Chapter 3.1, Problem 36E

To determine

To calculate: .The circle for the value $x$ .

Answer to Problem 36E

The required value of $x$ is $2$ .

Explanation of Solution

Given Information:

The area of circle is $25\pi$ and radius of the circle $x+3$ .

Formula Used:

Area of rectangle:

  $Area=\pi r^2$

Where $r$ is radius of circle.

Quadratic formula:

If $a{x}^2+bx+c=0$ then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where $a,b,c$ are constant and $a\ne 0$ .

Calculation:

Consider the area of circle is $25\pi$ .

To find the value of $x$ ; use area of circle formula.

  $Area=\pi r^2$

Substitute $25\pi$ for $Area$ and $x+3$ for $r$ in the above equation.

  $25\pi =\pi {\left(x+3\right)}^2$

Simplify the above equation.

  $\begin{array}{l}25\pi =\pi {\left(x+3\right)}^2\\ 25=x^2+6x+9\\ x^2+6x+9-25=0\\ x^2+6x-16=0\end{array}$

Now, use quadratic formula in the above equation.

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $1$ for $a$ , $6$ for $b$ and $-16$ for $c$ in the above equation.

  $\begin{array}{c}x=\frac{-6\pm \sqrt{6^2-4\left(1\right)\left(-16\right)}}{2\left(1\right)}\\ =\frac{-6\pm \sqrt{36+64}}{2}\\ =\frac{-6\pm \sqrt{100}}{2}\\ =\frac{-6\pm 10}{2}\end{array}$

Simplify the equation.

  $\begin{array}{c}x=\frac{-6\pm 10}{2}\\ =\frac{-6+10}{2},\frac{-6-10}{2}\\ =\frac{4}{2},\frac{-16}{2}\\ =2,-8\end{array}$

Since, $x$ is measurement and it is known that measurement cannot be negative.

Thus, the value of $x$ is $2$ .

Hence, the required real value of $x$ is $2$ .