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 | ||
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</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>. | + | 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 === | ||
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\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>. | ||
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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
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
Subtracting from both sides of the first equation and squaring both sides, we get
Now we substitute in as well as into the equation to get
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:
Rewrite equation 3 as . Substitute in equations 1 and 2 to get .
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.