NCERT Solutions for Class 8 Maths Chapter 5 Exercise 5.3 in Hindi and English medium updated for CBSE 20232024 exams. The solutions of ex. 5.3 class 8 math are revised on the basis of new rationalised syllabus and latest NCERT books for academic session 202324.
8th Maths Exercise 5.3 Solution in Hindi and English Medium
Class: 8  Mathematics 
Chapter: 5  Exercise: 5.3 
Topic Name:  Square and Square Roots 
Content Type:  Video and PDF Solutions 
Session:  CBSE 20232024 
Medium:  Hindi and English 
Class 8 Maths Chapter 5 Exercise 5.3 Solution
Class VIII Mathematics Ex. 5.3 of Square and Square Roots in Hindi and English free to use or download updated for 202324. Do you know that when the exponent (power) of a natural number is 2, the number so obtained is called a square number or a perfect square? Thus, when a number is multiplied by itself, the product is a perfect square. In 8th Maths NCERT Book exercise 5.3, we will go through the concepts of square and square roots or natural numbers.
Pythagorean Triplet
If a, b, c are three numbers where c > a and c > b such that a² + b² = c². Then (a, b, c) is called Pythagorean triplet.
Example:
(3, 4, 5), (6, 8, 10) and (5, 12, 13) and Pythagorean triplets, because
(i) (3)² + (4)² = (5)²
(ii) (6)² + (8)² = (10)²
(iii) (5)² + (12)² = (13)²
This, for any number m < 1, 2m, m² –1 and, m² + 1 is a Pythagorean triplet.
Example 1. Write a Pythagorean triplet whose one member is:
(i) 14 (ii) 16
We know that 2m, m² –1 and, m² + 1 is a Pythagorean triplet.
(i) Let 2m = 14
m = 7
So, m² – 1 = 72 – 1 = 48 and m² + 1 = 72 + 1 = 50
Hence, 14, 48, 50 is a Pythagorean triplet.
(ii) Let 2m = 16
Or, m = 8
So, m² – 1 = 82 – 1 = 63 and m² + 1 = 8² + 1 = 65
Hence, 16, 63, 65 is a Pythagorean triplet.
ShortCut Methods for Squaring a Number
Column method for squaring a twodigit number Let the given number have the tens digit = a and the units digit = b. Then, we have to square this number.
Class 8 Maths Exercise 5.3 Extra Questions
Give reason to show that none of the numbers given below is a perfect square: (i) 2162 (ii) 6843 (iii) 9637 (iv) 6598
We know that a number ending in 2, 3, 7 or 8 is never a perfect square.
Hence, none of the numbers 2162, 6843, 9637 and 6598 is a perfect square.
Give reason to show that none of the numbers 640, 81000 and 3600000 is perfect square.
We know that a number ending in an odd number of zeros is never a perfect square. So, none of the numbers 640, 81000 and 3600000 is a perfect square.
Express: (i) 64 as the sum of 8 odd natural numbers. (ii) 121 as the sum of 11 odd natural numbers.
We have
(i) 64 = 8² = sum of 8 odd natural numbers = (1 + 3 + 5 + 7 + 9 + 11 + 13 +15).
(ii) 121 = (11)² = sum of first 11 odd natural numbers = (1 + 3 + 5 + 7 + 9 + 11 + 13 +15 + 17 + 19 + 21).
Find the Pythagorean triplet whose smallest members is 12.
For every natural number m < 1, (2m, m 1, m +1) is a Pythagorean triplet. Putting 2m = 12, we get m = 6. Thus, we get the triplet (12, 35, 37). Product of Two Consecutive Odd or Consecutive Even Numbers
Evaluate (i) 49 x 51 (ii) 30 x 32.
We have:
(i) 49 x 51 = (50 – 1) x (50 + 1)
= [(50)² 1²] = (2500 – 1) = 2499.
(ii) 30 x 32 = (31 – 1) x (31 + 1)
= [(31)² 1²] = (961 – 1) = 960.
Step 1
Make three columns, I, II and III, headed by a, (2 x a x b) and b respectively. Write the values of a², (2 x a x b) and b2 in columns I, II and III respectively.
Step 2
In Column III, underline the unit digit of b and carry over the tens digit of it to Column II and add it to the value of (2 x a x b).
Class 8 Maths Exercise 5.3 Important Questions
What is the square of an even number?
Squares of even numbers are even (and in fact divisible by 4), since (2n)² = 4n².
What is the square of an odd number?
Squares of odd numbers are odd, since (2n + 1)² = 4(n² + n) + 1.
Express 81 as the sum of 9 odd numbers.
The sum of first nine odd numbers are 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81 = 9²
Find the square of 47..
Given number = 47. a = 4 and b = 7
I  II  III 
a^{2}  (2 x a x b)  b^{2} 
16 + 6 = 22

56 + 4 = 60  49 
So, (47)^{2 }= 2209
Step 3
In Column II, underline the unit digit of the number obtained in Step 2 and carry over the tens digit of it to Column I and add it to the value of a.
Step 4
Underline the number obtained in Step 3 in Column 1. The underlined digits give the required square number.