Problem with triangle in a circle

Question

We are given that $ABC$ is equilateral so $AB=BC=CA$, and that the length of the circle is $24\pi$. What is the area of the triangle?

Since $24\pi=2\pi r$ we get that $r=12$ and $OB=OC=OA=r$; $OAB$, $OAC$, $OCB$ is isosceles and because $ABC$ is equilateral the area of the triangle is $\frac{a^2\sqrt{3}}{4}$, how to find $AB=BC=CA=a$?

Answer 1

Draw in the midpoint of $\overline{AB}$ and call it Point M. $\triangle$AOM is a $30^\circ-60^\circ-90^\circ$ right triangle, which means that since the circumradius is 12, $|\overline{AO}|$ = 12, $|\overline{OM}|$ = 6, and $|\overline{AM}| = 6\sqrt{3}$ which means $|\overline{AB}| = 12\sqrt{3}$. The formula for the area of an equilateral triangle is A = $\frac{s^2\sqrt{3}}{4}$, so the area of $\triangle$ABC is $\frac{(12\sqrt{3})^2\sqrt{3}}{4} = \frac{432\sqrt{3}}{4} \approx 187$un$^2$.

Answer 2

We use Euler formula for distance between center of circucircle O and incircle I:

$d^2=R(R-2r)$

where R is the radius of circucircle(you found it equal to $R=2\sqrt 6$) and r is the radius on incircle.In our case $d=0$ because these centers are coincident, so we have:

$d^2=R(R-2r)=0$

$R\neq 0\Rightarrow R-2r=0\rightarrow r=\frac R 2=\sqrt 6$

Also :

$r=\frac a 2 tan 30^o=\frac a2\frac 1{\sqrt 3}=\sqrt 6$

Finally:

$a=6\sqrt 2$

Answer 3

Let $M$ be the midpoint of $AB$. Then, by Viviani's theorem, $|CM|=18$.

$\triangle CMB$ is right-angled, and so apply Pythagoras's theorem to find

$$18^2 + \frac{a}{2}^2 = a^2$$

$$a^2=432$$

$$a=12\sqrt{3}$$