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Revision as of 10:53, 4 July 2013

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

$a+b &= \frac{80 + 32^2}{64} = \frac{69}{4}.$ (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 $h = 16 - \frac{5}{4} = \frac{59}{4}$.

Solution 3

From the problem, we know that

$\begin{align*}

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

\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

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. Unknown error_msg)

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. Unknown error_msg)

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

Solution 4

Let $a$ and $b$ be the legs of the triangle, and $c$ the hypotenuse.

Since the area is 20, we have $\frac{1}{2}ab = 20 => ab=40$.

Since the perimeter is 32, we have $a + b + c = 32$.

The Pythagorean Theorem gives $c^2 = a^2 + b^2$.

This gives us three equations with three variables:

$ab = 40 \\ a + b + c = 32 \\ c^2 = a^2 + b^2$

Rewrite equation 3 as $c^2 = (a+b)^2 - 2ab$. Substitute in equations 1 and 2 to get $c^2 = (32-c)^2 - 80$.

$c^2 = (32-c)^2 - 80 \\ c^2 = 1024 - 64c + c^2 - 80 \\ 64c = 944 \\ c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4}$.

The answer is choice (B).

See also

2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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