Chapter 16, Problem 13RP

To determine

To find: the area of circle from the equilateral triangle which is inscribed in circle.

Answer to Problem 13RP

The area of circle is $48\pi$ .

Explanation of Solution

Given:

  

Concept used:

The equilateral triangle has all sidesame and angle ${60}^0$ each.

Area of the circle:

  $\pi r^2$ .

Calculation:

The equilateral triangle bisects each line from the vertices of equilateral makes the intersection of the point through that angle ${60}^0$ can be divided to ${30}^0$ . Which the chord is given here $12$ which is divided by $2$ two to give $6$ .

Hypotenuse according to the diagram will be $2$ and that of base is $\sqrt{3}$ .

Chords here is base and radius is hypotenuse so by comparing:

  $\begin{array}{l}r:\frac{12}{2}=2:\sqrt{3}.\\ \frac{r}{6}=\frac{2}{\sqrt{3}}.\\ r=\frac{12}{\sqrt{3}}=\mathrm{6.92820.}\end{array}$

Area of the circle is:

  $\pi r^2=\pi \times {\left(6.92820\right)}^2=47.9999\pi \approx 48\pi .$

Hence, the area of circle is $48\pi$ .