Class 8 Squares and Square Roots Ex 5.2

NCERT Solutions For Class 8 Maths Chapter 5 Ex 5.2

ncert solutions for class 8 maths chapter 5, Exercise 5.2 involve complete  answers for each question in the exercise 5.2. The solutions provide students a  strategic methods  to prepare for their exam. Class 8 Maths Chapter 8 Squares and Square Roots Exercise 5.2 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 5.2 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) \(32^2 = 32 \times 32-1024\)

(ii) \(35^2 = 35 \times 35 = 1225\)

(iii) \(86^2 = 86 \times 86 = 7396\)

(iv) \(93^2 =93 \times 93= 8649\)

(v) \(71^2 = 71 \times 71= 5041\)

(vi) \(46^2 = 46 \times 46= 2116\)

Question 2:

Write a Pythagorean triplet whose one member is

(i) 6 (ii) 14 (iii) 16 (iv) 18

Answer:

For any natural number \(m > 1, 2m, m^2− 1, m^2+ 1\) forms a Pythagorean triplet.

(i) Let 2m = 6

m = 3

Therefore, the Pythagorean triplets are \(2m=2 × 3, m^2 − 1 =3^2 − 1, m^2 + 1 =3^2 + 1 = 6, 8, 10\).

(ii) Let 2m = 14

m = 7

Thus, \(m^2 − 1 = 49 − 1 = 48\)

\(m^2+ 1 = 49 + 1 = 50\)

Therefore, the required triplet is 14, 48, 50.

(iii) Let 2m = 16

m = 8

Thus, \(m^2 − 1 = 64 − 1 = 63 \)

\(m^2 + 1 = 64 + 1 = 65\)

Therefore, the Pythagorean triplet is 16, 63, and 65.

(iv) Let 2m =18

m = 9

\( 2m =18\)

\(m^2 − 1 = 81 − 1 = 80\)

\(m^2 + 1 = 81 + 1 = 82\)

Therefore, the Pythagorean triplet is 18, 80, and 82.