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Difference between revisions of "2007 AMC 10B Problems/Problem 13"
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==Solution==
 
==Solution==
  
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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 quarter of the circle with radius <math>2,</math> and the isosceles triangle is a right triangle. Therefore, the area of half the intersection is
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You can find the area of half the intersection by subtracting the isosceles triangle in the sector from the whole sector. This sector is 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>

Latest revision as of 19:42, 15 February 2024

Problem

Two circles of radius $2$ are centered at $(2,0)$ and at $(0,2).$ What is the area of the intersection of the interiors of the two circles?

$\textbf{(A) } \pi -2 \qquad\textbf{(B) } \frac{\pi}{2} \qquad\textbf{(C) } \frac{\pi \sqrt{3}}{3} \qquad\textbf{(D) } 2(\pi -2) \qquad\textbf{(E) } \pi$

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 one-fourth of the area of the circle with radius $2,$ and the isosceles triangle is a right triangle. Therefore, the area of half the intersection is \[\frac{1}{4} 4\pi - \frac{1}{2}(2)(2) = \pi - 2\] That means the area of the whole intersection is $\boxed{\mathrm{(D) \ } 2(\pi-2)}$

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

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