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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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