m2 – 1 = 14
m2 = 15
m = √15
The value of m is not an integer
Again for m2 + 1 = 14
m2 = 13
m = √13
Hence, the value of m is not an integer
Now take 2m = 14
m = 7
m2 – 1 = 72 – 1 = 49 – 1 = 48
m2 + 1 = 72 + 1 = 49 + 1 = 50
2m = 2 × 7 = 14
∴ Required triplet is (14, 48, 50)

(iii) Take m2 – 1 = 16
m2 = 17
m = √17
The value of m is not an integer.
Again, for m2 + 1 = 16
m2 = 15
m = √15
Hence, the value of m is not an integer
Now, take 2m = 16
m = 8
m2 – 1 = 82 – 1 = 64 – 1 = 63
m2 + 1 = 82 + 1 = 64 + 1 = 65
2m = 2 × 8 = 16
∴ Required triplet is (16, 63, 65) (iv) Take m2 – 1 = 18
m2 = 19
m = √19
The value of m is not an integer.
For m2 + 1 = 18
m2 = 17
m = √17
The value of m is not an integer
For 2m = 18
m = 9
∴ Required triplet
m2 – 1 = 92 – 1 = 81 – 1 = 80
m2 + 1 = 92 + 1 = 81 + 1 = 82
2m = 2 × 9 = 18
∴ Required triplet is (18, 80, 82)