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?
Answer Choices:
A. 3π+1−3
B. 2π(2−3)
C. π(2−3)
D. 6π+23−1
E. 3π−1+3
Solution:
Since OC=1 and OE=2, it follows that ∠EOC=60∘ and ∠EOA=30∘. The area of the shaded region is the area of the 30∘ sector DOE minus the area of congruent triangles OBD and OBE. First note that
Area ( Sector DOE)=121(4π)=3π.
In right triangle OCE, we have CE=3, so BE=3−1. Therefore
Area(△OBE)=21(3−1)(1).
The required area is consequently
3π−2(23−1)=(A)3π+1−3
OR
Let F be the point where ray OA intersects the circle, and let G be the point where ray OC intersects the circle.
Let a be the area of the shaded region described in the problem, and b be the area of the region bounded by AD,AF, and the minor arc from D to F. Then b is also the area of the region bounded by CE,CG, and the minor arc from G to E. By the Inclusion-Exclusion Principle,
2b−a= Area ( Quartercircle OFG)− Area (Square OABC)=π−1
Since b is the area of a 60∘ sector from which the area of △OAD has been deleted, we have
b=32π−23
Hence the area of the shaded region described in the problem is
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