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>
  
__TOC__
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== Solution ==
==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
<center><math>
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<center>
a+b+\sqrt{a^2+b^2} = 32 \\
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<math>a+b+\sqrt{a^2+b^2} = 32 \\\\
\frac{1}{2}ab = 20
+
\frac{1}{2}ab = 20</math>
</math></center>
+
</center>
 
Re-arranging the first equation and squaring,  
 
Re-arranging the first equation and squaring,  
<center><math> =
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<center>
\sqrt{a^2+b^2} &= 32-(a+b)\\
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<math> \sqrt{a^2+b^2} = 32-(a+b)\\\\
a^2 + b^2 &= 32^2 - 64(a+b) + (a+b)^2\\
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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\\
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a^2 + b^2 + 64(a+b) = a^2 + b^2 + 2ab + 32^2\\\\
a+b &= \frac{2ab+32^2}{64}</math></center>
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a+b = \frac{2ab+32^2}{64}</math>
 +
</center>
 
From <math>(2)</math> we have <math>2ab = 80</math>, so
 
From <math>(2)</math> we have <math>2ab = 80</math>, so
<center><math>a+b &= \frac{80 + 32^2}{64} = \frac{69}{4}.</math></center>
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<center>
The length of the hypotenuse is <math>p - a - b = 32 - \frac{69}{4} = \frac{59}{4}\ \mathrm{(B)}</math>.
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<math>a+b = \frac{80 + 32^2}{64} = \frac{69}{4}.</math>
 +
</center>
 +
 
 +
The length of the hypotenuse is <math>p - a - b = 32 - \frac{69}{4} = \frac{59}{4}\boxed{\ \mathrm{(B)}}</math>.
  
 
=== Solution 2 ===
 
=== Solution 2 ===
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=== Solution 3 ===
 
=== Solution 3 ===
 
From the problem, we know that  
 
From the problem, we know that  
<center>begin{align*}
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<center><cmath>\begin{align*}
 
a+b+c &= 32 \\
 
a+b+c &= 32 \\
 
2ab &= 80. \\
 
2ab &= 80. \\
\end{align*}</center>
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\end{align*}</cmath></center>
  
 
Subtracting <math>c</math> from both sides of the first equation and squaring both sides, we get
 
Subtracting <math>c</math> from both sides of the first equation and squaring both sides, we get
<center><math>\begin{align*}
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<center><cmath>\begin{align*}
 
(a+b)^2 &= (32 - c)^2\\
 
(a+b)^2 &= (32 - c)^2\\
 
a^2 + b^2 + 2ab &= 32^2 + c^2 - 64c.\\
 
a^2 + b^2 + 2ab &= 32^2 + c^2 - 64c.\\
\end{align*}</math></center>
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\end{align*}</cmath></center>
  
 
Now we substitute in <math>a^2 + b^2 = c^2</math> as well as <math>2ab = 80</math> into the equation to get
 
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><math>\begin{align*}
+
<center><cmath>\begin{align*}
 
80 &= 1024 - 64c\\
 
80 &= 1024 - 64c\\
 
c &= \frac{944}{64}.
 
c &= \frac{944}{64}.
\end{align*}</math></center>
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\end{align*}</cmath></center>
  
 
Further simplification yields the result of <math>\frac{59}{4}</math>.
 
Further simplification yields the result of <math>\frac{59}{4}</math>.
  
 
=== Solution 4 ===
 
=== Solution 4 ===
 
 
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.
  
Line 65: Line 67:
 
Substitute in equations 1 and 2 to get <math>c^2 = (32-c)^2 - 80</math>.
 
Substitute in equations 1 and 2 to get <math>c^2 = (32-c)^2 - 80</math>.
  
<center><math>c^2 = (32-c)^2 - 80 \\
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<center><math>c^2 = (32-c)^2 - 80 \\\\
c^2 = 1024 - 64c + c^2 - 80 \\
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c^2 = 1024 - 64c + c^2 - 80 \\\\
64c = 944 \\
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64c = 944 \\\\
 
c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4} </math>. </center>
 
c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4} </math>. </center>
 
The answer is choice (B).
 
The answer is choice (B).
  
==See also==
+
=== Solution 5 ===
 +
Let <math>a</math>, <math>b</math>, and <math>c</math> be the sides of the triangle, with <math>c</math> as the hypotenuse.
 +
 
 +
We know that <math>a + b + c =32</math>.
 +
 
 +
According to the Pythagorean Theorem, we have <math>a^2 + b^2 = c^2</math>.
 +
 
 +
We also know that <math>ab</math> = 40, since the area of the triangle is 20.
 +
 
 +
We substitute <math>2ab</math> into <math>a^2 + b^2 = c^2</math> to get <math>(a+b)^2 = c^2 + 80</math>.
 +
 
 +
Moving the <math>c^2</math> to the left, we again rewrite to get <math>(a+b+c)(a+b-c) = 80</math>.
 +
 
 +
We substitute our value of 32 for <math>a+b+c</math> twice into our equation and subtract to get <math>a + b = \frac{69}{4}</math>.
 +
 
 +
Finally, subtracting this from our original value of 32, we get <math>\frac{59}{4}</math>, or <math>B</math>.
 +
 
 +
== 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 00:32, 19 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 10 Problems and Solutions

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