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

- Class 8 Maths Squares and Square Roots Exercise 6.1
- Class 8 Maths Squares and Square Roots Exercise 6.2
- Class 8 Maths Squares and Square Roots Exercise 6.3
- Class 8 Maths Squares and Square Roots Exercise 6.4
- Squares and Square Roots Class 8 Extra Questions

**NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6.2**

Question 1.

Find the square of the following numbers.

(i) 32

(ii) 35

(iii) 86

(iv) 93

(v) 71

(vi) 46

Solution:

(i) 32 = 30 + 2

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

= 30(30 + 2) + 2(30 + 2)

= 30^{2} + 30 × 2 + 2 × 30 + 2^{2}

= 900 + 60 + 60 + 4

= 1024

Thus (32)^{2} = 1024

(ii) 35 = (30 + 5)

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

= 30(30 + 5) + 5(30 + 5)

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

= 900 + 150 + 150 + 25

= 1225

Thus (35)^{2} = 1225

(iii) 86 = (80 + 6)

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

= 80(80 + 6) + 6(80 + 6)

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

= 6400 + 480 + 480 + 36

= 7396

Thus (86)^{2} = 7396

(iv) 93 = (90+ 3)

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

= 90 (90 + 3) + 3(90 + 3)

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

= 8100 + 270 + 270 + 9

= 8649

Thus (93)^{2} = 8649

(v) 71 = (70 + 1)

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

= 70 (70 + 1) + 1(70 + 1)

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

= 4900 + 70 + 70 + 1

= 5041

Thus (71)^{2} = 5041

(vi) 46 = (40+ 6)

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

= 40 (40 + 6) + 6(40 + 6)

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

= 1600 + 240 + 240 + 36

= 2116

Thus (46)^{2} = 2116

Question 2.

Write a Pythagorean triplet whose one member is

(i) 6

(ii) 14

(iii) 16

(iv) 18

Solution:

(i) Let m^{2} – 1 = 6

[Triplets are in the form 2m, m^{2} – 1, m^{2} + 1]

m^{2} = 6 + 1 = 7

So, the value of m will not be an integer.

Now, let us try for m^{2} + 1 = 6

⇒ m^{2} = 6 – 1 = 5

Also, the value of m will not be an integer.

Now we let 2m = 6 ⇒ m = 3 which is an integer.

Other members are:

m^{2} – 1 = 3^{2} – 1 = 8 and m^{2} + 1 = 3^{2} + 1 = 10

Hence, the required triplets are 6, 8 and 10

(ii) Let m^{2} – 1 = 14 ⇒ m^{2} = 1 + 14 = 15

The value of m will not be an integer.

Now take 2m = 14 ⇒ m = 7 which is an integer.

The member of triplets are 2m = 2 × 7 = 14

m^{2} – 1 = (7)^{2} – 1 = 49 – 1 = 48

and m^{2} + 1 = (7)^{2} + 1 = 49 + 1 = 50

i.e., (14, 48, 50)

(iii) Let 2m = 16 m = 8

The required triplets are 2m = 2 × 8 = 16

m^{2} – 1 = (8)^{2} – 1 = 64 – 1 = 63

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

i.e., (16, 63, 65)

(iv) Let 2m = 18 ⇒ m = 9

Required triplets are:

2m = 2 × 9 = 18

m^{2} – 1 = (9)^{2} – 1 = 81 – 1 = 80

and m^{2 }+ 1 = (9)^{2} + 1 = 81 + 1 = 82

i.e., (18, 80, 82)

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