Chapter 13.3, Problem 35HP

To determine

To calculate:

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

  

Answer to Problem 35HP

The probability of the point chosen at the random lies in shaded region is $14.3\%$ .

Explanation of Solution

Given information:

  

Calculation:

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

Consider:

Area of shaded region: $SR$

Area of big rectangle: $BR$

The area of the shaded region is equal to the subtraction of the big rectangle’s area and total area (i.e. addition of circle area + Two half circle’s area + Two small rectangle’s area).

  

Now the area of all the parts is calculated as:

Area of the big rectangle:

  $\begin{array}{l}{A}_{BR}=\left(6r\right)\left(2r\right)\\ A_{BR}=12r^2\end{array}$

Area of circle:

  $A_C=\pi r^2$

Area of half circle:

  $A_{HC}=\frac{\pi r^2}{2}$

Area of small rectangles:

  $\begin{array}{l}{A}_{SR}=\left(r\right)\left(2r\right)\\ A_{SR}=2r^2\end{array}$

The probability of the point chosen at the random lies in shaded region is calculated as:

  $\begin{array}{l}P=\frac{12r^2-\left[\pi r^2+\frac{1}{2}\left(\frac{\pi r^2}{2}\right)+2\left(2r^2\right)\right]}{12r^2}\\ P=\frac{12r^2-2\pi r^2-4r^2}{12r^2}\\ P=\frac{r^2\left(4-\pi \right)}{12r^2}\\ P=\frac{4-\pi }{6}\\ P\approx 14.3\%\end{array}$