# Squares And Square Roots

## Exercise 5.2

1). Find the square of the following numbers.

(i) 32 (ii) 35

(iii) 86 (iv) 93

(v) 71 (vi) 46

Solution:

(i) 32

32 = 30 + 2

(a+b)^{2} = a^{2} + 2ab + b^{2}

32^{2} = (30 + 2)^{2}

= (30)^{2} + 2 X 30 X 2 + (2)^{2}

= 900 + 120 + 4

= 1024

(ii) 35

35 = 30 + 5

(a+b)^{2} = a^{2} + 2ab + b^{2}

35^{2} = (30 + 5)^{2}

= (30)^{2} + 2 X 30 X 5 + (5)^{2}

= 900 + 300 + 25

= 1225

(iii) 86

86 = 80 + 6

(a+b)^{2} = a^{2} + 2ab + b^{2}

86^{2} = (80 + 6)^{2}

= (80)^{2} + 2 X 80 X 6 + (6)^{2}

= 6400 + 960 + 36

= 7396

(iv) 93

93 = 90 + 3

(a+b)^{2} = a^{2} + 2ab + b^{2}

93^{2} = (90 + 3)^{2}

= (90)^{2} + 2 X 90 X 3 + (3)^{2}

= 8100 + 540 + 9

= 8649

(v) 71

71 = 70 + 1

(a+b)^{2} = a^{2} + 2ab + b^{2}

71^{2} = (70 + 1)^{2}

= (70)^{2} + 2 X 70 X 1 + (1)^{2}

= 4900 + 140 + 1

= 5041

(vi) 46

46 = 40 + 6

(a+b)^{2} = a^{2} + 2ab + b^{2}

46^{2} = (40 + 6)^{2}

= (40)^{2} + 2 X 40 X 6 + (6)^{2}

= 1600 + 480 + 36

= 2116

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

(i) 6 (ii) 14

(iii) 16 (iv) 18

Solution:

(i) 6

We can get Pythagorean triplets by using general form 2m, m^{2} – 1,

m^{2} + 1.

Let m^{2} – 1 = 6

m^{2} = 6 + 1

m^{2} = 7

m cannot be an integer.

Let m^{2} + 1 = 6

m^{2 }= 6 – 1

m^{2} = 5

m cannot be an integer

therefore, let 2m = 6

m = 3

m^{2} – 1 = 3^{2} – 1 = 9 – 1 = 8

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

Ans: The Pythagorean triplets are 6, 8, 10.

(ii) 14

We can get Pythagorean triplets by using general form 2m, m^{2} – 1,

m^{2} + 1.

Let m^{2} – 1 = 14

m^{2} = 14 + 1

m^{2} = 15

m cannot be an integer.

Let m^{2} + 1 = 14

m^{2 }= 14 – 1

m^{2} = 13

m cannot be an integer

therefore, let 2m = 14

m = 7

m^{2} – 1 = 7^{2} – 1 = 49 – 1 = 48

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

Ans: the Pythagorean triplets are 14, 48, 50.

(iii) 16

We can get Pythagorean triplets by using general form 2m, m^{2} – 1,

m^{2} + 1.

let 2m = 16

m = 8

m^{2} – 1 = 8^{2} – 1 = 64 – 1 = 63

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

Ans: the Pythagorean triplets are 16, 63, 65.

(iv) 18

We can get Pythagorean triplets by using general form 2m, m^{2} – 1,

m^{2} + 1.

let 2m = 18

m = 9

m^{2} – 1 = 9^{2} – 1 = 81 – 1 = 80

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

Ans: the Pythagorean triplets are 18, 80, 82

Click here for the solutions of Std 8 Maths

1). Rational Numbers

2). Linear Equations in One Variable

3). Understanding Quadrilaterals

4). Data Handling

5). Squares and Square Roots