A [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle) has [perimeter](https://artofproblemsolving.com/wiki/index.php/Perimeter) $32$ and area $20$. What is the length of its [hypotenuse](https://artofproblemsolving.com/wiki/index.php/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 1
Let the legs of the triangle have lengths $a,b$. Then, by the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/Inradius), and $s$ is the [semiperimeter](https://artofproblemsolving.com/wiki/index.php/Semiperimeter), we can find that $r = \frac{20}{32/2} = \frac{5}{4}$. In a right triangle, $r = s - h$, where $h$ is the hypotenuse, so $h = 16 - \frac{5}{4} = \frac{59}{4}$ $\mathrm{(B)}$.
Note: If the reader is unfamiliar with the inradius being equal to the semiperimeter minus the hypotenuse for a right triangle, it should not be too difficult to prove the relationship for themselves.~mobius247
Why $r=s-h$. Draw a right triangle and inscribe a circle. Connect the center of the circle with tangency points; there are 3, one from each side of the triangle. Now, connect the center with each vertice and you will see 3 pairs of congruent triangles by HL (Hypotenuse-Leg Congruence). From these congruent triangles, mark equal lengths and you will find that $r=s-h$. 
~BakedPotato66

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 $\frac{59}{4} \rightarrow \mathrm{(B)}$.

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 $a + b = \frac{69}{4}$.
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 can be observed 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
$(5) \phantom{a} ab=40.$
Equation 3 is already simplified.
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

Solution 8
This solution is very similar to Solution 1, except instead of subtracting $a+b$ from both sides in the first part, we subtract $\sqrt{a^2+b^2}$ from both sides, which gets us:
\begin{align*} a+b&=32-\sqrt{a^2+b^2}\\ (a+b)^2&=(32-\sqrt{a^2+b^2})^2\\ a^2+2ab+b^2&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}+a^2+b^2\\ 2ab&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}.\\ \end{align*}
We know that 
\[\frac{1}{2}ab=20\]
so 
\[ab=40.\]
We can then substitute $40$ for $ab$ to get us:
\begin{align*} 80&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}\\ 2\cdot32\cdot\sqrt{a^2+b^2}&=32^2-80\\ 64\cdot\sqrt{a^2+b^2}&=32^2-80\\ \sqrt{a^2+b^2}&=\frac{32^2-80}{64}\\ \end{align*}
We know that $\sqrt{a^2+b^2}$ is the hypotenuse, so we only have to solve the right-hand side now.
\begin{align*} \sqrt{a^2+b^2}&=\frac{32^2-80}{64}\\ &=\frac{32^2}{64}-\frac{80}{64}\\ &=\frac{32\cdot32}{32\cdot2}-\frac{32\cdot\frac{5}{2}}{32\cdot2}\\ &=\frac{32}{2}-\frac{\frac{5}{2}}{2}\\ &=16-\frac{5}{4}\\ &=\frac{64}{4}-\frac{5}{4}\\ &=\boxed{\mathrm{(B)} \frac{59}{4}}\\ \end{align*}
~zlrara01