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

 

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



Question 1. Find the square of the following numbers.


(i) 32

Solution:


32 = 30 + 2

(32)² = (30 + 2)²

= (30)² + (2)² + 2 × 30 × 2

= 900 + 4 + 120

= 1024

Thus (32)² = 1024


(ii) 35

Solution:

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

Solution:

86 = (80 + 6)

862 = (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

Solution:

 93 = (90+ 3)

932 = (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

Solution:

71 = (70 + 1)

712 = (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

Solution:

46 = (40+ 6)

462 = (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

Solution:

Let m2 – 1 = 6

[Triplets are in the form 2m, m2 – 1, m2 + 1]

m2 = 6 + 1 = 7

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

Now, let us try for m2 + 1 = 6

⇒ m2 = 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:

m2 – 1 = 32 – 1 = 8 and m2 + 1 = 32 + 1 = 10

Hence, the required triplets are 6, 8 and 10


(ii) 14

Solution:

Let m2 – 1 = 14 ⇒ m2 = 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

m2 – 1 = (7)2 – 1 = 49 – 1 = 48

and m2 + 1 = (7)2 + 1 = 49 + 1 = 50

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


(iii) 16

Solution:

 Let 2m = 16 m = 8

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

m2 – 1 = (8)2 – 1 = 64 – 1 = 63

m2 + 1 = (8)2 + 1 = 64 + 1 = 65

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


(iv) 18

Solution:

Let 2m = 18 ⇒ m = 9

Required triplets are:

2m = 2 × 9 = 18

m2 – 1 = (9)2 – 1 = 81 – 1 = 80

and m2 + 1 = (9)2 + 1 = 81 + 1 = 82

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



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