Question

A point in the figure is selected at random. Find the probability that the point will be in the shaded region.

A. about 25%
B. about 60%
C. about 50%
D. about 75%

Answer 1

B. about 60%

Explanation:

Let us assume that the radius of the semi-circle is d units.

So the area of the semi-circle will be,

`=(\pi d^2)/(2)`

In total there are 4 semi circles, so the net area will be,

`=4* (\pi d^2)/(2)`

`=2\pi d^2`

The side length of the square is twice the radius of the semi circle. So the side length of the square is 2d units.

Area of the square is,

`=(2d)^2\\\\=4d^2`

The total area will be,

`=2\pi d^2+4d^2`

So that the random point will be in the shaded region is,

`=\frac{\text{Area of semi circles}}{\text{Area of square}}`

`=(2\pi d^2)/(2\pi d^2+4d^2)`

`=(2\pi)/(2\pi+4)`

`=0.6`

`=60\%`

Answer 2

Area of the figure:
Af = a² + 2 a²/4 π = 2.57 a²
Area of the shaded region:
As = 2 a²/4 π = 1.57 a²
x = (1.57 · 100) / 2.57 = 61 % ≈ 60 %
Answer: B) about 60 %