Difference between revisions of "2008 AMC 10A Problems/Problem 18"
(→Solution 1) |
m (→Solution 1) |
||
| Line 13: | Line 13: | ||
</center> | </center> | ||
Re-arranging the first equation and squaring, | Re-arranging the first equation and squaring, | ||
| − | <center><math> = | + | <center> |
| − | \sqrt{a^2+b^2} &= 32-(a+b)\\ | + | <math> = \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}</math></center> | 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> | ||
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 23:09, 27 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,
$= \sqrt{a^2+b^2} &= 32-(a+b)\\ a^2 + b^2 &= 32^2 - 64(a+b) + (a+b)^2\\ a^2 + b^2 + 64(a+b) &= a^2 + b^2 + 2ab + 32^2\\
a+b &= \frac{2ab+32^2}{64}$ (Error compiling LaTeX. Unknown error_msg)From 
we have 
, so
$a+b &= \frac{80 + 32^2}{64} = \frac{69}{4}.$ (Error compiling LaTeX. Unknown error_msg)
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:
Rewrite equation 3 as 
.
Substitute in equations 1 and 2 to get 
.
. 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. 
