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

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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.
 
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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== See Also ==
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{{AMC12 box|year=2012|ab=B|num-b=9|num-a=11}}

Revision as of 22:20, 12 January 2013

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.


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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