Exercise 6.2 Page: 98

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

**i. 32**

**ii. 35**

**iii. 86**

**iv. 93**

**v. 71**

**vi. 46**

Solution:

i. (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)^{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)^{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)^{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)^{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)^{2}

= (50 -4 )^{2}

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

= 2500 + 16 – 400

= 2116

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

**i. 6**

**ii. 14**

**iii. 16**

**iv. 18**

Solution:

For any natural number m, we know that 2m, m^{2}–1, m^{2}+1 is a Pythagorean triplet.

i. 2m = 6

⇒ m = 6/2 = 3

m2–1= 32 – 1 = 9–1 = 8

m2+1= 32+1 = 9+1 = 10

∴ (6, 8, 10) is a Pythagorean triplet.

ii. 2m = 14

⇒ m = 14/2 = 7

m2–1= 72–1 = 49–1 = 48

m2+1 = 72+1 = 49+1 = 50

∴ (14, 48, 50) is not a Pythagorean triplet.

iii. 2m = 16

⇒ m = 16/2 = 8

m^{2}–1 = 82–1 = 64–1 = 63

m^{2}+ 1 = 82+1 = 64+1 = 65

∴ (16, 63, 65) is a Pythagorean triplet.

iv. 2m = 18

⇒ m = 18/2 = 9

m^{2}–1 = 92–1 = 81–1 = 80

m^{2}+1 = 92+1 = 81+1 = 82

∴ (18, 80, 82) is a Pythagorean triplet.