Difference between revisions of "2008 AMC 10A Problems/Problem 18"
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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> | + | <math> \sqrt{a^2+b^2} = 32-(a+b)\\\\ |
− | a^2 + 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 + 64(a+b) = a^2 + b^2 + 2ab + 32^2\\\\ |
− | a+b | + | 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 | + | <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>. | ||
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=== Solution 3 === | === Solution 3 === | ||
From the problem, we know that | From the problem, we know that | ||
− | <center>< | + | <center><cmath>\begin{align*} |
a+b+c &= 32 \\ | a+b+c &= 32 \\ | ||
2ab &= 80. \\ | 2ab &= 80. \\ | ||
− | \end{align*}</ | + | \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>< | + | <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*}</ | + | \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>< | + | <center><cmath>\begin{align*} |
80 &= 1024 - 64c\\ | 80 &= 1024 - 64c\\ | ||
c &= \frac{944}{64}. | c &= \frac{944}{64}. | ||
− | \end{align*}</ | + | \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>. | ||
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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 and area
. What is the length of its hypotenuse?
Contents
Solution
Solution 1
Let the legs of the triangle have lengths . Then, by the Pythagorean Theorem, the length of the hypotenuse is
, and the area of the triangle is
. So we have the two equations
Re-arranging the first equation and squaring,
From we have
, so
The length of the hypotenuse is .
Solution 2
From the formula , where
is the area of a triangle,
is its inradius, and
is the semiperimeter, we can find that
. It is known that in a right triangle,
, where
is the hypotenuse, so
.
Solution 3
From the problem, we know that
![\begin{align*} a+b+c &= 32 \\ 2ab &= 80. \\ \end{align*}](http://latex.artofproblemsolving.com/9/b/1/9b1ad9609b79bdcfc10cd8052a111b659e5b0f13.png)
Subtracting 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*}](http://latex.artofproblemsolving.com/8/f/1/8f13a9ea7ec9417ab1fab36f5ae1d9ecb5ef4179.png)
Now we substitute in as well as
into the equation to get
![\begin{align*} 80 &= 1024 - 64c\\ c &= \frac{944}{64}. \end{align*}](http://latex.artofproblemsolving.com/a/a/1/aa184abe19e53f4bb2e8712ad5b4e6a2e0aaf82d.png)
Further simplification yields the result of .
Solution 4
Let and
be the legs of the triangle, and
the hypotenuse.
Since the area is 20, we have .
Since the perimeter is 32, we have .
The Pythagorean Theorem gives .
This gives us three equations with three variables:
![$ab = 40 \\ a + b + c = 32 \\ c^2 = a^2 + b^2$](http://latex.artofproblemsolving.com/a/b/3/ab34477dc4e0b22ad043a9d2ae5228ade3611d17.png)
Rewrite equation 3 as .
Substitute in equations 1 and 2 to get
.
![$c^2 = (32-c)^2 - 80 \\\\ c^2 = 1024 - 64c + c^2 - 80 \\\\ 64c = 944 \\\\ c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4}$](http://latex.artofproblemsolving.com/b/b/0/bb0955d038919472bbefb6e672b0b6c7192fb561.png)
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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