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 | ||
Line 27: | 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 48: | Line 53: | ||
\end{align*}</cmath></center> | \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} \rightarrow \mathrm{(B)}</math>. |
=== Solution 4 === | === Solution 4 === | ||
− | + | 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 | ||
Since the area is 20, we have <math>\frac{1}{2}ab = 20 => ab=40</math>. | Since the area is 20, we have <math>\frac{1}{2}ab = 20 => ab=40</math>. | ||
Line 73: | 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 === | ||
+ | Let <math>a</math>, <math>b</math>, and <math>c</math> be the sides of the triangle with <math>c</math> as the hypotenuse. | ||
+ | |||
+ | We know that <math>a + b + c =32</math>. | ||
+ | |||
+ | According to the Pythagorean Theorem, we have <math>a^2 + b^2 = c^2</math>. | ||
+ | |||
+ | We also know that <math>ab = 40</math>, since the area of the triangle is <math>20</math>. | ||
+ | |||
+ | We substitute <math>2ab</math> into <math>a^2 + b^2 = c^2</math> to get <math>(a+b)^2 = c^2 + 80</math>. | ||
+ | |||
+ | Moving the <math>c^2</math> to the left, we again rewrite to get <math>(a+b+c)(a+b-c) = 80</math>. | ||
+ | |||
+ | We substitute our value of <math>32</math> for <math>a+b+c</math> twice into our equation and subtract to get <math>a + b = \frac{69}{4}</math>. | ||
+ | |||
+ | Finally, subtracting this from our original value of <math>32</math>, we get <math>\frac{59}{4}</math>, or <math>B</math>. | ||
+ | |||
+ | === Solution 6 === | ||
+ | |||
+ | Let the legs be <math>a, b</math>. Then the hypotenuse is <math>\sqrt{a^2 + b^2}.</math> | ||
+ | |||
+ | We know that <math>ab = 40</math> and <math>a + b + \sqrt{a^2 + b^2} = 32</math>. | ||
+ | |||
+ | The first equation gives <cmath>b = \frac{40}{a}</cmath> and we can plug this into the second equation, yielding: <cmath>a + \frac{40}{a} + \sqrt{ a^2 + \frac{1600}{a^2}} = 32.</cmath> | ||
+ | |||
+ | Letting <math>X = a + \frac{40}{a}</math>, the equation becomes: <cmath>X + \sqrt{X^2 - 80} = 32.</cmath> | ||
+ | |||
+ | We can bring the <math>X</math> to the right side and square which yields: <cmath>X^2 - 80 = (X - 32)^2 = X^2 - 64X + 1024.</cmath> | ||
+ | |||
+ | So, <cmath>-80 = -64X + 1024 \rightarrow X = \frac{69}{4}.</cmath> | ||
+ | |||
+ | Now, we know that <math>a + \frac{40}{a} = \frac{69}{4}.</math> | ||
+ | |||
+ | Multiplying both sides by <math>4a</math> gives: <cmath>4a^2 - 69a + 160.</cmath> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ~conantwiz2023 | ||
+ | |||
+ | === Solution 7 === | ||
+ | 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. | ||
+ | |||
+ | Onto the calculating process. | ||
+ | |||
+ | --------------------------------- | ||
+ | |||
+ | 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{(B) \frac{59}{4}}}</math>. | ||
+ | |||
+ | ~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== | ||
+ | https://youtu.be/mm1fGEKhQSA | ||
+ | |||
+ | ~savannahsolver | ||
− | ==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}} |
Latest revision as of 13:15, 1 August 2022
Contents
[hide]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.