Chapter 13.3, Problem 40STP

To determine

To calculate:

The probability that randomly chosen point in the interior of the square will also lie in the interior of the circle.

Answer to Problem 40STP

The probability that randomly chosen point in the interior of the square will also lie in the interior of the circle is $\frac{\pi }{9}$ .

Explanation of Solution

Given information:

A circle with radius $3$ is contained in a square with side length $9$ .

Calculation:

Consider the figure as shown below:

  

The area of the square having the side length $9$ is:

  $\begin{array}{l}s=9^2\\ s=81\end{array}$

The area of the circle having radius $3$ is:

  $\begin{array}{l}A=\pi \left(3^2\right)\\ A=9\pi \end{array}$

The probability that randomly chosen point in the interior of the square will also lie in the interior of the circle is $P$ .

  $\begin{array}{l}P=\frac{A\left(C\right)}{A\left(S\right)}\\ P=\frac{9\pi }{81}\\ P=\frac{\pi }{9}\end{array}$

The probability that randomly chosen point in the interior of the square will also lie in the interior of the circle is $\frac{\pi }{9}$ .

Thus, the answer is matched with the option D.