Chapter 16.2, Problem 3PS

To determine

To find: Thevalue of $a+b+c$ .

Answer to Problem 3PS

The value of $a+b+c$ is $81$ .

Explanation of Solution

Given information:

The given figure is,

  

Calculate the diameter of the circle as,

  $\begin{array}{c}D=\frac{16}{\sqrt{4^2-x^2}}\\ =\frac{16}{\sqrt{16-x^2}}\text{unit}\end{array}$

The area of the circle is calculated as,

  $\begin{array}{c}A=\frac{\pi }{4}{D}^2\\ =\frac{\pi }{4}{\left(\frac{16}{\sqrt{16-x^2}}\right)}^2\\ =\frac{64\pi }{16-x^2}{\text{unit}}^2\end{array}$

The given expression for the area is,

  $A=\frac{a\pi }{b-c{x}^2}$

Compare the given and calculated area to obtain,

  $\begin{array}{l}a=64\\ b=16\\ c=1\end{array}$

The value of $a+b+c$ is calculated as,

  $\begin{array}{c}a+b+c=64+16+1\\ =81\end{array}$

Thus, the value of $a+b+c$ is $81$ .