Difference between revisions of "2008 AMC 10A Problems/Problem 18"
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=== Solution 4 === | === Solution 4 === |
Revision as of 19:16, 4 June 2021
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
![\begin{align*} a+b+c &= 32 \\ 2ab &= 80. \\ \end{align*}](http://latex.artofproblemsolving.com/9/b/1/9b1ad9609b79bdcfc10cd8052a111b659e5b0f13.png)
Subtracting from both sides of the first equation and squaring both sides, we get
![\begin{align*} (a+b)^2 &= (32 - c)^2\\ a^2 + b^2 + 2ab &= 32^2 + c^2 - 64c.\\ \end{align*}](http://latex.artofproblemsolving.com/8/f/1/8f13a9ea7ec9417ab1fab36f5ae1d9ecb5ef4179.png)
Now we substitute in as well as
into the equation to get
![\begin{align*} 80 &= 1024 - 64c\\ c &= \frac{944}{64}. \end{align*}](http://latex.artofproblemsolving.com/a/a/1/aa184abe19e53f4bb2e8712ad5b4e6a2e0aaf82d.png)
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:
![$ab = 40 \\ a + b + c = 32 \\ c^2 = a^2 + b^2$](http://latex.artofproblemsolving.com/a/b/3/ab34477dc4e0b22ad043a9d2ae5228ade3611d17.png)
Rewrite equation 3 as .
Substitute in equations 1 and 2 to get
.
![$c^2 = (32-c)^2 - 80 \\\\ c^2 = 1024 - 64c + c^2 - 80 \\\\ 64c = 944 \\\\ c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4}$](http://latex.artofproblemsolving.com/b/b/0/bb0955d038919472bbefb6e672b0b6c7192fb561.png)
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 legs be . Then the hypotenuse is
We know that and
.
The first equation gives and we can plug this into the second equation, yielding:
Letting , the equation becomes:
We can bring the to the right side and square which yields:
So,
Now, we know that
Multiplying both sides by gives:
It's not hard to see that the roots of this equation are and
. We want the hypotenuse which is
We can now apply Vieta's Formula which gives:
~conantwiz2023
Solution 7
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.
Onto the calculating process.
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
Video Solution
~savannahsolver
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.