Question

30 Points !!!!!

Answer 1

Let $x$ and $y$ be two legs of a right triangle. By Pythagoras' theorem, the hypotenuse will have length $\sqrt{x^2+y^2}$:

$x^2+y^2=(\sqrt{x^2+y^2})^2$

What you should do is read the first two values in the given lists as $x$ and $y$, and the third as $\sqrt{x^2+y^2}$.

So for example, if I gave you the list 3, 4, 5, we take

$x=3$
$y=4$
$\implies\sqrt{x^2+y^2}=\sqrt{9+16}=\sqrt{25}=5$

Since $\sqrt{x^2+y^2}=x^2+y^2$, it follows that (3, 4, 5) is indeed a Pythagorean triple.

Let's check with the first problem: If $x=65$ and $y=72$, we have

$\sqrt{x^2+y^2}=\sqrt{9409}=97$

which means (65, 72, 97) is also a Pythagorean triple.