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
Ex 6.2 Class 8 Maths 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
Ex 6.2 Class 8 Maths 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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