Problem:
A square of side length 1 and a circle of radius 3β/3 share the same center. What is the area inside the circle, but outside the square?
Answer Choices:
A. 3Οββ1
B. 92Οββ33ββ
C. 18Οβ
D. 41β
E. 2Ο/9
Solution:
Let O be the common center of the circle and the square. Let M be the midpoint of a side of the square and P and Q be the vertices of the square on the side containing M. Since
OM2=(21β)2<(33ββ)2<(22ββ)2=OP2=OQ2,
the midpoint of each side is inside the circle and the vertices of the square are outside the circle. Therefore the circle intersects the square in two points along each side.
Let A and B be the intersection points of the circle with PQβ. Then M is also the midpoint of AB and β³OMA is a right triangle. By the Pythagorean Theorem AM=23β1β, so β³OMA is a 30β60β90β right triangle. Then β AOB=60β, and
the area of the sector corresponding to β AOB is 61ββ Οβ (33ββ)2=18Οβ. The area of β³AOB is 2β 21ββ 21ββ 23β1β=123ββ. The area outside the square but inside the circle is 4β (18Οββ123ββ)=(B)92Οββ33βββ.
