Difference between revisions of "2008 AMC 10A Problems/Problem 18"
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Further simplification yields the result of <math>\frac{59}{4}</math>. | Further simplification yields the result of <math>\frac{59}{4}</math>. | ||
+ | |||
+ | === Solution 4 === | ||
+ | |||
+ | Let <math>a</math> and <math>b</math> be the legs of the triangle, and <math>c</math> the hypotenuse. | ||
+ | |||
+ | Since the area is 20, we have <math>\frac{1}{2}ab = 20 => ab=40</math>. | ||
+ | |||
+ | Since the perimeter is 32, we have <math>a + b + c = 32</math>. | ||
+ | |||
+ | The Pythagorean Theorem gives <math>c^2 = a^2 + b^2</math>. | ||
+ | |||
+ | This gives us three equations with three variables: | ||
+ | |||
+ | <center><math>ab = 40 \ | ||
+ | a + b + c = 32 \ | ||
+ | c^2 = a^2 + b^2</math></center> | ||
+ | |||
+ | Rewrite equation 3 as <math>c^2 = (a+b)^2 - 2ab</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 \ | ||
+ | c^2 = 1024 - 64c + c^2 - 80 \ | ||
+ | 64c = 944 \ | ||
+ | c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4} </math>. </center> | ||
+ | The answer is choice (D). | ||
==See also== | ==See also== |
Revision as of 21:40, 5 February 2012
Problem
A right triangle has perimeter and area
. What is the length of its hypotenuse?
Contents
[hide]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
a+b+\sqrt{a^2+b^2} &= 32 \ \frac{1}{2}ab &= 20
\end{align}$ (Error compiling LaTeX. Unknown error_msg)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}\end{align*}$ (Error compiling LaTeX. Unknown error_msg)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
a+b+c &= 32 \ 2ab &= 80. \
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)Subtracting from both sides of the first equation and squaring both sides, we get
(a+b)^2 &= (32 - c)^2\ a^2 + b^2 + 2ab &= 32^2 + c^2 - 64c.\
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)Now we substitute in as well as
into the equation to get
80 &= 1024 - 64c\ c &= \frac{944}{64}.
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)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/7/7/5/7759f060a61f35fc1dfd668f8e691e10c28f4b5c.png)
The answer is choice (D).
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 |