Segments formed by an inscribed equilateral triangle

Hello. This is a problem from a book I'm studying on, I can't understand the solution.

"Find each segment formed by an inscribed equilateral triangle if the radius of the circle is 8"

Solution
R=8. Since $\displaystyle s=R\sqrt{3}=8\sqrt{3}$, the area of $\displaystyle \vartriangle ABC$ is $\displaystyle \tfrac{1}{4}s^{2}\sqrt{3}=48\sqrt{3}$.
Also, area of circle O = $\displaystyle \pi R^{2}=64\pi $.
Hence area of segment BDC = $\displaystyle \tfrac{1}{3}\left( 64\pi -48\sqrt{3} \right)$

These are my questions:
1.- Why is $\displaystyle s=R\sqrt{3}=8\sqrt{3}$?

2.And why is the area of $\displaystyle \vartriangle ABC$ equal to $\displaystyle \tfrac{1}{4}s^{2}\sqrt{3}=48\sqrt{3}$? I know the area of a triangle is equal to $\displaystyle \tfrac{1}{2}bh$, how do you get the height?
Thanks.

Inscribed equilateral triangle:

Draw apothem from center perpendicular to a side.

Draw radius from center to vertex of same side.

Triangle formed is 30-60-90.

If hypotenuse(radius) is 8, then side opposite the 30 deg. angle is 4 (apothem).

Side opposite 60 degree angle is $\displaystyle 4\sqrt3$

Thus, the length of a side is $\displaystyle 2(4\sqrt3)=8\sqrt3$

Perimeter = $\displaystyle 3(8\sqrt3)=24\sqrt3$

$\displaystyle Area=\frac{1}{2}Pa$, where P=perimeter, a=apothem

Finally, $\displaystyle A=\frac{1}{2}(24\sqrt3)(4)=48\sqrt3$

Thanks a lot.