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

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

Re-arranging the first equation and squaring,

$\sqrt{a^2+b^2} = 32-(a+b)\\\\ a^2 + b^2 = 32^2 - 64(a+b) + (a+b)^2\\\\ a^2 + b^2 + 64(a+b) = a^2 + b^2 + 2ab + 32^2\\\\ a+b = \frac{2ab+32^2}{64}$

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

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

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

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

$\begin{align*} a+b+c &= 32 \\ 2ab &= 80. \\ \end{align*}$

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

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

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

$\begin{align*} 80 &= 1024 - 64c\\ c &= \frac{944}{64}. \end{align*}$

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).

Solution 5

Let $a$ , $b$ , and $c$ be the sides of the triangle, with $c$ as the hypotenuse.

We know that $a + b + c =32$ .

According to the Pythagorean Theorem, we have $a^2 + b^2 = c^2$ .

We also know that $ab$ = 40, since the area of the triangle is 20.

We substitute $2ab$ into $a^2 + b^2 = c^2$ to get $(a+b)^2 = c^2 + 80$ .

Moving the $c^2$ to the left, we again rewrite to get $(a+b+c)(a+b-c) = 80$ .

We substitute our value of 32 for $a+b+c$ twice into our equation and subtract to get $a+b = <cmath>\frac{69}{4}</cmath>.

Finally, subtracting this from our original value of 32, we get <cmath>\frac{59}{4}</cmath>, or$ (Error compiling LaTeX. Unknown error_msg)B$.

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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