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 $r_1$ and $r_2$ be the radii of the $1^{\text{st }}$ and $2^{\text{nd }}$ circles.
$h=\sqrt{3}\Rightarrow r_1=\frac{\sqrt{3}}{3}\Rightarrow r_1=\frac{1}{\sqrt{3}}\left[\tan 30^{\circ }=\frac{r_1}{1}\right][13t]$
$A_1=\frac{\pi }{3}.\text{ Now }{r}_1+r_2=2\left(r_1-r_2\right)\Rightarrow r_2=\frac{1}{3\sqrt{3}}$
$A_2=\frac{\pi }{27}\text{ and so on }$
Total area $(A)=\pi \left(\frac{1}{3}+\frac{1}{27}+\frac{1}{243}+\dots \dots .\right)=\frac{\pi /3}{1-(1/9)}=\frac{3\pi }{8}$