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.