Question

Infinite number of triangles are formed as shown in figure. If total area of these triangles is A then $8A$ is equal to

• A. 3
• B. 4
• C. 1
• D. 2

Answer 1

Answer: (A)

We have,
$A=\left(\frac{1}{2}\times \frac{1}{3}\times 1\right)+\left(\frac{1}{2}+\frac{1}{9}\times 2\right)+\left(\frac{1}{2}\times \frac{1}{27}+3\right)+\dots \infty$
$=\underset{s}{\underbrace{\frac{1}{2}\left(\frac{1}{3}+\frac{2}{9}+\frac{3}{27}+\dots \infty \right)}}=\frac{S}{2}$
Let $S=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\dots \infty$, then
$\frac{S}{3}=0+\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\dots \infty$
(Subtracting)
$\Rightarrow \frac{2S}{3}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\dots \infty$
$=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{2}$
$\Rightarrow S=\frac{3}{4}$
$\therefore A=\frac{S}{2}=\frac{3}{8}$
$\Rightarrow 8A=3$