Chapter 13.3, Problem 24PPS

To determine

To calculate:

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

Answer to Problem 24PPS

The probability of the point chosen at the random lies in shaded region is $0.755$ .

Explanation of Solution

Given information:

  

Calculation:

This given figure consists of equilateral triangle with the side length of $14$ units and height $7\sqrt{3}$ units that contains three un-shaded equilateral triangles with side length $4$ units and height $2\sqrt{3}$ units.

The probability of a point chosen at the random lies in shaded region is equal to the area of the shaded region which divided by the area of given entire figure.

The area of larger 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 larger shaded triangle

  $\begin{array}{l}=\frac{1}{2}\times 14\times 7\sqrt{3}\\ =49\sqrt{3}{\text{unit}}^2\end{array}$

The area of larger shaded triangle is $49\sqrt{3}{\text{unit}}^2$ .

The area of smaller un-shaded triangle is:

Put the value in equation (1):

Area of smaller un-shaded triangle

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

The area of smaller un-shaded triangle is $4\sqrt{3}{\text{unit}}^2$ .

The area of larger triangle is $49\sqrt{3}{\text{unit}}^2$ , and the area of three smaller triangles is $4\sqrt{3}{\text{unit}}^2$ .

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.

  $\begin{array}{l}P\left(\text{Point}\text{lies}\text{in}\text{shaded}\text{region}\right)=\frac{\text{area}\text{of}\text{shaded}\text{region}}{\text{area}\text{of}\text{entire}\text{region}}\\ P\left(\text{Point}\text{lies}\text{in}\text{shaded}\text{region}\right)=\frac{49\sqrt{3}-3\left(4\sqrt{3}\right)}{49\sqrt{3}}\\ P\left(\text{Point}\text{lies}\text{in}\text{shaded}\text{region}\right)=\frac{37}{49}\\ P\left(\text{Point}\text{lies}\text{in}\text{shaded}\text{region}\right)=0.755\end{array}$