Difference between revisions of "2008 AMC 10A Problems/Problem 18"

Revision as of 23:04, 27 September 2015

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}$ (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*}

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).

@@ Line 27: / Line 27: @@
 === Solution 3 ===
 From the problem, we know that
-<center>
+<center>\begin{align*}
-<math>\begin{align*}
 a+b+c &= 32 \\
 ab &= 80. \\
-\end{align*}</math></center>
+\end{align*}</center>
 Subtracting <math>c</math> from both sides of the first equation and squaring both sides, we get
-<center>
+<center><math>\begin{align*}
-<math>\begin{align*}
 (a+b)^2 &= (32 - c)^2\\
 a^2 + b^2 + 2ab &= 32^2 + c^2 - 64c.\\
@@ Line 41: / Line 39: @@
 Now we substitute in <math>a^2 + b^2 = c^2</math> as well as <math>2ab = 80</math> into the equation to get
-<center>
+<center><math>\begin{align*}
-<math>\begin{align*}
 &= 1024 - 64c\\
 c &= \frac{944}{64}.

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