NCERT Solutions For Class 8 Maths Chapter 6 Ex 6.2
ncert solutions for class 8 maths chapter 6, Exercise 6.2 involve complete answers for each question in the exercise 6.2. The solutions provide students a strategic methods to prepare for their exam. Class 8 Maths Chapter 8 Squares and Square Roots Exercise 6.2 questions and answers helps students to perform better in exam and it will clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems. NCERT Solutions for Class 8 Maths Chapter 8 Squares and Square Roots Exercise 6.2 prepared by our subject matter experts in very delicate, easy and creative way.
Question 1:
Find the square of the following numbers
(i) 32 (ii) 35 (iii) 86 (iv) 93 (v) 71 (vi) 46
Answer:
(i) 322 = (30 + 2)2
= 30 (30 + 2) + 2 (30 + 2)
= 302 + 30 × 2 + 2 × 30 + 22
= 900 + 60 + 60 + 4
= 1024
(ii) The number 35 has 5 in its unit’s place. Therefore,
352 = (3) (3 + 1) hundreds + 25
= (3 × 4) hundreds + 25
= 1200 + 25 = 1225
(iii) 862 = (80 + 6)2
= 80 (80 + 6) + 6 (80 + 6)
= 802 + 80 × 6 + 6 × 80 + 62
= 6400 + 480 + 480 + 36
= 7396
(iv) 932 = (90 + 3)2
= 90 (90 + 3) + 3 (90 + 3)
= 902 + 90 × 3 + 3 × 90 + 32
= 8100 + 270 + 270 + 9
= 8649
(v) 712 = (70 + 1)2
= 70 (70 + 1) + 1 (70 + 1)
= 702 + 70 × 1 + 1 × 70 + 12
= 4900 + 70 + 70 + 1
= 5041
(vi) 462 = (40 + 6)2
= 40 (40 + 6) + 6 (40 + 6)
= 402 + 40 × 6 + 6 × 40 + 62
= 1600 + 240 + 240 + 36
= 2116
Question 2:
Write a Pythagorean triplet whose one member is
(i) 6 (ii) 14 (iii) 16 (iv) 18
Answer:
For any natural number m > 1, 2m, m2− 1, m2+
1 forms a Pythagorean triplet.
(i) If we take m2 + 1 = 6, then m2 = 5
The value of m will not be an integer.
If we take m2− 1 = 6, then m2= 7
Again the value of m is not an integer.
Let 2m = 6
m = 3
Therefore, the Pythagorean triplets are 2 × 3, 32 − 1, 32 + 1 or 6, 8, and 10.
(ii) If we take m2+ 1 = 14, then m2= 13
The value of m will not be an integer.
If we take m2− 1 = 14, then m2= 15
Again the value of m is not an integer.
Let 2m = 14
m = 7
Thus, m2 − 1 = 49 − 1 = 48 and m2+ 1 = 49 + 1 = 50
Therefore, the required triplet is 14, 48, and 50.
(iii) If we take m2 + 1 = 16, then m2 =
15
The value of m will not be an integer.
If we take m2 − 1= 16, then m2 = 17
Again the value of m is not an integer.
Let 2m = 16
m = 8
Thus, m2 − 1 = 64 − 1 = 63 and m2 + 1 = 64 + 1 = 65
Therefore, the Pythagorean triplet is 16, 63, and 65.
(iv) If we take m2+ 1 = 18,
m2 = 17
The value of m will not be an integer.
If we take m2− 1 = 18, then m2 = 19
Again the value of m is not an integer.
Let 2m =18
m = 9
Thus, m2 − 1 = 81 − 1 = 80 and m2 + 1 = 81 + 1 = 82
Therefore, the Pythagorean triplet is 18, 80, and 82.
