Show that the length of the hypotenuse of a right triangle is $\ge$ the length of a leg.

Question

This question is in chapter 5 $\S$3 of Serge Lang's Basic Mathematics.

Am I wrong in thinking there is no right triangle that exists such that the length of the hypotenuse is equal to the length of a leg? It seems to me that for a right triangle, the length of the hypotenuse will always be greater than the length of a leg, but never equal.

Answer 1

An equality can happen in degenerate triangle. However, it wouldn't be a right triangle anymore.

By the pythagorean theorem,

$$c^2 = a^2 + b^2$$

If $a=0$ or $b=0$, then it is the case that the hypothenuse is the same length as a leg.

So it only is true, if we include degenerate triangles.

In the case that $a$ and $b$ must be positive, then $c^2$ is greater than $a^2$ or $b^2$.

Suppose that $c^2$ = $a^2$,

By definition, we know that $c^2 = a^2 + b^2$, and since $b^2$ is non-zero then $a^2 \lt a^2 + b^2$, meaning that $c^2$ cannot be equal to $a^2$. The same argument applies to $b^2$.

Answer 2

By the Pythagorean Theorem, we know if a right triangle has legs of length $a,b$, then it's hypotenous has length $$ c:=\sqrt{a^2+b^2}. $$ Since these lengths $a,b$ are both positive, as they connect two distinct points in a plane, it follows that $$ c > \sqrt{\max\{a^2,b^2\}} = max\{a,b\}. $$

Hope this help. Thanks for the awesome question! :D