Chapter 8.7, Problem 58PFA

To determine

To calculate the radius of the circle.

Answer to Problem 58PFA

  $(5x-4)$

Explanation of Solution

Given:

  $\text{Area of the circle, }A=\pi (25x^2-40x+16)$

Formula used:

  $\text{Area of the circle}=\pi r^2$

Roots of the polynomial of the form $a{x}^2+bx+c=a(x-x_1)(x-x_2)$ are

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Let x be the radius of the circle.

Given that the area of the circle $\pi (25x^2-40x+16),$

Substituting in the given equation for $a=25,b=-40,c=16$ and the formula,

  $\begin{array}{l}\Rightarrow x=\frac{-(-40)\pm \sqrt{{(-40)}^2-4(25)(16)}}{2(25)}\\ \Rightarrow x=\frac{40\pm \sqrt{1600-1600}}{50}\\ \Rightarrow x=\frac{40}{50}=\frac{4}{5}\end{array}$

Therefore, the roots are

  $\Rightarrow x_1=x_2=\frac{4}{5}$

Therefore, the expression can be factorised as

  $\begin{array}{l}\Rightarrow 25x^2-40x+16=(x-\frac{4}{5})(x-\frac{4}{5})\\ \Rightarrow 25x^2-40x+16=(5x-4)(5x-4)\\ $ \Rightarrow 25x^2-40x+16={(5x-4)}^2$\end{array}$

Equating the general equation of area of circle with the given area,

  $\begin{array}{l}\text{Area of the circle}=\pi r^2\\ \Rightarrow \pi (25x^2-40x+16)=\pi r^2\\ \Rightarrow (25x^2-40x+16)=r^2\\ $ \Rightarrow {(5x-4)}^2=r^2$\Rightarrow (5x-4)=r\end{array}$

Conclusion:

Therefore, the radius of the circle is $(5x-4)$ .