Chapter 13.3, Problem 29PPS

To determine

To calculate:

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

Answer to Problem 29PPS

The probability of the point chosen at the random lies in shaded region is $0.24$ .

Explanation of Solution

Given information:

  

Calculation:

The given figure consists of shaded circle with radius $5$ and interior un-shaded pentagon with the line of the length $5$ between center of the pentagon and its vertices.

The probability that the point chosen at random lies in shaded region is same to the area of shaded regions divided by the area of the given figure.

Let,

  $x$ : The pentagon’s apothem.

The area of pentagon is equal to half of apothem and the perimeter of the pentagon.

Since the triangle within the hexagon is the right triangle, the apothem is equal to $5\left(\sin \left({54}^{\circ }\right)\right)$ and the length of the side of pentagon is $10\left(cos\left({54}^{\circ }\right)\right)$ .

The area of pentagon:

  $\begin{array}{l}=\frac{1}{2}\times 5\times 5\left(\sin \left({54}^{\circ }\right)\right)\times 10\left(cos\left({54}^{\circ }\right)\right)\\ =59.4\end{array}$

The area of circle is given by $\pi r^2$ .

Where,

  $r$ : Radius of the circle

The area of circle:

  $\begin{array}{l}=\pi r^2\\ =\pi \left(5^2\right)\\ =25\pi \end{array}$

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

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

  $A$ : Area of shaded region

  $S$ : Area of entire region

  $\begin{array}{l}P\left(A\right)=\frac{A}{S}\\ P\left(A\right)=\frac{25\pi -\frac{1}{2}\times 5\times 5\left(\sin \left({54}^{\circ }\right)\right)\times 10\left(cos\left({54}^{\circ }\right)\right)}{25\pi }\\ P\left(A\right)=\frac{25\pi -59.4}{25\pi }\\ P\left(A\right)=0.24\end{array}$

The probability of the point chosen at the random lies in shaded region is $0.24$ .