Difference between revisions of "2008 AMC 10A Problems/Problem 18"

Latest revision as of 13:15, 1 August 2022

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

Why $r=s-h$ . 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 $r=s-h$ .

~BakedPotato66

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

Solution 8

This solution is very similar to Solution 1, except instead of subtracting $a+b$ from both sides in the first part, we subtract $\sqrt{a^2+b^2}$ from both sides, which gets us:

$\begin{align*} a+b&=32-\sqrt{a^2+b^2}\\ (a+b)^2&=(32-\sqrt{a^2+b^2})^2\\ a^2+2ab+b^2&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}+a^2+b^2\\ 2ab&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}.\\ \end{align*}$

We know that $\frac{1}{2}ab=20$ so $ab=40.$

We can then substitute $40$ for $ab$ to get us:

$\begin{align*} 80&=32^2-2\cdot32\cdot\sqrt{a^2+b^2}\\ 2\cdot32\cdot\sqrt{a^2+b^2}&=32^2-80\\ 64\cdot\sqrt{a^2+b^2}&=32^2-80\\ \sqrt{a^2+b^2}&=\frac{32^2-80}{64}\\ \end{align*}$

We know that $\sqrt{a^2+b^2}$ is the hypotenuse, so we only have to solve the right-hand side now.

$\begin{align*} \sqrt{a^2+b^2}&=\frac{32^2-80}{64}\\ &=\frac{32^2}{64}-\frac{80}{64}\\ &=\frac{32\cdot32}{32\cdot2}-\frac{32\cdot\frac{5}{2}}{32\cdot2}\\ &=\frac{32}{2}-\frac{\frac{5}{2}}{2}\\ &=16-\frac{5}{4}\\ &=\frac{64}{4}-\frac{5}{4}\\ &=\boxed{\mathrm{(B)} \frac{59}{4}}\\ \end{align*}$

~zlrara01

Video Solution

https://youtu.be/mm1fGEKhQSA

~savannahsolver

2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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