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2006 AMC 10B Problems/Problem 19

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Problem

A circle of radius $2$ is centered at $O$ . Square $OABC$ has side length $1$ . Sides $AB$ and $CB$ are extended past $B$ to meet the circle at $D$ and $E$ , respectively. What is the area of the shaded region in the figure, which is bounded by $BD$ , $BE$ , and the minor arc connecting $D$ and $E$ ?

$\mathrm{(A) \ } \frac{\pi}{3}+1-\sqrt{3}\qquad \mathrm{(B) \ } \frac{\pi}{2}(2-\sqrt{3})\qquad \mathrm{(C) \ } \pi(2-\sqrt{3}...$

Solution

The shaded area is equivalent to the area of sector $DOE$ , minus the area of triangle $DOE$ plus the area of triangle $DBE$ .

Using the Pythagorean Theorem:

$(DA)^2=(CE)^2=2^2-1^2=3$

$DA=CE=\sqrt{3}$

Clearly, $DOA$ and $EOC$ are $30-60-90$ triangles with $\angle EOC = \angle DOA = 60^\circ$ .

Since $OABC$ is a square, $\angle COA = 90^\circ$ .

$\angle DOE$ can be found by doing some subtraction of angles.

$\angle COA - \angle DOA = \angle EOA$

$90^\circ - 60^\circ = \angle EOA = 30^\circ$

$\angle DOA - \angle EOA = \angle DOE$

$60^\circ - 30^\circ = \angle DOE = 30^\circ$

So, the area of sector $DOE$ is $\frac{30}{360} \cdot \pi \cdot 2^2 = \frac{\pi}{3}$ .

The area of triangle $DOE$ is $\frac{1}{2}\cdot 2 \cdot 2 \cdot \sin 30^\circ = 1$ .

Since $AB=CB=1$ , $DB=ED=(\sqrt{3}-1)$ .

So, the area of triangle $DBE$ is $\frac{1}{2} \cdot (\sqrt{3}-1)^2 = 2-\sqrt{3}$ .

Therefore, the shaded area is $(\frac{\pi}{3}) - (1) + (2-\sqrt{3}) = \frac{\pi}{3}+1-\sqrt{3} \Rightarrow A$

See Also

2006 AMC 10B Problems

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