If the difference between the areas of the circumcircle and the incircle of an equilateral triangle is 66 $$\text{cm}^2$$​, then the area of the triangle is:

Correct option is C

Given:
The triangle is equilateral.
Difference in areas:

Area of Circumcircle - Area of Incircle = 66 $$\text{ cm}^2$$​​
Formula Used:​
For an equilateral triangle with side length a:
Circumradius (R):
R = $$\frac{a}{\sqrt{3}}$$​​
Inradius (r):
r = $$\frac{a}{2\sqrt{3}}$$​​
Area of Circumcircle:
​$$\pi R^2 = \pi \left(\frac{a}{\sqrt{3}}\right)^2 = \frac{\pi a^2}{3}$$​​

Area of Incircle:

​$$\pi r^2 = \pi \left(\frac{a}{2\sqrt{3}}\right)^2 = \frac{\pi a^2}{12}$$​​

Area of Equilateral Triangle:
​$$\text{Area} = \frac{\sqrt{3}}{4} a^2$$​​
Solution:

​$$\text{Area of Circumcircle} - \text{Area of Incircle} = 66 \text{ cm}^2 \\ \ \\\frac{\pi a^2}{3} - \frac{\pi a^2}{12} = 66 \\ \ \\\frac{4\pi a^2 - \pi a^2}{12} = 66 \\ \ \\\frac{3\pi a^2}{12} = 66 \\ \ \\\frac{\pi a^2}{4} = 66 \\ \ \\\pi a^2 = 264 \\ \ \\a^2 = \frac{264}{\pi} \\ \ \\ \text{Area} = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times \frac{264}{\pi} = \frac{66\sqrt{3}}{\pi} \text{ cm}^2 = \frac {66\sqrt{3}}{\frac{22}{7}} = 21 \sqrt3$$​​