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) \(18\)

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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