2008 AMC 10A Problems/Problem 18

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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. ! Package amsmath Error: \begin{align} allowed only in paragraph mode.)

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. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)

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

$a+b &= \frac{80 + 32^2}{64} = \frac{69}{4}.$ (Error compiling LaTeX. ! Misplaced alignment tab character &.)

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

Solution 3

From the problem, we know that

$\begin{align*}

a+b+c &= 32 \\ 2ab &= 80. \\

\end{align*}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)

Subtracting $c$ from both sides of the first equation and squaring both sides, we get

$\begin{align*}

(a+b)^2 &= (32 - c)^2\\ a^2 + b^2 + 2ab &= 32^2 + c^2 - 64c.\\

\end{align*}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)

Now we substitute in $a^2 + b^2 = c^2$ as well as $2ab = 80$ into the equation to get

$\begin{align*}

80 &= 1024 - 64c\\ c &= \frac{944}{64}.

\end{align*}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)

Further simplification yields the result of $\frac{59}{4}$.

See also

2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 10 Problems and Solutions
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