Difference between revisions of "2008 AMC 10A Problems/Problem 18"
(→Solution 1) |
(Latex fix) |
||
Line 9: | Line 9: | ||
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> | <center> | ||
− | <math>a+b+\sqrt{a^2+b^2} = 32 \\ | + | <math>a+b+\sqrt{a^2+b^2} = 32 \\\\ |
\frac{1}{2}ab = 20</math> | \frac{1}{2}ab = 20</math> | ||
</center> | </center> | ||
Re-arranging the first equation and squaring, | Re-arranging the first equation and squaring, | ||
<center> | <center> | ||
− | <math> | + | <math> \sqrt{a^2+b^2} = 32-(a+b)\\\\ |
− | a^2 + 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 + 64(a+b) = a^2 + b^2 + 2ab + 32^2\\\\ |
− | a+b | + | a+b = \frac{2ab+32^2}{64}</math> |
</center> | </center> | ||
From <math>(2)</math> we have <math>2ab = 80</math>, so | From <math>(2)</math> we have <math>2ab = 80</math>, so | ||
<center> | <center> | ||
− | <math>a+b | + | <math>a+b = \frac{80 + 32^2}{64} = \frac{69}{4}.</math> |
</center> | </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>. | ||
Line 30: | Line 31: | ||
=== Solution 3 === | === Solution 3 === | ||
From the problem, we know that | From the problem, we know that | ||
− | <center>< | + | <center><cmath>\begin{align*} |
a+b+c &= 32 \\ | a+b+c &= 32 \\ | ||
2ab &= 80. \\ | 2ab &= 80. \\ | ||
− | \end{align*}</ | + | \end{align*}</cmath></center> |
Subtracting <math>c</math> from both sides of the first equation and squaring both sides, we get | Subtracting <math>c</math> from both sides of the first equation and squaring both sides, we get | ||
− | <center>< | + | <center><cmath>\begin{align*} |
(a+b)^2 &= (32 - c)^2\\ | (a+b)^2 &= (32 - c)^2\\ | ||
a^2 + b^2 + 2ab &= 32^2 + c^2 - 64c.\\ | a^2 + b^2 + 2ab &= 32^2 + c^2 - 64c.\\ | ||
− | \end{align*}</ | + | \end{align*}</cmath></center> |
Now we substitute in <math>a^2 + b^2 = c^2</math> as well as <math>2ab = 80</math> into the equation to get | Now we substitute in <math>a^2 + b^2 = c^2</math> as well as <math>2ab = 80</math> into the equation to get | ||
− | <center>< | + | <center><cmath>\begin{align*} |
80 &= 1024 - 64c\\ | 80 &= 1024 - 64c\\ | ||
c &= \frac{944}{64}. | c &= \frac{944}{64}. | ||
− | \end{align*}</ | + | \end{align*}</cmath></center> |
Further simplification yields the result of <math>\frac{59}{4}</math>. | Further simplification yields the result of <math>\frac{59}{4}</math>. | ||
Line 68: | Line 69: | ||
Substitute in equations 1 and 2 to get <math>c^2 = (32-c)^2 - 80</math>. | Substitute in equations 1 and 2 to get <math>c^2 = (32-c)^2 - 80</math>. | ||
− | <center><math>c^2 = (32-c)^2 - 80 \\ | + | <center><math>c^2 = (32-c)^2 - 80 \\\\ |
− | c^2 = 1024 - 64c + c^2 - 80 \\ | + | c^2 = 1024 - 64c + c^2 - 80 \\\\ |
− | 64c = 944 \\ | + | 64c = 944 \\\\ |
c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4} </math>. </center> | c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4} </math>. </center> | ||
The answer is choice (B). | The answer is choice (B). |
Revision as of 17:32, 29 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,
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
Subtracting from both sides of the first equation and squaring both sides, we get
Now we substitute in as well as into the equation to get
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.