Difference between revisions of "2008 AMC 10A Problems/Problem 18"
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− | ==Problem== | + | == Problem == |
A [[right triangle]] has [[perimeter]] <math>32</math> and area <math>20</math>. 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> | ||
− | + | == Solution == | |
− | ==Solution== | ||
=== 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 | ||
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=== Solution 4 === | === Solution 4 === | ||
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Let <math>a</math> and <math>b</math> be the legs of the triangle, and <math>c</math> the hypotenuse. | Let <math>a</math> and <math>b</math> be the legs of the triangle, and <math>c</math> the hypotenuse. | ||
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The answer is choice (B). | The answer is choice (B). | ||
− | ===Solution 5=== | + | === Solution 5 === |
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Let <math>a</math>, <math>b</math>, and <math>c</math> be the sides of the triangle, with <math>c</math> as the hypotenuse. | Let <math>a</math>, <math>b</math>, and <math>c</math> be the sides of the triangle, with <math>c</math> as the hypotenuse. | ||
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Finally, subtracting this from our original value of 32, we get <math>\frac{59}{4}</math>, or <math>B</math>. | Finally, subtracting this from our original value of 32, we get <math>\frac{59}{4}</math>, or <math>B</math>. | ||
− | ==See | + | == 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]] | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:32, 19 October 2020
Contents
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
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).
Solution 5
Let , , and be the sides of the triangle, with as the hypotenuse.
We know that .
According to the Pythagorean Theorem, we have .
We also know that = 40, since the area of the triangle is 20.
We substitute into to get .
Moving the to the left, we again rewrite to get .
We substitute our value of 32 for twice into our equation and subtract to get .
Finally, subtracting this from our original value of 32, we get , or .
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.