Problem:
A square of side length 1 and a circle of radius 3​/3 share the same center. What is the area inside the circle, but outside the square?
Answer Choices:
A. 3π​−1
B. 92π​−33​​
C. 18π​
D. 41​
E. 2Ï€/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=(21​)2<(33​​)2<(22​​)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 PQ​. Then M is also the midpoint of AB and △OMA is a right triangle. By the Pythagorean Theorem AM=23​1​, so △OMA is a 30−60−90∘ right triangle. Then ∠AOB=60∘, and
the area of the sector corresponding to ∠AOB is 61​⋅π⋅(33​​)2=18π​. The area of △AOB is 2⋅21​⋅21​⋅23​1​=123​​. The area outside the square but inside the circle is 4⋅(18π​−123​​)=(B)92π​−33​​​.
