Chapter 13.3, Problem 23PPS

To determine

To calculate:

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

Answer to Problem 23PPS

The probability of the point chosen at the random lies in shaded region is $\frac{1}{2}$ .

Explanation of Solution

Given information:

  

Calculation:

The given figure consists of shaded square with the side length of $8$ units contains un-shaded square with the side length is

The side length:

  $\begin{array}{l}=\sqrt{4^2+4^2}\\ =4\sqrt{2}\text{units}\end{array}$

The area of larger shaded square is:

Area of square: $l\times h\mathrm{...}\left(1\right)$

Put the value in equation (1):

Area of larger shaded square

  $\begin{array}{l}=8\times 8\\ =64{\text{unit}}^2\end{array}$

The area of larger shaded triangle is $64{\text{unit}}^2$ .

The area of smaller un-shaded square is:

Put the value in equation (1):

Area of smaller un-shaded square

  $\begin{array}{l}=4\sqrt{2}\times 4\sqrt{2}\\ =32{\text{unit}}^2\end{array}$

The area of smaller un-shaded square is $32{\text{unit}}^2$ .

Area of the larger shaded square is $64{\text{unit}}^2$ and the area of small Square is $32{\text{unit}}^2$ .

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

  $\begin{array}{l}P\left(\text{Point}\text{lies}\text{in}\text{shaded}\text{region}\right)=\frac{\text{area}\text{of}\text{shaded}\text{region}}{\text{area}\text{of}\text{entire}\text{region}}\\ P\left(\text{Point}\text{lies}\text{in}\text{shaded}\text{region}\right)=\frac{64-32}{64}\\ P\left(\text{Point}\text{lies}\text{in}\text{shaded}\text{region}\right)=\frac{32}{64}\\ P\left(\text{Point}\text{lies}\text{in}\text{shaded}\text{region}\right)=\frac{1}{2}\end{array}$

The probability of point chosen at the random lies in shaded region is $\frac{1}{2}$ .