Difference between revisions of "2012 AMC 12B Problems/Problem 10"
(Created page with "==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 i...") |
Username222 (talk | contribs) |
||
| Line 8: | Line 8: | ||
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. | ||
| + | |||
| + | == See Also == | ||
| + | |||
| + | {{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 (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 | |
