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 | + | 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> |
== See Also == | == See Also == | ||
Revision as of 23:48, 13 January 2015
Problem
What is the area of the polygon whose vertices are the points of intersection of the curves 
and 
Solution
The first curve is a circle with radius 
centered at the origin, and the second curve is an ellipse with center 
and end points of 
and 
. Finding points of intersection, we get , 
, and 
, forming a triangle with height of 
and base of 
So the area of this triangle is 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
