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

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

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.

@@ Line 10: / Line 10: @@
 ==Solution 2==
-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>
+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>
 ~dragnin
+~minor edits by [https://artofproblemsolving.com/wiki/index.php/User:Kevinchen_yay KevinChen_Yay]
+==Solution 3==
+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>.
+~PeterDoesPhysics
+More information on the shoelace theorem:
+https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem
 == See Also ==
 {{AMC12 box|year=2012|ab=B|num-b=9|num-a=11}}
+extremely misplaced problem.
 [[Category:Introductory Geometry Problems]]
 {{MAA Notice}}

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