Question

What is the area of an equilateral triangle whose vertices lie on a circle with radius 2?

Answer 1

`3*sqrt(3)~=5.196`

Explanation:

Refer to figure below

The figure represents an equilateral triangle inscribed in a circle, where `s` stands for the triangle's sides, `h` stands for the height of the triangle, and `R` stands for the circle's radius.

We can see that triangles ABE, ACE and BCE are congruents, that's why we can say that angle `E hat C D=(A hat C D)/2=60^@/2=30^@`.

We can see in `triangle_(CDE)` that
`cos 30^@=(s/2)/R` => `s=2*R*cos 30^@=cancel(2)*R*sqrt(3)/cancel(2)` => `s=sqrt(3)*R`

In `triangle_(ACD)` we cant see that
`tan 60^@=h/(s/2)` => `h=s*tan 60^@/2` => `h=sqrt(3)/2*s=sqrt(3)/2*sqrt(3)*R` => `h=(3R)/2`

From the formula of the area of the triangle:
`S_triangle=(base*height)/2`
We get
`S_triangle=(s*h)/2=(sqrt(3)R*(3R)/2)/2=(3*sqrt(3)*R^2)/4=(3*sqrt(3)*cancel(2^2))/cancel(4)=3*sqrt(3)`