Chapter 13.3, Problem 25PPS

To determine

To calculate:

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

Answer to Problem 25PPS

The probability of the point chosen at the random lies in shaded region is $0.535$ .

Explanation of Solution

Given information:

  

Calculation:

The given figure consists of the shaded circle with the radius of $2.5$ units contains $2$ un-shaded right triangles, one with the base length of $2.5$ units and height of $2.5$ units, and other with base length of $3$ units and height of $4$ units.

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

The area of largest un-shaded triangle is:

Area of triangle: $\frac{1}{2}\times l\times h\mathrm{...}\left(1\right)$

Put the value in equation (1):

Area of largest un-shaded triangle

  $\begin{array}{l}=\frac{1}{2}\times 3\times 4\\ =6{\text{unit}}^2\end{array}$

The area of largest shaded triangle is $6{\text{unit}}^2$ .

The area of smallest un-shaded triangle is:

Put the value in equation (1):

Area of smallest un-shaded triangle

  $\begin{array}{l}=\frac{1}{2}\times 2.5\times 2.5\\ =3.125{\text{unit}}^2\end{array}$

The area of smallest un-shaded triangle is $3.125{\text{unit}}^2$ .

The area of shaded circle is:

  $\begin{array}{l}=\pi {\left(2.5\right)}^2\\ =6.25\pi {\text{units}}^2\end{array}$

The area of circle is $6.25\pi {\text{units}}^2$ .

The area of circle is $6.25\pi {\text{units}}^2$ , the area of the largest triangle is $6{\text{unit}}^2$ , and the area of smallest triangles is $3.125{\text{unit}}^2$ .

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

  $\begin{array}{l}P\left(\text{Point}\text{lies}\text{in}\text{shaded}\text{region}\right)=\frac{\text{area}\text{of}\text{shaded}\text{region}}{\text{area}\text{of}\text{entire}\text{region}}\\ P\left(\text{Point}\text{lies}\text{in}\text{shaded}\text{region}\right)=\frac{6.25\pi -\left(6+3.125\right)}{6.25\pi }\\ P\left(\text{Point}\text{lies}\text{in}\text{shaded}\text{region}\right)=\frac{6.25\pi -9.125}{6.25\pi }\\ P\left(\text{Point}\text{lies}\text{in}\text{shaded}\text{region}\right)=0.535\end{array}$

The probability of the point chosen at the random lies in shaded region is $0.535$ .