Difference between revisions of "2008 AMC 10A Problems/Problem 18"
(→Solution 1) |
|||
Line 8: | Line 8: | ||
=== 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
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. ! Misplaced alignment tab character &.)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. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)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. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)Now we substitute in as well as into the equation to get
80 &= 1024 - 64c\\ c &= \frac{944}{64}.
\end{align*}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.)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.