# Ex.6.2 Q1 Squares and Square Roots Solutions - NCERT Maths Class 8

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## Question

Find the squares of the following numbers.

(i) $$32$$

(ii) $$35$$

(iii) $$86$$

(iv) $$93$$

(v) $$71$$

(vi) $$46$$

## Text Solution

What is known?

Numbers

What is unknown?

Squares of the numbers.

Reasoning 1:

There is a way to find this by without multiplication.

Use identity:

$$(a+b)^2=a^2+2ab+b^2$$

$$(a+b)^2=a^2-2ab+b^2$$

Steps (i):

\begin{align}32 &= 30 + 2\\{32^2} &= {(30 + 2)^2}\\ &= 30(30 + 2) + 2(30 + 2)\\ &= {30^2} + 30 \times 2 + 2 \times 30 + {2^2}\\ &= 900 + 60 + 60 + 4\\ &= 1024\end{align}

Reasoning 2:

If a number have its unit digit $$5$$ i.e. $$a5$$, its square number will be [$$a (a+1)\; \rm hundred+25$$].

Steps (ii):

Hence $$a = 3$$

Square of the number $$35$$

\begin{align}&= [3(3+1) \text{hundreds}+25]\\&= [ (3\times 4) \;\text{hundreds}+25]\\&= 1200+25\\&= 1225 \end{align}

Reasoning:

There is a way to find this by without multiplication

Steps (iii):

\begin{align}86 &= 80 + {86^2}\\ &= {(80 + 6)^2}{\rm{ }}\\ &= 80(80 + 6) + 6(80 + 6)\\ &= {80^2} + 80 \times 6 + 6 \times 80 + {6^2}\\ &= 6400 + 480 + 480 + 36\\&= 7396 \end{align}

Reasoning:

There is a way to find this by without multiplication.

Steps (iv):

\begin{align}93 &= 90 + 3\\{93^2} &= {(90 + 3)^2}\\ &= 90(90 + 3) + 3(90 + 3)\\ &= {90^2} + 90 \times 3 + 3 \times 90 + {3^2}\\ &= 8100 + 270 + 270 + 9\\ &= 8649 \end{align}

Reasoning

There is a way to find this by without multiplication.

Steps (v):

\begin{align}71 &= 70 + 1\\{71^2} &= {(70 + 1)^2}\\ &= 70(70 + 1) + 1(70 + 1)\\ &= {70^2} + 70 \times 1 + 1 \times 70 + {1^2}\\ &= 4900 + 70 + 70 + 1\\ &= 5041 \end{align}

Reasoning

There is a way to find this by without multiplication.

Steps (vi):

\begin{align} 46 &= 40 + 6\\{46^2} &= {(40 + 6)^2}\\ &= 40(40 + 6) + 6(40 + 6)\\ &= {40^2} + 40 \times 6 + 6 \times 40 + {6^2}\\ &= 1600 + 240 + 240 + 36\\ &= 2116 \end{align}

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