An equilateral triangle is inscribed in a circle of radius . Show that the ratio of the area of the triangle to the area of the circle is equal to .

Understand the Question An equilateral triangle of side length is inscribed in a circle of radius . The circumradius connects the center of the circle (which coincides with the centroid) to each vertex, meaning , where is the height of the triangle. We find both the area of the circle () and the area of the equilateral triangle () in terms of , and then compute their ratio. Step 1 · Find the Area of the Circle Given radius of the circle . Step 2 · Find the Side Length of the Triangle Let be the side length and be the height of the equilateral triangle. The centroid divides the median (height) in a ratio, so the circumradius is: For an equilateral triangle of side : Equating both expressions for : Step 3 · Find the Area of the Triangle The area of an equilateral triangle with side is: Substitute : Step 4 · Calculate the Ratio of the Areas Taking the ratio of the area of the triangle to the area of the circle: Approximating with and : Answer > Common Mistakes Circumradius vs. Inradius: For an inscribed triangle, (circumradius). Confusing it with the inradius will give incorrect side dimensions. Inverting the Ratio: Make sure to divide the area of the triangle by the area of the circle (), not the other way around. Squaring Error: Ensure is simplified properly to before multiplying by .

Measuring Space: Perimeter and Area | Exercise 6.3

Question 8

An equilateral triangle is inscribed in a circle of radius $r$ . Show that the ratio of the area of the triangle to the area of the circle is equal to $\dfrac{3\sqrt{3}}{4\pi} \approx 0.413$ .

Solution

Understand the Question

An equilateral triangle of side length $s$ is inscribed in a circle of radius $r$ .
The circumradius $r$ connects the center of the circle (which coincides with the centroid) to each vertex, meaning $r = \dfrac{2}{3}h$ , where $h = \dfrac{\sqrt{3}}{2}s$ is the height of the triangle.
We find both the area of the circle ( $\pi r^2$ ) and the area of the equilateral triangle ( $\dfrac{\sqrt{3}}{4}s^2$ ) in terms of $r$ , and then compute their ratio.

Step 1 · Find the Area of the Circle

Given radius of the circle $= r$ . Diagram 1

$\text{Area of circle} = \pi r^2$

Step 2 · Find the Side Length of the Triangle

Let $s$ be the side length and $h$ be the height of the equilateral triangle. Diagram 2

The centroid divides the median (height) in a $2:1$ ratio, so the circumradius $r$ is: $r = \dfrac{2}{3}h \implies h = \dfrac{3}{2}r$

For an equilateral triangle of side $s$ : $h = \dfrac{\sqrt{3}}{2}s$

Equating both expressions for $h$ : $\dfrac{\sqrt{3}}{2}s = \dfrac{3}{2}r$

$\sqrt{3}s = 3r$

$s = \dfrac{3}{\sqrt{3}}r = \sqrt{3}r$

Step 3 · Find the Area of the Triangle

The area of an equilateral triangle with side $s$ is: $\text{Area of triangle} = \dfrac{\sqrt{3}}{4}s^2$

Substitute $s = \sqrt{3}r$ :

\begin{aligned} \text{Area of triangle} &= \dfrac{\sqrt{3}}{4}(\sqrt{3}r)^2 \\[0.6em] &= \dfrac{\sqrt{3}}{4}(3r^2) \\[0.6em] &= \dfrac{3\sqrt{3}}{4}r^2 \end{aligned}

Step 4 · Calculate the Ratio of the Areas

Taking the ratio of the area of the triangle to the area of the circle:

\begin{aligned} \text{Ratio} &= \dfrac{\text{Area of triangle}}{\text{Area of circle}} \\[0.6em] &= \dfrac{\dfrac{3\sqrt{3}}{4}r^2}{\pi r^2} \\[0.6em] &= \dfrac{3\sqrt{3}}{4\pi} \end{aligned}

Approximating with $\sqrt{3} \approx 1.732$ and $\pi \approx 3.14159$ :

\begin{aligned} \text{Ratio} &= \dfrac{3 \times 1.732}{4 \times 3.14159} \\[0.6em] &= \dfrac{5.196}{12.56636} \\[0.6em] &\approx 0.413 \end{aligned}

Answer

$\dfrac{3\sqrt{3}}{4\pi} \approx 0.413$

Common Mistakes

Circumradius vs. Inradius: For an inscribed triangle, $r = \dfrac{2}{3}h = \dfrac{s}{\sqrt{3}}$ (circumradius). Confusing it with the inradius $r_{\text{in}} = \dfrac{1}{3}h = \dfrac{s}{2\sqrt{3}}$ will give incorrect side dimensions.
Inverting the Ratio: Make sure to divide the area of the triangle by the area of the circle ( $\text{Triangle} : \text{Circle}$ ), not the other way around.
Squaring Error: Ensure $(\sqrt{3}r)^2$ is simplified properly to $3r^2$ before multiplying by $\dfrac{\sqrt{3}}{4}$ .

Question 8

More questions in Exercise 6.3