If an equilateral triangle has an altitude of length $$12\sqrt5$$​ cm, then the difference between the areas of the circumscribed circle and the inscribed circle is:

Correct option is D

Given:

Altitude of an equilateral triangle = $$12\sqrt{5} \, \text{cm}$$​​

Find the difference between the areas of the circumscribed and inscribed circles.

Formula Used
In an equilateral triangle of side a:

Height h = $$\frac{\sqrt{3}}{2}a$$​

a =$$\frac{2h}{\sqrt{3}}$$​​

Radius of incircle: $$r = \frac{a}{2\sqrt{3}}$$​​

Radius of circumcircle: $$R = \frac{a}{\sqrt{3}}$$​​

Area of circle $$= \pi r^2$$​​

Solution:

a = $$\frac{2h}{\sqrt{3}} = \frac{2 \times 12\sqrt{5}}{\sqrt{3}} = \frac{24\sqrt{5}}{\sqrt{3}} = 24\sqrt{\frac{5}{3}}$$​

R =$$\frac{a}{\sqrt{3}} = \frac{24\sqrt{\frac{5}{3}}}{\sqrt{3}} = 24\sqrt{\frac{5}{9}} = 24 \times \frac{\sqrt{5}}{3} = 8\sqrt{5}$$​

Inradius:

r = $$\frac{a}{2\sqrt{3}} = \frac{24\sqrt{\frac{5}{3}}}{2\sqrt{3}} = 12 \times \sqrt{\frac{5}{9}} = 12 \times \frac{\sqrt{5}}{3} = 4\sqrt{5}$$​

Difference in areas:

​$$\pi(R^2 - r^2) = \pi\left((8\sqrt{5})^2 - (4\sqrt{5})^2\right) = \pi(320 - 80) = 240\pi$$​