Question

Let $C$ be a circle with centre $P_0$ and $AB$ 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 $C_1,C_2,C_3,\dots$. be circles with diameters $P_0{P}_1,P_1{P}_2,P_2{P}_3\dots ..$. respectively. Suppose the circles $C_1,C_2,C_3,\dots ..$. 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 $C_1=\frac{\pi }{4}{\left(\frac{r}{2}\right)}^2($ area of circle $=\frac{\pi d^2}{4})$
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 $=\frac{\pi }{4}\left[\frac{r^2}{4}+\frac{r^2}{16}+\frac{r^2}{64}+\dots \dots ...\right]=\frac{\pi r^2}{4}\cdot \frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{\pi r^2}{12}$
Hence ratio $=\frac{\pi r^2-\frac{\pi r^2}{12}}{\pi r^2}=\frac{11}{12}$