06/18/2020 (2011: National Target Round, Problem 5)

Q: Square ABCD is inscribed in a circle with diameter BD of length 2 units. Segment AB is the diameter of the half-circle on top of the square. What is the area of the shaded region, in square units? Express your answer as a common fraction.

A: ½ units²

The area of the shaded section is the area of the red area minus the green area

The length of AB is 2 ÷ √(2) = √(2). That means the length of the radius of the half-circle AB is √(2) ÷ 2, so the area of the half-circle (the top half of the red portion) is (π)(√(2) ÷ 2)² ÷ 2 = (¼) π.

The area of the triangle (the bottom half of the red portion) is ¼ of the area of ABCD, and the area of ABCD is AB² = 2. , so the bottom half of the red portion is 2 * ¼ = ½.

That means the area of the red area is (¼)π + ½.

The area of green area is (¼)(π)(2/2)² = (¼) π.

We subtract the green from the red and we get: ((¼)π + ½) - (¼) π = ½.