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

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==Problem==
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== Problem ==
 
A [[right triangle]] has [[perimeter]] <math>32</math> and area <math>20</math>. What is the length of its [[hypotenuse]]?
 
A [[right triangle]] has [[perimeter]] <math>32</math> and area <math>20</math>. What is the length of its [[hypotenuse]]?
  
 
<math>\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}</math>
 
<math>\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}</math>
  
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__TOC__
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== Solution ==
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==Solution==
 
 
=== Solution 1 ===
 
=== Solution 1 ===
 
Let the legs of the triangle have lengths <math>a,b</math>. Then, by the [[Pythagorean Theorem]], the length of the hypotenuse is <math>\sqrt{a^2+b^2}</math>, and the area of the triangle is <math>\frac 12 ab</math>. So we have the two equations
 
Let the legs of the triangle have lengths <math>a,b</math>. Then, by the [[Pythagorean Theorem]], the length of the hypotenuse is <math>\sqrt{a^2+b^2}</math>, and the area of the triangle is <math>\frac 12 ab</math>. So we have the two equations
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=== Solution 4 ===
 
=== Solution 4 ===
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Let <math>a</math> and <math>b</math> be the legs of the triangle, and <math>c</math> the hypotenuse.
 
Let <math>a</math> and <math>b</math> be the legs of the triangle, and <math>c</math> the hypotenuse.
  
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The answer is choice (B).
 
The answer is choice (B).
  
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===Solution 5===
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=== Solution 5 ===
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Let <math>a</math>, <math>b</math>, and <math>c</math> be the sides of the triangle, with <math>c</math> as the hypotenuse.
 
Let <math>a</math>, <math>b</math>, and <math>c</math> be the sides of the triangle, with <math>c</math> as the hypotenuse.
  
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Finally, subtracting this from our original value of 32, we get <math>\frac{59}{4}</math>, or <math>B</math>.
 
Finally, subtracting this from our original value of 32, we get <math>\frac{59}{4}</math>, or <math>B</math>.
  
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==See also==
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== See Also ==
 
{{AMC10 box|year=2008|ab=A|num-b=17|num-a=19}}
 
{{AMC10 box|year=2008|ab=A|num-b=17|num-a=19}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:32, 18 October 2020

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

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

See Also

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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