Q. 16 - Precalculus Enhanced with Graphing Utilities

Question

An equilateral triangle is inscribed in a circle

of radius r. See the figure. Express the circumference C of the circle as a function of the length x of a side of the triangle.

[Hint: First show that $r^2=\frac{x^2}{3}$.]

Step-by-Step Solution

Answer

The circumference C of the circle as a function of the length x of a side of the triangle is $C\left(x\right)=\frac{2\mathrm{\pi x}}{\sqrt{3}}$.

Step 1. Consider the following figure.

Step 2. Using Pythagorean Theorem.

$$\begin{gathered} r^2=h^2+{\left(\frac{x}{2}\right)}^2{r}^2-h^2=\frac{x^2}{4} \\ 4(r^2-h^2)=x^2 \end{gathered}$$

Step 3. Using Pythagorean Theorem.

$x^2={(r+h)}^2+{\left(\frac{x}{2}\right)}^2{x}^2-\frac{x^2}{4}={(r+h)}^2\frac{3}{4}{x}^2={(r+h)}^2{x}^2=\frac{4}{3}{(r+h)}^2$

Step 4. Substitute values of x 2 obtained in above steps.

$$\begin{gathered} \frac{4}{3}{(r+h)}^2=4(r^2-h^2) \\ \frac{r+h}{3}=r-h \\ r+h=3r-3h \\ 4h=2r \\ h=\frac{r}{2} \end{gathered}$$

Step 5. Substitute h = r 2 in first Pythagorean theorem step and find the circumference of the circle.

$$\begin{gathered} x^2=4(r^2-{\left(\frac{r}{2}\right)}^2)x^2=4\left(\frac{3}{4}{r}^2\right) \\ {x}^2=3r^2 \\ {r}^2=\frac{x^2}{3} \\ r=\frac{x}{\sqrt{3}} \end{gathered}$$

The circumference of the circle, C $=2\mathrm{\pi r}$

$$\begin{gathered} C=2\mathrm{\pi }r \\ C=2\mathrm{\pi }\frac{\mathrm{x}}{\sqrt{3}} \\ \mathrm{C}=\frac{2\mathrm{\pi x}}{\sqrt{3}} \end{gathered}$$