Question

Let $C$ be a circle with centre $P_0$ and $A B$ be a diameter of $C$. Suppose $P_1$ is the mid point of the line segment $P_0 B, P_2$ is the mid point of the line segment $P_1 B$ and so on. Let $C_1, C_2, C_3, \ldots$. be circles with diameters $P _0 P _1, P _1 P _2, P _2 P _3 \ldots . .$. respectively. Suppose the circles $C _1, C _2, C _3, \ldots . .$. are all shaded. The ratio of the area of the unshaded portion of $C$ to that of the original circle $C$ is

• A. $8: 9$
• B. $9: 10$
• C. $10: 11$
• D. $11: 12$

Answer 1

Answer: (D)

area of circle $C _1=\frac{\pi}{4}\left(\frac{ r }{2}\right)^2 \left(\right.$ area of circle $\left.=\frac{\pi d ^2}{4}\right)$
area of circle $C_2=\frac{\pi}{4}\left(\frac{r}{4}\right)^2$

area of circle $C _3=\frac{\pi}{4}\left(\frac{ r }{8}\right)^2$ and so on
$\therefore $ shaded area $=\frac{\pi}{4}\left[\frac{r^2}{4}+\frac{r^2}{16}+\frac{r^2}{64}+\ldots \ldots . ..\right]=\frac{\pi r^2}{4} \cdot \frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{\pi r^2}{12}$
Hence ratio $=\frac{\pi r ^2-\frac{\pi r ^2}{12}}{\pi r ^2}=\frac{11}{12} $