Problem:
Circles with centers at 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?
Answer Choices:
A. 183​
B. 242​
C. 36
D. 243​
E. 322​
Solution:
Through O draw a line parallel to AD intersecting PD at F.
Then AOFD is a rectangle and OPF is a right triangle. Thus DF=2,FP=2, and OF=42​. The area of trapezoid AOPD is 122​, and the area of hexagon AOBCPD is 2⋅122​=(B)242​​.
OR
Lines AD,BC, and OP intersect at a common point H.
Because ∠PDH=∠OAH=90∘, triangles PDH and OAH are similar with ratio of similarity 2. Thus 2HO=HP=HO+OP=HO+6, so HO=6 and AH=HO2−OA2​=42​. Hence the area of △OAH is (1/2)(2)(42​)=42​, and the area of △PDH is (22)(42​)=162​. The area of the hexagon is twice the area of △PDH minus twice the area of △OAH, so it is (B)242​​.
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