What is the approximate area of the shaded region?

swerve wrote:

Square ABCD is perfectly inscribed in the circle pictured above. If minor arc AD measure 2π, what is the approximate area of the shaded region?

A. 110
B. 72
C. 36
D. 18
E. 9

The OA is D.

Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.

The area of the shaded region = Area of the circle – Area of the square

Since the square is perfectly inscribed, Arc AD = Arc DC = Arc CB = Arc BA

Thus, length of arc AD = 1/4 of circumference of the circle

=> 2π = (2πr)/4 => r = 4

=> Area of the circle = πr^2 = π.4^2 = 16π

Side of the square would be given by √2r

Area of the square = (Side)^2 = (√2r)^2 = 2r^2 = 2.4^2 = 32

Thus, the area of the shaded region = Area of the circle – Area of the square = 16π – 32 = 16.(3.14) – 32 = ~50 – 32 = ~18.

The correct answer: D

Hope this helps!

-Jay
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swerve wrote:

Square ABCD is perfectly inscribed in the circle pictured above. If minor arc AD measure 2π, what is the approximate area of the shaded region?

A. 110
B. 72
C. 36
D. 18
E. 9

The OA is D.

Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.

Since minor arc AD is 1/4 of the circle, the circumference of the circle is 4 x 2π = 8π.

Now we can find the radius of the circle using the circumference formula C = 2πr:

8π = 2πr

4 = r

Thus, the area of the circle is π * 4^2 = 16π and the diameter of circle = 8. Since the diameter of circle = the diagonal of the square and the diagonal of the square = side√2, we have:

8 = side√2

8/√2 = side

Thus, the area of the square is (8/√2)^2 = 64/2 = 32.

So, the area of the shaded region is approximately 16π - 32 = 16(π - 2). Since π is approximately 3.14, we have 16(3.14 - 2) = 16(1.14), which is approximately 18.

Answer: D