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>
  
__TOC__
+
== 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
 
<center>
 
<center>
<math>a+b+\sqrt{a^2+b^2} = 32 \\
+
<math>a+b+\sqrt{a^2+b^2} = 32 \\\\
 
\frac{1}{2}ab = 20</math>
 
\frac{1}{2}ab = 20</math>
 
</center>
 
</center>
 
Re-arranging the first equation and squaring,  
 
Re-arranging the first equation and squaring,  
 
<center>
 
<center>
<math> =\sqrt{a^2+b^2} &= 32-(a+b)\\
+
<math> \sqrt{a^2+b^2} = 32-(a+b)\\\\
a^2 + b^2 &= 32^2 - 64(a+b) + (a+b)^2\\
+
a^2 + b^2 = 32^2 - 64(a+b) + (a+b)^2\\\\
a^2 + b^2 + 64(a+b) &= a^2 + b^2 + 2ab + 32^2\\
+
a^2 + b^2 + 64(a+b) = a^2 + b^2 + 2ab + 32^2\\\\
a+b &= \frac{2ab+32^2}{64}</math>
+
a+b = \frac{2ab+32^2}{64}</math>
 
</center>
 
</center>
 
From <math>(2)</math> we have <math>2ab = 80</math>, so
 
From <math>(2)</math> we have <math>2ab = 80</math>, so
 
<center>
 
<center>
<math>a+b &= \frac{80 + 32^2}{64} = \frac{69}{4}.</math>
+
<math>a+b = \frac{80 + 32^2}{64} = \frac{69}{4}.</math>
 
</center>
 
</center>
The length of the hypotenuse is <math>p - a - b = 32 - \frac{69}{4} = \frac{59}{4}\ \mathrm{(B)}</math>.
+
 
 +
The length of the hypotenuse is <math>p - a - b = 32 - \frac{69}{4} = \frac{59}{4}\boxed{\ \mathrm{(B)}}</math>.
  
 
=== 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>. It is known that 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>.
+
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
  
 
=== Solution 3 ===
 
=== Solution 3 ===
 
From the problem, we know that  
 
From the problem, we know that  
<center><math>\begin{align*}
+
<center><cmath>\begin{align*}
 
a+b+c &= 32 \\
 
a+b+c &= 32 \\
 
2ab &= 80. \\
 
2ab &= 80. \\
\end{align*}</math></center>
+
\end{align*}</cmath></center>
  
 
Subtracting <math>c</math> from both sides of the first equation and squaring both sides, we get
 
Subtracting <math>c</math> from both sides of the first equation and squaring both sides, we get
<center><math>\begin{align*}
+
<center><cmath>\begin{align*}
 
(a+b)^2 &= (32 - c)^2\\
 
(a+b)^2 &= (32 - c)^2\\
 
a^2 + b^2 + 2ab &= 32^2 + c^2 - 64c.\\
 
a^2 + b^2 + 2ab &= 32^2 + c^2 - 64c.\\
\end{align*}</math></center>
+
\end{align*}</cmath></center>
  
 
Now we substitute in <math>a^2 + b^2 = c^2</math> as well as <math>2ab = 80</math> into the equation to get
 
Now we substitute in <math>a^2 + b^2 = c^2</math> as well as <math>2ab = 80</math> into the equation to get
<center><math>\begin{align*}
+
<center><cmath>\begin{align*}
 
80 &= 1024 - 64c\\
 
80 &= 1024 - 64c\\
 
c &= \frac{944}{64}.
 
c &= \frac{944}{64}.
\end{align*}</math></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, and <math>c</math> the hypotenuse.
 
  
 
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 68: Line 69:
 
Substitute in equations 1 and 2 to get <math>c^2 = (32-c)^2 - 80</math>.
 
Substitute in equations 1 and 2 to get <math>c^2 = (32-c)^2 - 80</math>.
  
<center><math>c^2 = (32-c)^2 - 80 \\
+
<center><math>c^2 = (32-c)^2 - 80 \\\\
c^2 = 1024 - 64c + c^2 - 80 \\
+
c^2 = 1024 - 64c + c^2 - 80 \\\\
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 (B).
  
==See also==
+
=== 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
 +
 
 +
==Video Solution==
 +
https://youtu.be/mm1fGEKhQSA
 +
 
 +
~savannahsolver
 +
 
 +
== 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 19:29, 4 June 2021

Problem

A right triangle has perimeter $32$ and area $20$. What is the length of its hypotenuse?

$\mathrm{(A)}\ \frac{57}{4}\qquad\mathrm{(B)}\ \frac{59}{4}\qquad\mathrm{(C)}\ \frac{61}{4}\qquad\mathrm{(D)}\ \frac{63}{4}\qquad\mathrm{(E)}\ \frac{65}{4}$

Solution

Solution 1

Let the legs of the triangle have lengths $a,b$. Then, by the Pythagorean Theorem, the length of the hypotenuse is $\sqrt{a^2+b^2}$, and the area of the triangle is $\frac 12 ab$. So we have the two equations

$a+b+\sqrt{a^2+b^2} = 32 \\\\ \frac{1}{2}ab = 20$

Re-arranging the first equation and squaring,

$\sqrt{a^2+b^2} = 32-(a+b)\\\\ a^2 + b^2 = 32^2 - 64(a+b) + (a+b)^2\\\\ a^2 + b^2 + 64(a+b) = a^2 + b^2 + 2ab + 32^2\\\\ a+b = \frac{2ab+32^2}{64}$

From $(2)$ we have $2ab = 80$, so

$a+b = \frac{80 + 32^2}{64} = \frac{69}{4}.$

The length of the hypotenuse is $p - a - b = 32 - \frac{69}{4} = \frac{59}{4}\boxed{\ \mathrm{(B)}}$.

