2008 AMC 10A Problems/Problem 18

Problem

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

Solution

Solution 1

Let the legs of the triangle have lengths . Then, by the Pythagorean Theorem, the length of the hypotenuse is , and the area of the triangle is . So we have the two equations

Re-arranging the first equation and squaring,

$=

\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}$ (Error compiling LaTeX. Unknown error_msg)

From we have , so

$a+b &= \frac{80 + 32^2}{64} = \frac{69}{4}.$ (Error compiling LaTeX. Unknown error_msg)

The length of the hypotenuse is .

Solution 2

From the formula , where is the area of a triangle, is its inradius, and is the semiperimeter, we can find that . It is known that in a right triangle, , where is the hypotenuse, so .

Solution 3

From the problem, we know that

\begin{align*}

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

\end{align*}

Subtracting 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 as well as 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 .

Solution 4

Let and be the legs of the triangle, and the hypotenuse.

Since the area is 20, we have .

Since the perimeter is 32, we have .

The Pythagorean Theorem gives .

This gives us three equations with three variables:

Rewrite equation 3 as . Substitute in equations 1 and 2 to get .

.

The answer is choice (B).

See also

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.