NCERT Solutions Class 8 Mathematics
Chapter – 6 (Squares and Square)
The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 6 Squares and Square Exercise 6.2 has been provided here to help the students in solving the questions from this exercise.
Chapter 6: Squares and Square Roots
- NCERT Solution Class 8 Maths Ex – 6.1
- NCERT Solution Class 8 Maths Ex – 6.3
- NCERT Solution Class 8 Maths Ex – 6.4
Exercise – 6.2
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)2
= (30)2 + (2)2 + 2×30×2 [Since, (a+b)2 = a2+b2 +2ab]
= 900 + 4 + 120
= 1024
(ii) 35
= (30 + 5 )2
= (30)2 + (5)2 + 2×30×5 [Since, (a+b)2 = a2+b2 +2ab]
= 900 + 25 + 300
= 1225
(iii) 86
= (90 – 4)2
= (90)2 + (4)2 – 2×90×4 [Since, (a+b)2 = a2+b2 +2ab]
= 8100 + 16 – 720
= 8116 – 720
= 7396
(iv) 93
= (90 + 3 )2
= (90)2 + (3)2 + 2×90×3 [Since, (a+b)2 = a2+b2 +2ab]
= 8100 + 9 + 540
= 8649
(v) 71
= (70 + 1 )2
= (70)2 + (1)2 +2×70×1 [Since, (a+b)2 = a2+b2 +2ab]
= 4900 + 1 + 140
= 5041
(vi) 46
= (50 – 4 )2
= (50)2 + (4)2 – 2×50×4 [Since, (a+b)2 = a2+b2 +2ab]
= 2500 + 16 – 400
= 2116
2. Write a Pythagorean triplet whose one member is.
(i) 6
(ii) 14
(iii) 16
(iv) 18
Solution –
(i) 6
2m = 6
⇒ m = = 3
m2 – 1 = 32 – 1 = 9 – 1 = 8
m2 + 1 = 32 + 1 = 9 + 1 = 10
∴ (6, 8, 10) is a Pythagorean triplet.
(ii) 14
2m = 14
⇒ m = = 7
m2 – 1 = 72 – 1 = 49 – 1 = 48
m2 + 1 = 72 + 1 = 49 + 1 = 50
∴ (14, 48, 50) is not a Pythagorean triplet.
(iii) 16
2m = 16
⇒ m = = 8
m2 – 1 = 82 – 1 = 64 – 1 = 63
m2 + 1 = 82 + 1 = 64 + 1 = 65
∴ (16, 63, 65) is a Pythagorean triplet.
(iv) 18
2m = 18
⇒ m = = 9
m2 – 1 = 92 – 1 = 81 – 1 = 80
m2 + 1 = 92 + 1 = 81 + 1 = 82
∴ (18, 80, 82) is a Pythagorean triplet.
