Question

Write the Pythagorean triplet whose one member is:
$\left( i \right)\text{ 6}$ $\left( ii \right)\text{ 14}$ $\left( iii \right)\text{ 16}$

Answer 1

Hint: Since, Pythagorean triplets are three positive integers and also sides of a right angle triangle where the square of hypotenuse $\left( h \right)$ is equal to the sum of square of perpendicular $\left( p \right)$ and base $\left( b \right)$as ${{h}^{2}}={{p}^{2}}+{{b}^{2}}$ . Here, we can write this sides as a variable in the form of $2a$ , ${{a}^{2}}-1$ and $,{{a}^{2}}+1$ where we can put any values for $a$ that will satisfy all three sides and then it must be a Pythagorean triplet.

Complete step-by-step solution:
Since, we have the three subparts in the given question; we will solve all the three one by one.
$\left( i \right)\text{ 6}$
Since, we have given one member of Pythagorean triplet that means it has two more members and here we will find them. So, we will assume this member is $2a$ as:
$\Rightarrow 2a=6$
Now, we will divide by $2$ both sides in the above step as:
$\Rightarrow \dfrac{2a}{2}=\dfrac{6}{2}$
Here, we need to find the quotient of the division as:
$\Rightarrow a=3$
So, we got the value for $a$ . Now, we will use this value to get the other member of Pythagorean triplet as:
$\Rightarrow {{a}^{2}}-1$
$\Rightarrow {{3}^{2}}-1$
Now, we will put the value of square as:
$\Rightarrow 9-1$
After subtracting we will get the value for second member of Pythagorean triplet as:
$\Rightarrow 8$
Here, we will apply the formula form third member as:
$\Rightarrow {{a}^{2}}+1$
Now, we will put the value of $a$in the above step as:
$\Rightarrow {{3}^{2}}+1$
Here, we will simplify as:
$\Rightarrow 9+1$
$\Rightarrow 10$
Since, we got the third member that is $10$, we have the Pythagorean triplet $6$ , $8$ and $10$ .

$\left( ii \right)\text{ 14}$
Now, we will use the similar method as we did for the previous subparts as:
Let us consider the given member is $2a$ as:
$\Rightarrow 2a=14$
Now, we will divide by $2$ both sides of the above step as:
$\Rightarrow \dfrac{2a}{2}=\dfrac{14}{2}$
After solving the above step, we will get as:
$\Rightarrow a=7$
Here, we will use this value of $a$ to get the second member as:
$\Rightarrow {{a}^{2}}-1$
$\Rightarrow {{7}^{2}}-1$
Now, we will solve above step as:
$\Rightarrow 49-1$
$\Rightarrow 48$
So, we got the second member. Now, we will use the formula for third member as:
$\Rightarrow {{a}^{2}}+1$
Here, we will apply the value of $a$ and will do the further process of calculation as:
$\Rightarrow {{7}^{2}}+1$
$\Rightarrow 49+1$
$\Rightarrow 50$
Since, here we got the third member also. Thus, all three members of the Pythagorean triplet are $14$ , $48$ and $50$ .

$\left( iii \right)\text{ 16}$
Here, we will repeat the same process as we did in two previous subparts to get the Pythagorean triplet as:
$\Rightarrow 2a=16$
After dividing by $2$ both sides, we will have the value of $a$ as:
$\Rightarrow \dfrac{2a}{2}=\dfrac{16}{2}$
$\Rightarrow a=8$
Now, we will use this value to get the second and third member of Pythagorean triplet as:
For second member:
$\Rightarrow {{a}^{2}}-1$
Here, we will use the value of $a$ and will do necessary calculation as:
$\Rightarrow {{8}^{2}}-1$
$\Rightarrow 64-1$
$\Rightarrow 63$
Thus, we got the second member; we will look for third member as:
$\Rightarrow {{a}^{2}}+1$
Again, we will use the value of $a$ to get the third member as:
$\Rightarrow {{8}^{2}}+1$
Now, we will complete the process by doing required calculation as:
$\Rightarrow 64+1$
$\Rightarrow 65$
Here, we got a third member also. So, the Pythagorean triplet is $8$ , $63$ and $65$.

Note: Since, we know that the square of hypotenuse $\left( h \right)$ is equal to the sum of square of perpendicular $\left( p \right)$ and base $\left( b \right)$as ${{h}^{2}}={{p}^{2}}+{{b}^{2}}$ in any Pythagorean triplet. So, we can use this method to verify our solution as:
Since, the largest number among the all three members is hypotenuse. So, the process will be:
$\left( i \right)\text{ 6}$
\[\text{Pythagorean triplet }=\text{ 6},\text{ 8},\text{ 1}0\]
$\Rightarrow {{10}^{2}}={{6}^{2}}+{{8}^{2}}$
Now, we will square and will have the above step as:
$\Rightarrow 100=36+64$
Here, we will complete the process as:
$\Rightarrow 100=100$
Since, $L.H.S.=R.H.S.$
Hence, the solution is correct.

$\left( ii \right)\text{ 14}$
Since, \[\text{Pythagorean triplet }=\text{ 14},\text{ 48},\text{ 50}\]
So, $\Rightarrow {{50}^{2}}={{14}^{2}}+{{48}^{2}}$
Now, we will complete the process by doing necessary calculations as:
$\Rightarrow 2500=196+2304$
$\Rightarrow 2500=2500$
We got, $L.H.S.=R.H.S.$
Hence, the solution is correct.

$\left( iii \right)\text{ 16}$
We already have \[\text{Pythagorean triplet }=\text{ 16},\text{ 63},\text{ 65}\]
Thus, $\Rightarrow {{65}^{2}}={{16}^{2}}+{{63}^{2}}$
Here, we will do the required calculations as:
$\Rightarrow 4225=256+3969$
$\Rightarrow 4225=4225$
Here, we got the same result as we got previously subparts as $L.H.S.=R.H.S.$
Hence, the solution is correct.