Solution 2

From the formula $A = rs$, where $A$ is the area of a triangle, $r$ is its inradius, and $s$ is the semiperimeter, we can find that $r = \frac{20}{32/2} = \frac{5}{4}$. In a right triangle, $r = s - h$, where $h$ is the hypotenuse, so $h = 16 - \frac{5}{4} = \frac{59}{4}$ $\mathrm{(B)}$.

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

Solution 3

From the problem, we know that

\begin{align*} a+b+c &= 32 \\ 2ab &= 80. \\ \end{align*}

Subtracting $c$ 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*}

Now we substitute in $a^2 + b^2 = c^2$ as well as $2ab = 80$ into the equation to get

\begin{align*} 80 &= 1024 - 64c\\ c &= \frac{944}{64}. \end{align*}

Further simplification yields the result of $\frac{59}{4} \rightarrow \mathrm{(B)}$.

Solution 4

Let $a$ and $b$ be the legs of the triangle and $c$ the hypotenuse.

Since the area is 20, we have $\frac{1}{2}ab = 20 => ab=40$.

Since the perimeter is 32, we have $a + b + c = 32$.

The Pythagorean Theorem gives $c^2 = a^2 + b^2$.

This gives us three equations with three variables:

$ab = 40 \\ a + b + c = 32 \\ c^2 = a^2 + b^2$

Rewrite equation 3 as $c^2 = (a+b)^2 - 2ab$. Substitute in equations 1 and 2 to get $c^2 = (32-c)^2 - 80$.

$c^2 = (32-c)^2 - 80 \\\\ c^2 = 1024 - 64c + c^2 - 80 \\\\ 64c = 944 \\\\ c = \frac{944}{64} = \frac{236}{16} = \frac{59}{4}$.

The answer is choice (B).

Solution 5

Let $a$, $b$, and $c$ be the sides of the triangle with $c$ as the hypotenuse.

We know that $a + b + c =32$.

According to the Pythagorean Theorem, we have $a^2 + b^2 = c^2$.

We also know that $ab = 40$, since the area of the triangle is $20$.

We substitute $2ab$ into $a^2 + b^2 = c^2$ to get $(a+b)^2 = c^2 + 80$.

Moving the $c^2$ to the left, we again rewrite to get $(a+b+c)(a+b-c) = 80$.

We substitute our value of $32$ for $a+b+c$ twice into our equation and subtract to get $a + b = \frac{69}{4}$.

Finally, subtracting this from our original value of $32$, we get $\frac{59}{4}$, or $B$.

Solution 6

Let the legs be $a, b$. Then the hypotenuse is $\sqrt{a^2 + b^2}.$

We know that $ab = 40$ and $a + b + \sqrt{a^2 + b^2} = 32$.

The first equation gives \[b = \frac{40}{a}\] and we can plug this into the second equation, yielding: \[a + \frac{40}{a} + \sqrt{ a^2 + \frac{1600}{a^2}} = 32.\]

Letting $X = a + \frac{40}{a}$, the equation becomes: \[X + \sqrt{X^2 - 80} = 32.\]

We can bring the $X$ to the right side and square which yields: \[X^2 - 80 = (X - 32)^2 = X^2 - 64X + 1024.\]

So, \[-80 = -64X + 1024 \rightarrow X = \frac{69}{4}.\]

Now, we know that $a + \frac{40}{a} = \frac{69}{4}.$

Multiplying both sides by $4a$ gives: \[4a^2 - 69a + 160.\]

It can be observed that the roots of this equation are $a$ and $b$. We want the hypotenuse which is $\sqrt{a^2 + b^2} = \sqrt{ (a+b)^2 - 2ab}.$

We can now apply Vieta's Formula which gives: \[c = \sqrt{ \left( \frac{69}{4} \right)^2 - 80} = \boxed{ \frac{59}{4}}.\]

~conantwiz2023

Solution 7

Let the sides be $a, b, c$ where $a$ and $b$ are the legs and $c$ is the hypotenuse.

Since the perimeter is 32, we have

$(1) \phantom{a} a+b+c=32$.

Since the area is 20 and the legs are $a$ and $b$, we have that

$(2) \phantom{a} \frac{a \cdot b}{2}=20$.

By the Pythagorean Theorem, we have that

$(3) \phantom{a} a^2+b^2=c^2$.

Since we want $c$, we will equations $1, 2, 3$ be in the form of $c.$

Equation 1 can be turned into

$(4) \phantom{a} a+b=32-c$.

Equation 2 can be simplified into

$(5) \phantom{a} ab=40.$

Equation 3 is already simplified.

Onto the calculating process.


Squaring the 1st equation we have

$(a+b+c)^2=32^2.$

Expanding and grouping, we have

$(a^2+b^2+c^2)+2(ab+ac+bc)=32^2.$

By equation 3 and substituting we get

$2(c^2+ab+ac+bc)=32^2.$

By equation 5 and substituting we get

$2(c^2+40+ac+bc)=32^2.$

Note that we can factor $c$ out in the inner expression, and we get

$2(c^2+40+c(a+b))=32^2.$

By equation 4 and substituting, we have

$2(c^2+40+c(32-c))=32^2.$

Expanding, we have

$2(c^2+40+32c-c^2)=32^2.$

Simplifying, we have

$2(40+32c)=32^2.$

Expanding again, we get

$80+64c=32^2.$

Dividing both sides by $16$ gets us

$4c+5=64$

Calculating gets us

$c=\boxed{\mathrm{(B) \frac{59}{4}}}$.

~mathboy282

Video Solution

https://youtu.be/mm1fGEKhQSA

~savannahsolver

See Also

2008 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 10 Problems and Solutions

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