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2006 AMC 10B Problems/Problem 24

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Problem

Circles with centers O and P have radii 2 and 4, respectively, and are externally tangent. Points A and B on the circle with center O and points C and D on the circle with center P are such that AD and BC are common external tangents to the circles. What is the area of the concave hexagon AOBCPD?

Image:2006amc10b24.gif

\mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 24\sqrt{3}\qquad \...

Solution

Since a tangent line is perpendicular to the radius containing the point of tangency, \angle OAD = \angle PDA = 90^\circ.

Construct a perpendicular to DP that goes through point O. Label the point of intersection X.

Clearly OADX is a rectangle.

Therefore DX=2 and PX=2.

By the Pythagorean Theorem: OX = \sqrt{6^2 - 2^2} = 4\sqrt{2}.

The area of OADX is 2\cdot4\sqrt{2}=8\sqrt{2}.

The area of OXP is \frac{1}{2}\cdot2\cdot4\sqrt{2}=4\sqrt{2}.

So the area of quadrilateral OADP is 8\sqrt{2}+4\sqrt{2}=12\sqrt{2}.

Using similar steps, the area of quadrilateral OBCP is also 12\sqrt{2}

Therefore, the area of hexagon AOBCPD is 2\cdot12\sqrt{2}= 24\sqrt{2} \Rightarrow B

See Also

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