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

(→‎Solution 1)
(Latex fix)
Line 9: Line 9:
 
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>
 
<center>
−
<math>a+b+\sqrt{a^2+b^2} = 32 \\
+
<math>a+b+\sqrt{a^2+b^2} = 32 \\\\
 
\frac{1}{2}ab = 20</math>
 
\frac{1}{2}ab = 20</math>
 
</center>
 
</center>
 
Re-arranging the first equation and squaring,  
 
Re-arranging the first equation and squaring,  
 
<center>
 
<center>
−
<math> =\sqrt{a^2+b^2} &= 32-(a+b)\\
+
<math> \sqrt{a^2+b^2} = 32-(a+b)\\\\
−
a^2 + b^2 &= 32^2 - 64(a+b) + (a+b)^2\\
+
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^2 + b^2 + 64(a+b) = a^2 + b^2 + 2ab + 32^2\\\\
−
a+b &= \frac{2ab+32^2}{64}</math>
+
a+b = \frac{2ab+32^2}{64}</math>
 
</center>
 
</center>
 
From <math>(2)</math> we have <math>2ab = 80</math>, so
 
From <math>(2)</math> we have <math>2ab = 80</math>, so
 
<center>
 
<center>
−
<math>a+b &= \frac{80 + 32^2}{64} = \frac{69}{4}.</math>
+
<math>a+b = \frac{80 + 32^2}{64} = \frac{69}{4}.</math>
 
</center>
 
</center>
 +
 
The length of the hypotenuse is <math>p - a - b = 32 - \frac{69}{4} = \frac{59}{4}\ \mathrm{(B)}</math>.
 
The length of the hypotenuse is <math>p - a - b = 32 - \frac{69}{4} = \frac{59}{4}\ \mathrm{(B)}</math>.
  
Line 30: Line 31:
 
=== Solution 3 ===
 
=== Solution 3 ===
 
From the problem, we know that  
 
From the problem, we know that  
−
<center><math>\begin{align*}
+
<center><cmath>\begin{align*}
 
a+b+c &= 32 \\
 
a+b+c &= 32 \\
 
2ab &= 80. \\
 
2ab &= 80. \\
−
\end{align*}</math></center>
+
\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*}
+
<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>
+
\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>
+
\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>.
Line 68: Line 69:
 
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 \\
+
<center><math>c^2 = (32-c)^2 - 80 \\\\
−
c^2 = 1024 - 64c + c^2 - 80 \\
+
c^2 = 1024 - 64c + c^2 - 80 \\\\
−
64c = 944 \\
+
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).

Revision as of 17:32, 29 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,

$\sqrt{a^2+b^2} = 32-(a+b)\\\\ a^2 + b^2 = 32^2 - 64(a+b) + (a+b)^2\\\\ a^2 + b^2 + 64(a+b) = a^2 + b^2 + 2ab + 32^2\\\\ a+b = \frac{2ab+32^2}{64}$

From $(2)$ we have $2ab = 80$, so

$a+b = \frac{80 + 32^2}{64} = \frac{69}{4}.$

The length of the hypotenuse is $p - a - b = 32 - \frac{69}{4} = \frac{59}{4}\ \mathrm{(B)}$.

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).

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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