Difference between revisions of "2007 AMC 10B Problems/Problem 13"
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==Solution== | ==Solution== | ||
| − | You can find the area of half the intersection by subtracting the isosceles triangle in the sector from the whole sector. This sector is a | + | You can find the area of half the intersection by subtracting the isosceles triangle in the sector from the whole sector. This sector is a one-fourth of the area of the circle with radius <math>2,</math> and the isosceles triangle is a right triangle. Therefore, the area of half the intersection is |
<cmath>\frac{1}{4} 4\pi - \frac{1}{2}(2)(2) = \pi - 2</cmath> | <cmath>\frac{1}{4} 4\pi - \frac{1}{2}(2)(2) = \pi - 2</cmath> | ||
That means the area of the whole intersection is <math>\boxed{\mathrm{(D) \ } 2(\pi-2)}</math> | That means the area of the whole intersection is <math>\boxed{\mathrm{(D) \ } 2(\pi-2)}</math> | ||
Revision as of 12:33, 4 June 2021
Problem
Two circles of radius 
are centered at 
and at 
What is the area of the intersection of the interiors of the two circles?
Solution
You can find the area of half the intersection by subtracting the isosceles triangle in the sector from the whole sector. This sector is a one-fourth of the area of the circle with radius 
and the isosceles triangle is a right triangle. Therefore, the area of half the intersection is
![\[\frac{1}{4} 4\pi - \frac{1}{2}(2)(2) = \pi - 2\]](http://latex.artofproblemsolving.com/9/7/9/97995745dee6977306ec1aeabf92f355e0f7e6d2.png)
That means the area of the whole intersection is 
See Also
| 2007 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
