# Difference between revisions of "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 $h = 16 - \frac{5}{4} = \frac{59}{4}$ $\mathrm{(B)}$.

### 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 $c^2 = (32-c)^2 - 80$.

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

### 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 $a + b = \frac{69}{4}$.

Finally, subtracting this from our original value of 32, we get $\frac{59}{4}$, or $B$.

### Solution 6

Let the legs be $a, b$. Then the hypotenuse is $\sqrt{a^2 + b^2}.$

We know that $ab = 40$ and $a + b + \sqrt{a^2 + b^2} = 32$.

The first equation gives $$b = \frac{40}{a}$$ and we can plug this into the second equation, yielding: $$a + \frac{40}{a} + \sqrt{ a^2 + \frac{1600}{a^2}} = 32.$$

Letting $X = a + \frac{40}{a}$, the equation becomes: $$X + \sqrt{X^2 - 80} = 32.$$

We can bring the $X$ to the right side and square which yields: $$X^2 - 80 = (X - 32)^2 = X^2 - 64X + 1024.$$

So, $$-80 = -64X + 1024 \rightarrow X = \frac{69}{4}.$$

Now, we know that $a + \frac{40}{a} = \frac{69}{4}.$

Multiplying both sides by $4a$ gives: $$4a^2 - 69a + 160.$$

It's not hard to see that the roots of this equation are $a$ and $b$. We want the hypotenuse which is $\sqrt{a^2 + b^2} = \sqrt{ (a+b)^2 - 2ab}.$

We can now apply Vieta's Formula which gives: $$c = \sqrt{ \left( \frac{69}{4} \right)^2 - 80} = \boxed{ \frac{59}{4}}.$$

~conantwiz2023

### Solution 7

Let the sides be $a, b, c$ where $a$ and $b$ are the legs and $c$ is the hypotenuse.

Since the perimeter is 32, we have

$(1) \phantom{a} a+b+c=32$.

Since the area is 20 and the legs are $a$ and $b$, we have that

$(2) \phantom{a} \frac{a \cdot b}{2}=20$.

By the Pythagorean Theorem, we have that

$(3) \phantom{a} a^2+b^2=c^2$.

Since we want $c$, we will equations $1, 2, 3$ be in the form of $c.$

Equation 1 can be turned into

$(4) \phantom{a} a+b=32-c$.

Equation 2 can be simplified into

$(5) \phantom{a} ab=40.$

Onto the calculating process.

Squaring the 1st equation we have

$(a+b+c)^2=32^2.$

Expanding and grouping, we have

$(a^2+b^2+c^2)+2(ab+ac+bc)=32^2.$

By equation 3 and substituting we get

$2(c^2+ab+ac+bc)=32^2.$

By equation 5 and substituting we get

$2(c^2+40+ac+bc)=32^2.$

Note that we can factor $c$ out in the inner expression, and we get

$2(c^2+40+c(a+b))=32^2.$

By equation 4 and substituting, we have

$2(c^2+40+c(32-c))=32^2.$

Expanding, we have

$2(c^2+40+32c-c^2)=32^2.$

Simplifying, we have

$2(40+32c)=32^2.$

Expanding again, we get

$80+64c=32^2.$

Dividing both sides by $16$ gets us

$4c+5=64$

Calculating gets us

$c=\boxed{\mathrm{(B) \frac{59}{4}}}$.

~mathboy282

~savannahsolver