2008 AMC 10A Problems/Problem 18

Problem

A right triangle has perimeter $32$ and area $20$ . What is the length of its hypotenuse?

Let the legs of the triangle have lengths $a,b$ . Then, by the Pythagorean Theorem, the length of the hypotenuse is $\sqrt{a^2+b^2}$ , and the area of the triangle is $\frac 12 ab$ . So we have the two equations

$\begin{align}a+b+\sqrt{a^2+b^2} &= 32 \\\frac{1}{2}ab &= 20\end{align}$

Re-arranging the first equation and squaring,

$\begin{align*}\sqrt{a^2+b^2} &= 32-(a+b)\\a^2 + b^2 &= 32^2 - 64(a+b) + (a+b)^2\\a^2 + b^2 + 64(a+b) &= a^2 + b^2...$

From $(2)$ we have $2ab = 80$ , so

$a+b &= \frac{80 + 32^2}{64} = \frac{69}{4}.$

From the formula $A = rs$ , where $A$ is the area of a triangle, $r$ is its inradius, and $s$ is the semiperimeter, we can find that $r = \frac{20}{32/2} = \frac{5}{4}$ . It is known that in a right triangle, $r = s - h$ , where $h$ is the hypotenuse, so $h = 16 - \frac{5}{4} = \frac{59}{4}$ .