2008 AMC 10A Problems/Problem 18: Difference between revisions
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a+b &= \frac{2ab+32^2}{64}\end{align*}</math></center> | a+b &= \frac{2ab+32^2}{64}\end{align*}</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{ | <center><math>a+b &= \frac{80 + 32^2}{64} = \frac{69}{4}.</math></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>. | ||
Revision as of 16:04, 23 May 2008
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 | |
| 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 | ||
