Problem:
An 8 by 22​ rectangle has the same center as a circle of radius 2. The area of the region common to both the rectangle and the circle is
Answer Choices:
A. 2Ï€
B. 2Ï€+2
C. 4π−4
D. 2Ï€+4
E. 4π−2
Solution:
Let O be the center of the circle and rectangle, and let the circle and rectangle intersect at A,B,C and D as shown. Since AO=OB=2 and AB=22​, the width of the rectangle, it follows that ∠AOB=90∘. Hence ∠AOD=∠DOC=∠COB=90∘. The sum of the areas of sectors AOB and DOC is 2(41​(π22))=2π. The sum of the areas of isosceles right triangles AOD and COB is 2(21​⋅22)=4. Thus, the area of the region common to both the rectangle and the circle is 2π+4.
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