**Question 1. Find the square of the following numbers.**

**(i) 32**

(32)

^{2}= (30 + 2)

^{2}= (30)

^{2}+ (2)^{2}+ 2 × 30 × 2 [Since, (a + b)^{2}= a^{2 }+ b^{2}+ 2ab]= 900 + 4 + 120

= 1024

**(ii) 35**

(35)

^{2}= (30 + 5)

^{2}= (30)

^{2}+ (5)^{2 }+ 2 × 30 × 5 [Since, (a + b)^{2}= a^{2}+ b^{2}+ 2ab]= 900 + 25 + 300

= 1225

**(iii) 86**

(86)

^{2}= (90 – 4)

^{2}= (90)

^{2}+ (4)^{2}– 2 × 90 × 4 [Since, (a – b)^{2}= a^{2 }+ b^{2}– 2ab]= 8100 + 16 – 720

= 8116 – 720

= 7396

**(iv) 93 **

(93)

^{2}= (90 + 3)

^{2}= (90)

^{2}+ (3)^{2}+ 2 × 90 × 3 [Since, (a + b)^{2}= a^{2}+ b^{2}+ 2ab]

= 8100 + 9 + 540

= 8649

**(v) 71**

(71)

^{2}= (70 + 1)

^{2}= (70)

^{2}+ (1)^{2}+ 2 × 70 × 1 [Since, (a + b)^{2}= a^{2 }+ b^{2}+ 2ab]= 4900 + 1 + 140

= 5041

**(vi) 46**

(46)

^{2}= (50 – 4)

^{2}= (50)

^{2}+ (4)^{2}– 2 × 50 × 4 [Since, (a – b)^{2}= a^{2 }+ b^{2}– 2 ab]= 2500 + 16 – 400

= 2116

**Question 2. Write a Pythagorean triplet whose one member is.**

**(i) 6 **

For any natural number m, we know that 2m, m

^{2 }– 1, m^{2 }+ 1 is a Pythagorean triplet.2m = 6

m = 6/2 = 3

m

^{2 }– 1= 3^{2}– 1 = 9 – 1 = 8m

^{2 }+ 1= 3^{2 }+ 1 = 9 + 1 = 10∴ (6, 8, 10) is a Pythagorean triplet.

**(ii) 14**

2m = 14

m = 14/2 = 7

m

^{2 }– 1= 7^{2 }– 1 = 49 – 1 = 48m

^{2 }+ 1 = 7^{2 }+ 1 = 49 + 1 = 50∴ (14, 48, 50) is not a Pythagorean triplet.

**(iii) 16 **

2m = 16

m = 16/2 = 8

m

^{2 }– 1 = 8^{2 }– 1 = 64 – 1 = 63m

^{2 }+ 1 = 8^{2 }+ 1 = 64 + 1 = 65∴ (16, 63, 65) is a Pythagorean triplet.

**(iv) 18**

2m = 18

m = 18/2 = 9

m

^{2}– 1 = 9^{2}– 1 = 81 – 1 = 80m

^{2}+ 1 = 9^{2}+ 1 = 81 + 1 = 82∴ (18, 80, 82) is a Pythagorean triplet.