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. 3Οβ+1β3β
B. 2Οβ(2β3β)
C. Ο(2β3β)
D. 6Οβ+23ββ1β
E. 3Οββ1+3β
Solution:
Since OC=1 and OE=2, it follows that β EOC=60β and β EOA=30β. The area of the shaded region is the area of the 30β sector DOE minus the area of congruent triangles OBD and OBE. First note that
Area ( Sector DOE)=121β(4Ο)=3Οβ.
In right triangle OCE, we have CE=3β, so BE=3ββ1. Therefore
Area(β³OBE)=21β(3ββ1)(1).
The required area is consequently
3Οββ2(23ββ1β)=(A)3Οβ+1β3ββ
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,
2bβa= Area ( Quartercircle OFG)β Area (Square OABC)=Οβ1
Since b is the area of a 60β sector from which the area of β³OAD has been deleted, we have
b=32Οββ23ββ
Hence the area of the shaded region described in the problem is
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