ML Aggarwal Class 8 Ex 3.1 Solutions

Ex 3.1 Class 8 ML Aggarwal

Question 1.
Which of the following natural numbers are perfect squares? Give reasons in support of your answer.
(i) 729
(ii) 5488
(iii) 1024
(iv) 243
Solution:
(i) 729
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 3 Squares and Square Roots Ex 3.1 Q1.1
= 3 × 3 × 3 × 3 × 3 × 3
729 is the product of pairs of equal prime factors.
∴ 729 is a perfect square.

(ii) 5488
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 3 Squares and Square Roots Ex 3.1 Q1.2
= 2 × 2 × 2 × 2 × 7 × 7 × 7
After pairing the same prime factors,
we see that one factor 7 is left unpaired.
So, 5488 is not a perfect square.

(iii) 1024
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 3 Squares and Square Roots Ex 3.1 Q1.3
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
After pairing the same prime factors,
we see that there is no factor left.
So, 1024 is a perfect square.

(iv) 243
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 3 Squares and Square Roots Ex 3.1 Q1.4
= 3 × 3 × 3 × 3 ×3
After pairing the same prime factors.
We see that factor 3 is left unpaired.
So, 243 is not a perfect square.

Question 2.
Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number.
(i) 1296
(ii) 1764
(iii) 3025
(iv) 3969
Solution:
(i) 1296
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 3 Squares and Square Roots Ex 3.1 Q2.1
= 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3
After pairing the same prime factors, we see that no factor is left.
So, 1296 is a perfect square and is the perfect square of 2 × 2 × 3 × 3 = 36

(ii) 1764
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 3 Squares and Square Roots Ex 3.1 Q2.2
= 2 × 2 × 3 × 3 × 7 × 7
After pairing the same factors, no factor is left.
So, 1764 is a perfect square and
1764 is the perfect square of 2 × 3 × 7 = 42

(iii) 3025
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 3 Squares and Square Roots Ex 3.1 Q2.3
= 5 × 5 × 11 × 11
After pairing the same prime factors, we see that no factor is left.
So, 3025 is a perfect square and is the perfect square of 5 × 11 = 55

(iv) 3969
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 3 Squares and Square Roots Ex 3.1 Q2.4
= 3 × 3 × 3 × 3 × 7 × 7
After pairing the same prime factors, we see that no factor is left.
So, 3969 is a perfect square and is the square of 3 × 3 × 7 = 63

Question 3.
Find the smallest natural number by which 1008 should be multiplied to make it a perfect square.
Solution:
1008
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 3 Squares and Square Roots Ex 3.1 Q3.1
= 2 × 2 × 2 × 2 × 3 × 3 × 7
After pairing the same kind of prime factor, one factor 7 is left.
So, by multiplying 1008 by 7
We shall get a perfect square
∴ Required smallest number = 7

Question 4.
Find the smallest natural number by which 5808 should be divided to make it a perfect square. Also, find the number whose square is the resulting number.
Solution:
5808
ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 3 Squares and Square Roots Ex 3.1 Q4.1
= 2 × 2 × 2 × 2 × 3 × 11 × 11
After pairing the same kind of prime factors,
we see that factor 3 is left.
So, by dividing the number by 3, we get a perfect square.
∴ The square root of the resulting number
= 2 × 2 × 11 = 44

ML Aggarwal Class 8 Solutions for ICSE Maths

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