# Ex.6.3 Q3 Squares and Square Root - NCERT Maths Class 8

## Question

Find the square roots of $$100$$ and $$169$$ by the method of repeated subtraction.

Video Solution
Squares And Square Roots
Ex 6.3 | Question 3

## Text Solution

What is known?

Perfect square

What is Unknown?

Square root of follownig square by using repeated subtraction method.

Reasoning:

The sum of the first $$n$$ odd natural numbers in$$\,{{\rm{n}}^2}$$ i.e. every square number can be expressed as a sum of successive odd numbers starting from $$1$$. (Page $$100$$ of Grade B NCERT Book)

Steps:

Consider $$\sqrt {100}$$ then

(i) $$100 – 1 = 99$$

(ii) $$99 – 3 = 96$$

(iii) $$96 – 5 = 91$$

(iv) $$91 – 7 = 84$$

(v) $$84 – 9 = 75$$

(vi) $$75 – 11 = 64$$

(vii) $$64 – 13 = 51$$

(viii) $$51 – 15 = 36$$

(ix) $$36 – 17 = 19$$

(x) $$19 – 19 = 0$$

We have subtracted successive odd numbers, and $$10$$ steps have been required for getting the result as $$0.$$

So the square root of $$100$$ is $$10$$.

Consider $$\sqrt {169}$$ , then

i) $$169 – 1= 168$$

ii)$$168 – 3= 165$$

iii) $$165 – 5=160$$

iv) $$160 – 7=153$$

v) $$153 – 9=144$$

vi)$$144 – 11=133$$

vii) $$133 – 13=120$$

viii) $$120 – 15=105$$

ix) $$105 – 17=88$$

x) $$88 – 19=69$$

xi) $$69 – 21=48$$

xii) $$48 – 23=25$$

xiii) $$25 – 25=0$$

We have subtracted successive odd numbers, and $$13$$ steps have been required for getting the result as $$0.$$

So the square root of $$169$$ is $$13$$.

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