# 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.

## 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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