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Difference between revisions of "2012 AMC 12B Problems/Problem 10"

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==Problem==
 
==Problem==
  
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What is the area of the polygon whose vertices are the points of intersection of the curves x^2 + y^2 =25 and (x-4)^2 + 9y^2 = 81.
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What is the area of the polygon whose vertices are the points of intersection of the curves <math>x^2 + y^2 =25</math> and <math>(x-4)^2 + 9y^2 = 81 ?</math>
  
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<math>\textbf{(A)}\ 24\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 42</math>
  
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==Solution 1==
  
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==Solution==
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The first curve is a circle with radius <math>5</math> centered at the origin, and the second curve is an ellipse with center <math>(4,0)</math> and end points of <math>(-5,0)</math> and <math>(13,0)</math>. Finding points of intersection, we get <math>(-5,0)</math>, <math>(4,3)</math>, and <math>(4,-3)</math>, forming a triangle with height of <math>9</math> and base of <math>6.</math> So the area of this triangle is <math>9 \cdot 6 \cdot 0.5 =27 \textbf{ (B)}.</math>
  
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The first curve is a circle with radius 5 centered at the origin, and the second curve is an ellipse with center (4,0) and end points of (-5,0) and (13,0). Finding points of intersection, we get (-5,0) (4,3) and (4,-3), forming a triangle with height of 9 and base of 6. So 9x6x0.5 =27 ; B.
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==Solution 2==
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Given the equations <math>x^2 + y^2 =25</math> and <math>(x-4)^2 + 9y^2 = 81</math>, we can substitute <math>y^2=25-x^2</math> from the first equation and plug it in to the 2nd equation, giving us <math>(x-4)^2+9(25-x^2)=81</math>. After rearranging, <math>8x^2+8x-160=0</math> or <math>x^2+x-20=0</math>. The solutions are <math>x=-5</math> and <math>x=4</math>. This gives us the points <math>(-5,0),(4,3)</math>,and <math>(4,-3)</math>. The area of the triangle formed by these points is <math>27=\fbox{B}</math>
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~dragnin
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~minor edits by [https://artofproblemsolving.com/wiki/index.php/User:Kevinchen_yay KevinChen_Yay]
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==Solution 3==
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Using our algebra skills we find the points of intersection to be <math>(-5,0),(4,3)</math>,and <math>(4,-3)</math>. Using the shoelace theorem, we can easily find the area of the triangle to be <math>27=\fbox{B}</math>.
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~PeterDoesPhysics
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More information on the shoelace theorem:
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https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem
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== See Also ==
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{{AMC12 box|year=2012|ab=B|num-b=9|num-a=11}}
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extremely misplaced problem.  
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 08:56, 7 August 2024

Problem

What is the area of the polygon whose vertices are the points of intersection of the curves $x^2 + y^2 =25$ and $(x-4)^2 + 9y^2 = 81 ?$

$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 42$

Solution 1

The first curve is a circle with radius $5$ centered at the origin, and the second curve is an ellipse with center $(4,0)$ and end points of $(-5,0)$ and $(13,0)$. Finding points of intersection, we get $(-5,0)$, $(4,3)$, and $(4,-3)$, forming a triangle with height of $9$ and base of $6.$ So the area of this triangle is $9 \cdot 6 \cdot 0.5 =27 \textbf{ (B)}.$

Solution 2

Given the equations $x^2 + y^2 =25$ and $(x-4)^2 + 9y^2 = 81$, we can substitute $y^2=25-x^2$ from the first equation and plug it in to the 2nd equation, giving us $(x-4)^2+9(25-x^2)=81$. After rearranging, $8x^2+8x-160=0$ or $x^2+x-20=0$. The solutions are $x=-5$ and $x=4$. This gives us the points $(-5,0),(4,3)$,and $(4,-3)$. The area of the triangle formed by these points is $27=\fbox{B}$

~dragnin

~minor edits by KevinChen_Yay

Solution 3

Using our algebra skills we find the points of intersection to be $(-5,0),(4,3)$,and $(4,-3)$. Using the shoelace theorem, we can easily find the area of the triangle to be $27=\fbox{B}$.

~PeterDoesPhysics

More information on the shoelace theorem: https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem

See Also

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 12 Problems and Solutions

extremely misplaced problem. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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