An equilateral triangle is inscribed in a circle of radius R. Find the area of the equilateral triangle.

Guest May 26, 2020

#1**0 **

A diagram would be of great assistance for this problem, so I created one for you. I added lines and points where I saw fit to add clarity to the diagram:

In order to find the area of any triangle, we need the length of the base and the length of the perpendicular height (also known as the altitude) of the triangle. Then, we can apply the formula \(A_{\triangle ABC}=\frac{1}{2}bh\) to find the area. In the diagram above, \(\overline{AB}\) represents the base, and \(\overline{EC}\) represents the perpendicular height of the triangle.

\(m\angle ADB=120^{\circ}\)as each central angle \(\triangle ABC\) forms divides equally in thirds. By inserting an angle bisector of \(\angle ADB\), this means that \(m\angle ADE=60^{\circ}\). This angle bisector also happens to be a perpendicular of chord \(\overline{AB}\). Combining all of this information together means that \(\triangle ADE\) is a 30-60-90 triangle.

A 30-60-90 triangle has a constant ratio of its sides of \(1:\sqrt{3}:2\). In this particular case, \(r\) is the length of the hypotenuse of \(\triangle ADE\). Let's use these ratios to find the length of the base and the height in terms of \(r\).

\(\frac{AE}{r}=\frac{\sqrt{3}}{2}\\ AE=\frac{\sqrt{3}}{2}r\) | \(\frac{DE}{r}=\frac{1}{2}\\ DE=\frac{1}{2}r\) |

By parallel reasoning, \(BE=\frac{\sqrt{3}}{2}r\).

We can find \(AB\), the length of the base of the triangle in terms of r now.

\(AB=AE+BE\\ AB=\frac{\sqrt{3}}{2}r+\frac{\sqrt{3}}{2}r\\ AB=\sqrt{3}r\)

Similarly, we can do the same for \(CE\), the length of the perpendicular height of the triangle.

\(CE=CD+DE\\ CE=r+\frac{1}{2}r\\ CE=\frac{3}{2}r\)

Finally, let's solve for the area of the inscribed triangle in terms of r.

\(A_{\triangle ABC}=\frac{1}{2}*AB*CE\\ A_{\triangle ABC}=\frac{1}{2}*\sqrt{3}r*\frac{3}{2}r\\ A_{\triangle ABC}=\frac{3\sqrt{3}}{4}r^2\)

TheXSquaredFactor May 26, 2020