Chapter 13.3, Problem 31PPS

To determine

To calculate:

The probability that point chosen at the random lies in a shaded region:

Answer to Problem 31PPS

The probability of the point chosen at the random lies in a shaded region is $0.33$ .

Explanation of Solution

Given information:

  

Calculation:

The figure consists of circle with diameter of $16$ units within is a shaded hexagon with side length of $8$ units and apothem of $4\sqrt{3}$ units and a rectangle with the side length $8$ units and $8\sqrt{3}$ units.

Area of un-shaded rectangle:

  $\begin{array}{l}=8\text{units}\times 8\sqrt{3}\text{units}\\ \text{=64}\sqrt{3}\text{units}\end{array}$

The area of shaded hexagon is same to half of the perimeter of hexagon multiplied by the apothem of hexagon:

  $\begin{array}{l}=\frac{1}{2}\times 6\times 8\text{units}\times 4\sqrt{3}\text{units}\\ =96\sqrt{3}{\text{units}}^2\end{array}$

Then the area of un-shaded circle:

Area of un-shaded circle: $\pi {\left(8\right)}^2=64\pi {\text{units}}^2$

The probability that point chosen at the random lies in the shaded region dividing the area of shaded region by the area of the given figure:

  $P\left(A\right)$ : P (Point lies in shaded region)

  $A$ : Area of shaded region

  $S$ : Area of entire region

  $\begin{array}{l}P\left(A\right)=\frac{A}{S}\\ P\left(A\right)=\frac{32\sqrt{3}+\frac{64\pi }{3}-32\sqrt{3}}{64\pi }\\ P\left(A\right)=\frac{64\pi }{3}\\ P\left(A\right)=0.33\end{array}$

The probability of the point chosen at the random lies in a shaded region is $0.33$ .