Question

find the area of a regular triangle with a radius of 8

Answer 1

48√3 u² (or 83.14 u²)

1) Since the question mentions a radius and a regular triangle, then we can sketch out an inscribed triangle:

2) So, we can write out the following about it, every angle is 60º. Also, we have two right triangles below.

The triangle below is a special right triangle 30º 60º 90º. And prolonging a line segment we can trace 6 right triangles:

3) Now let's find the height of that triangle:

$\sin (30)=\frac{x}{8}\Rightarrow\frac{1}{2}=\frac{x}{8}\Rightarrow2x=8\Rightarrow x=4$

To find out the base, another trig ratio:

$\begin{gathered} \cos (30)=\frac{y}{8} \\ \frac{\sqrt[]{3}}{2}=\frac{y}{8} \\ 2y=8\sqrt[]{3} \\ y=4\sqrt[]{3} \end{gathered}$

3) Finally let's plug into the Area of a Triangle, and then multiply it by 6. To find the area of that larger triangle:

$A=6(\frac{1}{2}\cdot4\sqrt[]{3}\cdot4)\Rightarrow A=48\sqrt[]{3}$

Hence, the answer is 48√3