Question

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

• A. $3$
• B. $4$
• C. $1$
• D. $2$

Answer 1

The correct option is A $3$

According to figure, Total area $A=\frac{1}{2}\left(\frac{1}{3}\right)\left(1\right)+\frac{1}{2}\left(\frac{1}{9}\right)\left(2\right)+\frac{1}{2}\left(\frac{1}{27}\right)\left(3\right)+...$

$A=\frac{1}{2}[\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...]$ Which is A.G.P series

$\frac{1}{3}.A=\frac{1}{2}[\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...]$

$\frac{2A}{3}=\frac{1}{2}[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...]$

$\frac{2A}{3}=\frac{1}{2}.\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}\times \frac{1}{3}\times \frac{3}{2}$ [sum of infinite G.P = $\frac{a}{1-r}$]

$\Rightarrow \frac{2A}{3}=\frac{1}{4}$

$8A=3$

$\therefore$ Option A is correct