# Excercise 6.2 Squares and Square Roots- NCERT Solutions Class 8

## Chapter 6 Ex.6.2 Question 1

Find the squares of the following numbers.

(i) $$32$$

(ii) $$35$$

(iii) $$86$$

(iv) $$93$$

(v) $$71$$

(vi) $$46$$

### 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}

## Chapter 6 Ex.6.2 Question 2

Write a Pythagorean triplet whose one member is.

(i) $$6$$

(ii) $$14$$

(iii) $$16$$

(iv) $$18$$

### Solution

What is known?

One of the members of Pythagorean triplet.

What is unknown?

Other two member of Pythagorean triplet.

Reasoning:

For any natural number $$''m''$$ where $$m>1$$, we have \begin{align} (2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2 \end{align} so, $$2m$$, $$m^2 - 1 \,\,{\rm{and}}\,\,m^2 + 1$$ forms a Pythagorean triplet

Steps (i):

$$6$$

If we take

\begin{align} m^2 + 1 &= 6\\m^2 &= 5\end{align}

the value of $$m$$ will not be an integer.

If we take

\begin{align} m^2 - 1 &= 6\\m^2 &= 7 \end{align}

Again,the value of $$m$$ will not be an integer.

Let ,$$2m=6$$

\begin{align}m &= \frac{6}{2} = 3\\\\m^2 - 1 &= (3)^2 - 1 \\&= 9 - 1 \\&= 8\\\\m^2 + 1 &= (3)^2 + 1 \\&= 9 + 1 \\&= 10 \end{align}

Therefore, pythagorean triplets are $$6, 8$$ and $$10$$

Steps (ii):

$$14$$

If we take

\begin{align} m^2 + 1 &= 14\\m^2 &= 13 \end{align}

the value of the integer will not be an integer

If we take

\begin{align} m^2 - 1 &= 14\\m^2& = 15 \end{align}

Again, the value of $$m$$ will not be an integer.

Let $$2m=14$$

\begin{align} m&= \frac{14}{2} \\&= 7\\\\m^2 - 1 &= 7^2 - 1 \\&= 49 - 1 \\&= 48\\\\m^2 + 1 &= 7^2 + 1 \\&= 49 + 1 \\&= 50 \end{align}

Therefore ,$$14, 48, 50$$ are pythagorean triplets.

Steps (iii):

$$16$$

If we take

\begin{align} m^2 + 1& = 16\\m^2& = 15 \end{align}

the value of $$m$$ will be an integer.

If we take

\begin{align} m^2 - 1& = 16\\m^2& = 17 \end{align}

The value of $$m$$ will be an Integer.

Let ,$$2m=6$$

\begin{align} m &= \frac{16}{2} \\&= 8\\\\m^2 - 1 &= 8^2 - 1 \\&= 64 - 1 \\&= 63\\\\m^2 + 1 &= 8^2 + 1 \\&= 64 + 1 \\&= 65 \end{align}

Therefore, $$16, 63$$ and $$65$$ are pythagorean triplets.

Steps (iv):

$$18$$

If we take

\begin{align} m^2 - 1 &= 18\\m^2& = 19 \end{align}

the value of $$m$$ will not be an integer.

If we take

\begin{align} m^2 + 1& = 18\\m^2& = 17 \end{align}

again the value of $$m$$ will not be an Integer.

Let $$2m=18$$

\begin{align}m &= \frac{18}{2} \\&= 9\\\\m^2 - 1 &= 9^2 - 1 \\&= 81 - 1 \\&= 80\\\\m^2 + 1 &= 9^2 + 1 \\&= 81 + 1 \\&= 82 \end{align}

Therefore, $$18, 80$$ and $$82$$ are pythagorean triplets.

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