**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

Hence, (32)^{2 }= 1024

**ii. (35)**^{ 2}

= (30+5)^{ 2}

= (30)^{ 2} + (5)^{ 2} + 2×30×5

= 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

= 8100 + 9 + 540

= 8649

**v. (71) ^{ 2}**

= (70+1 )^{ 2}

= (70)^{ 2} + (1)^{ 2} +2×70×1

= 4900 + 1 + 140

= 5041

**vi. (46) ^{ 2}**

= (50 -4 )^{ 2}

= (50)^{ 2} + (4)^{ 2} – 2×50×4

= 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 triplet are in the form of 2m, m^{2}–1, and m^{2}+1

Now,

**i.** 2m = 6

m = 6/2 = 3

m^{2}–1 = 3^{2} – 1 = 9–1 = 8

m^{2}+1= 3^{2}+1 = 9+1 = 10

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

**ii.** 2m = 14

m = 14/2 = 7

m^{2 }– 1= 7^{2 }– 1 = 49 – 1 = 48

m^{2 }+ 1 = 7^{2 }+ 1 = 49 + 1 = 50

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

**iii.** 2m = 16

m = 16/2 = 8

m^{2 }– 1 = 8^{2 }– 1 = 64 – 1 = 63

m^{2 }+ 1 = 8^{2 }+ 1 = 64+1 = 65

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

**iv.** 2m = 18

m = 18/2 = 9

m^{2 }– 1 = 9^{2 }– 1 = 81–1 = 80

m^{2 }+ 1 = 9^{2 }+ 1 = 81+1 = 82

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

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