2008 AMC 10A Problems/Problem 18
Problem
A right triangle has perimeter and area
. What is the length of its hypotenuse?
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
.
See also
2008 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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All AMC 10 Problems and Solutions |