Difference between revisions of "2008 AMC 10A Problems/Problem 18"
I like pie (talk | contribs) (Added problem, solution still needed) |
(solutions) |
||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | A right triangle has perimeter 32 and area 20. What is the length of its hypotenuse? | + | A [[right triangle]] has [[perimeter]] <math>32</math> and area <math>20</math>. What is the length of its [[hypotenuse]]? |
<math>\mathrm{(A)}\ \frac{57}{4}\qquad\mathrm{(B)}\ \frac{59}{4}\qquad\mathrm{(C)}\ \frac{61}{4}\qquad\mathrm{(D)}\ \frac{63}{4}\qquad\mathrm{(E)}\ \frac{65}{4}</math> | <math>\mathrm{(A)}\ \frac{57}{4}\qquad\mathrm{(B)}\ \frac{59}{4}\qquad\mathrm{(C)}\ \frac{61}{4}\qquad\mathrm{(D)}\ \frac{63}{4}\qquad\mathrm{(E)}\ \frac{65}{4}</math> | ||
| + | __TOC__ | ||
==Solution== | ==Solution== | ||
| − | {{ | + | === 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 | ||
| + | <center><math>\begin{align} | ||
| + | a+b+\sqrt{a^2+b^2} &= 32 \\ | ||
| + | \frac{1}{2}ab &= 20 | ||
| + | \end{align}</math></center> | ||
| + | Re-arranging the first equation and squaring, | ||
| + | <center><math>\begin{align*} | ||
| + | \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*}</math></center> | ||
| + | From <math>(2)</math> we have <math>2ab = 80</math>, so | ||
| + | <center><math>a+b &= \frac{80 + 32^2}{64} = \frac{59}{4}.</math></center> | ||
| + | The length of the hypotenuse is <math>p - a - b = 32 - \frac{69}{4} = \frac{59}{4}\ \mathrm{(B)}</math>. | ||
| + | |||
| + | === Solution 2 === | ||
| + | From the formula <math>A = rs</math>, where <math>A</math> is the area of a triangle, <math>r</math> is its [[inradius]], and <math>s</math> is the [[semiperimeter]], we can find that <math>r = \frac{20}{32/2} = \frac{5}{4}</math>. It is known that in a right triangle, <math>r = s - h</math>, where <math>h</math> is the hypotenuse, so <math>h = 16 - \frac{5}{4} = \frac{59}{4}</math>. | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=A|num-b=17|num-a=19}} | {{AMC10 box|year=2008|ab=A|num-b=17|num-a=19}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
Revision as of 10:04, 26 April 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 | ||
