In the following figure, ABC is an equilateral triangle which is inscribed inside a circle and whose radius is r. Which of the following is the area of the triangle ?

BD=DC = r
According to Pythagoras Theorem , BD$$^{2}$$ = r$$^{2}$$ = DE$$^{2}$$+ BE$$^{2}$$
$$BE^{2}$$= $$r$$$$^{2}$$ - $$ DE$$$$^{2}$$
$$BE$$=$$(r^{2} - DE^{2})$$ $$^{\frac{1}{2}}$$
$$r$$$$^{2}$$ - $$ DE$$$$^{2}$$= $$(r+DE) \times (r-DE)$$.
$$BC=2\times BE$$=
$$AE= (r+DE)$$
Area of $$\triangle$$ ABC = $$\frac{1}{2} \times BC \times AE$$ = $$\frac{1}{2} \times 2\times (r^{2}$$ - $$DE^{2})$$ $$\times$$ $$(r+DE)$$
$$\frac{1}{2} \times 2\times (r^{2}$$ - $$DE^{2})$$ $$\times$$ $$(r+DE)$$ = $$(r-DE) \times (r+DE) \times (r+DE)$$ = $$(r-DE)^\frac{1}{2}$$ $$(r+DE)^\frac{3}{2}$$

Hence Option C is the correct answer.