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

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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>
 
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>
  
 
==Solution==
 
==Solution==

Revision as of 23:46, 13 January 2015

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

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.

See Also

2012 AMC 12B (ProblemsAnswer KeyResources)
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

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