2008 AMC 10A Problems/Problem 18

Problem

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

$\mathrm{(A)}\ \frac{57}{4}\qquad\mathrm{(B)}\ \frac{59}{4}\qquad\mathrm{(C)}\ \frac{61}{4}\qquad\mathrm{(D)}\ \frac{63}{4}\qquad\mathrm{(E)}\ \frac{65}{4}$

Solution

Solution 1

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}$ (Error compiling LaTeX. Unknown error_msg)

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 + 2ab + 32^2\\

a+b &= \frac{2ab+32^2}{64}\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

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

$a+b &= \frac{80 + 32^2}{64} = \frac{59}{4}.$ (Error compiling LaTeX. Unknown error_msg)

The length of the hypotenuse is $p - a - b = 32 - \frac{69}{4} = \frac{59}{4}\ \mathrm{(B)}$ .

Solution 2

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 $h = 16 - \frac{5}{4} = \frac{59}{4}$ .

2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2008 AMC 10A Problems/Problem 18

Problem

Contents

Solution

Solution 1

Solution 2

See also