Difference between revisions of "2008 AMC 10A Problems/Problem 18"
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=== 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> | + | <center><math> |
− | a+b+\sqrt{a^2+b^2} | + | a+b+\sqrt{a^2+b^2} = 32 \\ |
− | \frac{1}{2}ab | + | \frac{1}{2}ab = 20 |
− | + | </math></center> | |
Re-arranging the first equation and squaring, | Re-arranging the first equation and squaring, | ||
− | <center><math> | + | <center><math>= |
\sqrt{a^2+b^2} &= 32-(a+b)\\ | \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 | + | 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> | <center><math>a+b &= \frac{80 + 32^2}{64} = \frac{69}{4}.</math></center> |
Revision as of 13:59, 26 June 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
![$a+b+\sqrt{a^2+b^2} = 32 \\ \frac{1}{2}ab = 20$](http://latex.artofproblemsolving.com/e/d/2/ed236d9e66ef3ee62eec6c86f37fa2f48bcc70b7.png)
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}$ (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 (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.