Question

Write a Pythagorean triplet whose one member is 16.

Answer 1

Hint: By Pythagorean triplet we mean that the Pythagoras theorem requires three numbers to be satisfied, therefore we have been asked to find a Pythagorean triplet.
Pythagoras theorem is generally applied in the right triangle and it states that the square of the longest side is equal to the sum of the square of the other two sides.
Using the Pythagoras theorem we will solve the given problem.

Complete step-by-step solution:
Let's discuss Pythagoras theorem in more detail and then we will solve the given question.
The Pythagorean equation relates the sides of a right triangle in a simple way, so that if the lengths of any two sides are known the length of the third side can be found. Statement of the Pythagoras theorem states that in any right triangle, the hypotenuse is greater than any one of the other sides but less than their sum.
Now, we will do the calculation part of the problem;
Let's take three numbers of the Pythagorean triplet as 2m, m2-1 and m2+1.
If we take 16 equal to 2m;
$ \Rightarrow 2m = 16$
$ \Rightarrow m = \dfrac{{16}}{2}$
$ \Rightarrow m = 8$ , we obtain m equal to 8.
On substituting the value of m in order to get other two numbers is;
$ \Rightarrow {m^2} - 1 = {8^2} - 1$
$ \Rightarrow 64 - 1 = 63$ (We obtain another number as 63)
$ \Rightarrow {m^2} + 1 = {8^2} + 1$
$ \Rightarrow 64 + 1 = 65$ (We obtained 65 as another number)

Therefore the set of triplets of the Pythagoras as, 16,63 and 65.

Note: Pythagoras theorem states that the sum of squares of the adjacent and opposite sides of a triangle is equal to the square of hypotenuse of the same triangle. Pythagoras formula is given by ${\text{opposite}}^2 + {\text{adjacent}}^2 = {\text{hypotenuse}}^2$.