Difference between revisions of "2008 AMC 10A Problems/Problem 18"
(→Solution 5) |
m (→Solution 8) |
||
(7 intermediate revisions by 3 users not shown) | |||
Line 26: | Line 26: | ||
=== Solution 2 === | === Solution 2 === | ||
− | From the formula <math>A = rs</math>, where <math>A</math> is the area of a triangle, <math>r</math> is its [[inradius]], and <math>s</math> is the [[semiperimeter]], we can find that <math>r = \frac{20}{32/2} = \frac{5}{4}</math>. | + | From the formula <math>A = rs</math>, where <math>A</math> is the area of a triangle, <math>r</math> is its [[inradius]], and <math>s</math> is the [[semiperimeter]], we can find that <math>r = \frac{20}{32/2} = \frac{5}{4}</math>. In a right triangle, <math>r = s - h</math>, where <math>h</math> is the hypotenuse, so <math>h = 16 - \frac{5}{4} = \frac{59}{4}</math> <math>\mathrm{(B)}</math>. |
+ | |||
+ | '''Note:''' If the reader is unfamiliar with the inradius being equal to the semiperimeter minus the hypotenuse for a right triangle, it should not be too difficult to prove the relationship for themselves.~mobius247 | ||
+ | |||
+ | '''Why <math>r=s-h</math>.''' Draw a right triangle and inscribe a circle. Connect the center of the circle with tangency points; there are 3, one from each side of the triangle. Now, connect the center with each vertice and you will see 3 pairs of congruent triangles by HL (Hypotenuse-Leg Congruence). From these congruent triangles, mark equal lengths and you will find that <math>r=s-h</math>. | ||
+ | |||
+ | ~BakedPotato66 | ||
=== Solution 3 === | === Solution 3 === | ||
Line 71: | Line 77: | ||
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 <math>(B)</math>. |
=== Solution 5 === | === Solution 5 === | ||
Line 108: | Line 114: | ||
Multiplying both sides by <math>4a</math> gives: <cmath>4a^2 - 69a + 160.</cmath> | Multiplying both sides by <math>4a</math> gives: <cmath>4a^2 - 69a + 160.</cmath> | ||
− | It | + | It can be observed that the roots of this equation are <math>a</math> and <math>b</math>. We want the hypotenuse which is <math>\sqrt{a^2 + b^2} = \sqrt{ (a+b)^2 - 2ab}.</math> |
We can now apply Vieta's Formula which gives: <cmath>c = \sqrt{ \left( \frac{69}{4} \right)^2 - 80} = \boxed{ \frac{59}{4}}.</cmath> | We can now apply Vieta's Formula which gives: <cmath>c = \sqrt{ \left( \frac{69}{4} \right)^2 - 80} = \boxed{ \frac{59}{4}}.</cmath> | ||
Line 190: | Line 196: | ||
~mathboy282 | ~mathboy282 | ||
+ | |||
+ | === Solution 8 === | ||
+ | This solution is very similar to Solution 1, except instead of subtracting <math>a+b</math> from both sides in the first part, we subtract <math>\sqrt{a^2+b^2}</math> from both sides, which gets us: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | a+b&=32-\sqrt{a^2+b^2}\\ | ||
+ | (a+b)^2&=(32-\sqrt{a^2+b^2})^2\\ | ||
+ | a^2+2ab+b^2&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}+a^2+b^2\\ | ||
+ | 2ab&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}.\\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | We know that | ||
+ | <cmath>\frac{1}{2}ab=20</cmath> | ||
+ | so | ||
+ | <cmath>ab=40.</cmath> | ||
+ | |||
+ | We can then substitute <math>40</math> for <math>ab</math> to get us: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 80&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}\\ | ||
+ | 2\cdot32\cdot\sqrt{a^2+b^2}&=32^2-80\\ | ||
+ | 64\cdot\sqrt{a^2+b^2}&=32^2-80\\ | ||
+ | \sqrt{a^2+b^2}&=\frac{32^2-80}{64}\\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | We know that <math>\sqrt{a^2+b^2}</math> is the hypotenuse, so we only have to solve the right-hand side now. | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \sqrt{a^2+b^2}&=\frac{32^2-80}{64}\\ | ||
+ | &=\frac{32^2}{64}-\frac{80}{64}\\ | ||
+ | &=\frac{32\cdot32}{32\cdot2}-\frac{32\cdot\frac{5}{2}}{32\cdot2}\\ | ||
+ | &=\frac{32}{2}-\frac{\frac{5}{2}}{2}\\ | ||
+ | &=16-\frac{5}{4}\\ | ||
+ | &=\frac{64}{4}-\frac{5}{4}\\ | ||
+ | &=\boxed{\mathrm{(B)} \frac{59}{4}}\\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | ~zlrara01 | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 14:15, 1 August 2022
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
. In a right triangle,
, where
is the hypotenuse, so
.
Note: If the reader is unfamiliar with the inradius being equal to the semiperimeter minus the hypotenuse for a right triangle, it should not be too difficult to prove the relationship for themselves.~mobius247
Why . Draw a right triangle and inscribe a circle. Connect the center of the circle with tangency points; there are 3, one from each side of the triangle. Now, connect the center with each vertice and you will see 3 pairs of congruent triangles by HL (Hypotenuse-Leg Congruence). From these congruent triangles, mark equal lengths and you will find that
.
~BakedPotato66
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 .
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 , since the area of the triangle is
.
We substitute into
to get
.
Moving the to the left, we again rewrite to get
.
We substitute our value of for
twice into our equation and subtract to get
.
Finally, subtracting this from our original value of , 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 can be observed 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
Solution 8
This solution is very similar to Solution 1, except instead of subtracting from both sides in the first part, we subtract
from both sides, which gets us:
We know that
so
We can then substitute for
to get us:
We know that is the hypotenuse, so we only have to solve the right-hand side now.
~zlrara01
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.