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 $C_1,C_2,C_3,...$ be circles with diameter $P_0{P}_1,P_1{P}_2,P_2{P}_3,...$, respectively. Suppose the circles $C_1,C_2,C_3,...$ 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

The correct option is D 11 : 12

Let $P_{o}B=R$

Diameter $P_o{P}_1=\frac{R}{2}$
Diameter $P_1{P}_2=\frac{R}{4}$

Diameter $P_2{P}_3=\frac{R}{8}$

Area of shaded $=\pi [{\left(\frac{R}{4}\right)}^2+{\left(\frac{R}{8}\right)}^2+{\left(\frac{R}{16}\right)}^2+....]$

Area of unshaded $=\pi R^2-\pi R^2[\frac{1}{4^2}+\frac{1}{8^2}+\frac{1}{{16}^2}+....]$

$A_1=\pi R^2[1-\frac{1/4^2}{(1-1/4)}]A_1=\pi R^2[1-\frac{1}{12}]=\frac{11}{12}\pi R^2$

Area of circle $=\pi R^2=A_2$

$\frac{A_1}{A_2}=\frac{11}{12}\frac{\pi R^2}{\pi R^2}=\frac{11}{12}$