Problem:
An 8 by 22β 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. 2Ο
B. 2Ο+2
C. 4Οβ4
D. 2Ο+4
E. 4Οβ2
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=22β, the width of the rectangle, it follows that β AOB=90β. Hence β AOD=β DOC=β COB=90β. The sum of the areas of sectors AOB and DOC is 2(41β(Ο22))=2Ο. The sum of the areas of isosceles right triangles AOD and COB is 2(21ββ 22)=4. Thus, the area of the region common to both the rectangle and the circle is 2Ο+4.