Difference between revisions of "2008 AMC 10A Problems/Problem 18"
(→Solution 2) |
Mathboy282 (talk | contribs) (→Solution 6) |
||
Line 89: | Line 89: | ||
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>. | ||
+ | |||
+ | == Solution 6 == | ||
+ | Let the sides be <math>a, b, c</math> where <math>a</math> and <math>b</math> are the legs and <math>c</math> is the hypotenuse. | ||
+ | |||
+ | Since the perimeter is 32, we have | ||
+ | |||
+ | <math>(1) \phantom{a} a+b+c=32</math>. | ||
+ | |||
+ | Since the area is 20 and the legs are <math>a</math> and <math>b</math>, we have that | ||
+ | |||
+ | <math>(2) \phantom{a} \frac{a \cdot b}{2}=20</math>. | ||
+ | |||
+ | By the Pythagorean Theorem, we have that | ||
+ | |||
+ | <math>(3) \phantom{a} a^2+b^2=c^2</math>. | ||
+ | |||
+ | Since we want <math>c</math>, we will equations <math>1, 2, 3</math> be in the form of <math>c.</math> | ||
+ | |||
+ | Equation 1 can be turned into | ||
+ | |||
+ | <math>(4) \phantom{a} a+b=32-c</math>. | ||
+ | |||
+ | Equation 2 can be simplified into | ||
+ | |||
+ | <math>(5) \phantom{a} ab=40.</math> | ||
+ | |||
+ | Equation 3 is already simplified. | ||
+ | |||
+ | Squaring the 1st equation we have | ||
+ | |||
+ | <math>(a+b+c)^2=32^2.</math> | ||
+ | |||
+ | Expanding and grouping, we have | ||
+ | |||
+ | <math>(a^2+b^2+c^2)+2(ab+ac+bc)=32^2.</math> | ||
+ | |||
+ | By equation 3 and substituting we get | ||
+ | |||
+ | <math>2(c^2+ab+ac+bc)=32^2.</math> | ||
+ | |||
+ | By equation 5 and substituting we get | ||
+ | |||
+ | <math>2(c^2+40+ac+bc)=32^2.</math> | ||
+ | |||
+ | Note that we can factor <math>c</math> out in the inner expression, and we get | ||
+ | |||
+ | <math>2(c^2+40+c(a+b))=32^2.</math> | ||
+ | |||
+ | By equation 4 and substituting, we have | ||
+ | |||
+ | <math>2(c^2+40+c(32-c))=32^2.</math> | ||
+ | |||
+ | Expanding, we have | ||
+ | |||
+ | <math>2(c^2+40+32c-c^2)=32^2.</math> | ||
+ | |||
+ | Simplifying, we have | ||
+ | |||
+ | <math>2(40+32c)=32^2.</math> | ||
+ | |||
+ | Expanding again, we get | ||
+ | |||
+ | <math>80+64c=32^2.</math> | ||
+ | |||
+ | Dividing both sides by <math>16</math> gets us | ||
+ | |||
+ | <math>4c+5=64</math> | ||
+ | |||
+ | Calculating gets us | ||
+ | |||
+ | <math>c=\boxed{\mathrm{\frac{59}{4}}}</math>. | ||
+ | |||
+ | ~mathboy282 | ||
== See Also == | == See Also == |
Revision as of 16:31, 22 December 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 (B).
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 .
Solution 6
Let the sides be where and are the legs and is the hypotenuse.
Since the perimeter is 32, we have
.
Since the area is 20 and the legs are and , we have that
.
By the Pythagorean Theorem, we have that
.
Since we want , we will equations be in the form of
Equation 1 can be turned into
.
Equation 2 can be simplified into
Equation 3 is already simplified.
Squaring the 1st equation we have
Expanding and grouping, we have
By equation 3 and substituting we get
By equation 5 and substituting we get
Note that we can factor out in the inner expression, and we get
By equation 4 and substituting, we have
Expanding, we have
Simplifying, we have
Expanding again, we get
Dividing both sides by gets us
Calculating gets us
.
~mathboy282
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.