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

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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 \( 13\).