Question

A circle is inscribed in an equilateral triangle whose side length is 2. Then another circle is inscribed externally tangent to the first circle but inside the triangle as shown, and then another and another. If this process continues indefinitely, the total area of all the circles is

• A. $\frac{3 \pi}{8}$
• B. $\frac{4 \pi}{8}$
• C. $\frac{5 \pi}{8}$
• D. $\frac{6 \pi}{8}$

Answer 1

Answer: (A)

Let $\quad r_1$ and $r_2$ be the radii of the $1^{\text {st }}$ and $2^{\text {nd }}$ circles.
$h =\sqrt{3} \Rightarrow r _1=\frac{\sqrt{3}}{3} \Rightarrow \quad r _1=\frac{1}{\sqrt{3}}\left[\tan 30^{\circ}=\frac{ r _1}{1}\right][13 t ] $
$A _1=\frac{\pi}{3} . \text { Now } r _1+ r _2=2\left( r _1- r _2\right) \Rightarrow r _2=\frac{1}{3 \sqrt{3}}$
$A _2=\frac{\pi}{27} \text { and so on }$
Total area $( A )=\pi\left(\frac{1}{3}+\frac{1}{27}+\frac{1}{243}+\ldots \ldots.\right)=\frac{\pi / 3}{1-(1 / 9)}=\frac{3 \pi}{8}$