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# Ex.6.2 Q2 Squares and Square Roots Solutions - NCERT Maths Class 8

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## Question

Write a Pythagorean triplet whose one member is.

(i) $$6$$

(ii) $$14$$

(iii) $$16$$

(iv) $$18$$

Video Solution
Squares And Square Roots
Ex 6.2 | Question 2

## Text 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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