Question

Write a Pythagorean triplet whose one member is 16.

Answer 1

Hint: This type of problem is based on the concept of Pythagorean triplet. First, we have to consider the given number. We know that Pythagorean triplets are of the form 2m, \[{{m}^{2}}-1\] and \[{{m}^{2}}+1\]. Let us substitute 2m=16. Divide the whole expression by 2 and find the value of m. We know that the square of 8 is 64. Using this, we can find \[{{m}^{2}}-1\] and \[{{m}^{2}}+1\]. Do necessary calculations and find the required Pythagorean triplet.

Complete step by step solution:
According to the question, we are asked to find the Pythagorean triplet whose one member is 16.
We have been given that one member is 16.
We know that a Pythagorean triplet is of the form form 2m, \[{{m}^{2}}-1\] and \[{{m}^{2}}+1\].
Let us substitute 16 as 2m.
\[\Rightarrow 2m=16\]
We can write the expression as
\[2m=2\times 8\]
Let us now divide the whole expression by 2.
\[\Rightarrow \dfrac{2m}{2}=\dfrac{2\times 8}{2}\]
We find that 2 are common in both the numerator and denominator of both LHS and RHS.
On cancelling 2, we get
m=8
Now, we have to find the other two members of the triplet.
Let us consider \[{{m}^{2}}-1\].
We know that the square of 8 is 64.
Therefore, we get \[{{m}^{2}}=64\].
\[\Rightarrow {{m}^{2}}-1=64-1\]
On further simplification, we get
\[{{m}^{2}}-1=63\]
Now, let us consider \[{{m}^{2}}+1\].
We know that \[{{m}^{2}}=64\]. On substituting in the considered expression, we get
\[{{m}^{2}}+1=64+1\]
On further simplification, we get
\[{{m}^{2}}+1=65\]
Therefore, the triplets are 16, 63 and 65.
Hence, the Pythagorean triplet whose one member is 16 is 16, 63 and 65.

Note:
We can check whether the obtained answer is correct or not.
We have to add the squares of the least two numbers in the triplet and check whether it is equal to the square of the largest number in the triplet.
\[\Rightarrow {{\left( 16 \right)}^{2}}+{{\left( 63 \right)}^{2}}={{\left( 65 \right)}^{2}}\]
Consider the LHS.
LHS= \[{{\left( 16 \right)}^{2}}+{{\left( 63 \right)}^{2}}\]
We know that \[{{\left( 16 \right)}^{2}}=256\] and \[{{\left( 63 \right)}^{2}}=3969\].
Therefore, we get
LHS=256+3969
On further simplification, we get
LHS=4225
Now consider RHS.
RHS= \[{{\left( 65 \right)}^{2}}\]
We know that \[{{\left( 65 \right)}^{2}}=4225\].
Therefore, we get
RHS=4225.
Here, we find that LHS=RHS.
Hence, the obtained answer is correct.