PSEB 8th Class Maths Solutions Chapter 6 Squares and Square Roots Ex 6.2

Punjab State Board PSEB 8th Class Maths Book Solutions Chapter 6 Squares and Square Roots Ex 6.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Ex 6.2

1. Find the square of the following numbers:

Question (i).
32
Solution:
Let us use: (a + b)² = a² + 2ab + b²
(32)² = (30 + 2)²
= (30)² + 2 (30) (2) + (2)²
= 900 + 120 + 4
= 1024

Question (ii).
35
Solution:
(35)² = (30 + 5)²
= (30)² + 2 (30) (5) + (5)²
= 900 + 300 + 25
= 1200 + 25
= 1225
Second method :
[Note : The unit digit of 35 is 5.]
(35)² = 3 × (3 + 1) × 100 + 25
= 3 × 4 × 100 + 25
= 1200 + 25
= 1225

Question (iii).
86
Solution:
(86)² = (80 + 6)²
= (80)² + 2 (80) (6) + (6)²
= 6400 + 960 + 36
= 7396

Question (iv).
93
Solution:
(93)² = (90 + 3)²
= (90)² + 2 (90) (3) + (3)²
= 8100 + 540 + 9
= 8649

Question (v).
71
Solution:
(71 )² = (70 + 1)²
= (70)² + 2 (70) (1) + (1)²
= 4900 + 140 + 1
= 5041

Question (vi).
46
Solution:
(46)² = (40 + 6)²
= (40)² + 2 (40) (6) + (6)²
= 1600 + 480 + 36
= 2116

2. Write a Pythagorean triplet whose one member is:

Question (i).
6
Solution:
Here, 2n = 6
∴ n = 3

Now, n² – 1 = 3² – 1
= 9 – 1
= 8
and n² + 1 = 3² + 1
= 9 + 1 = 10
Thus, the required Pythagorean triplet is 6, 8, 10.

Question (ii).
14
Solution:
Here, 2n = 14
∴ n = 7
Now, n² – 1 = 7² – 1 = 49 – 1 = 48
and n² + 1 = 7² + 1 = 49 + 1 = 50
Thus, the required Pythagorean triplet is 14, 48, 50.

Question (iii).
16
Solution:
Here, 2n = 16
∴ n = 8
Now, n² – 1 = 8² – 1 = 64 – 1 = 63
and n² + 1 = 8² + 1 = 64 + 1 = 65
Thus, the required Pythagorean triplet is 16, 63, 65.

Question (iv).
18
Solution:
Here, 2n = 18
∴ n = 9
Now n² – 1 = 9² – 1 = 81 – 1 = 80
and n² + 1 = 92 + 1 = 81 + 1 = 82
Thus, the required Pythagorean triplet is 18, 80, 82.