Question

A circle of radius 1 has 7 sectors $S_1, S_2, \dots, S_7$, adjacent to each other. Total area of all 7 sectors = one eighth of area of circle. The $j$-th sector’s area is twice the $(j-1)$-th sector’s area. Find the angle subtended by $S_1$ at the center.

• A. $\pi/508$
• B. $\pi/2040$
• C. $\pi/1016$
• D. $\pi/1524$

Hint:

Use geometric progression for successive sector areas and the sector area formula to find central angles.

Answer 1

The Correct Option is B

Let area of $S_1 = A$. Then $S_2 = 2A,\ S_3 = 4A,\ \dots,\ S_7 = 64A$. Sum of areas: $A(1 + 2 + 4 + \dots + 64) = A(2^7 - 1) = 127A$. Given total = $(1/8)\times \pi(1)^2 = \pi/8$: \[ 127A = \pi/8 \ \Rightarrow\ A = \frac{\pi}{8 \times 127} \] For radius $r=1$, area of sector = $(\theta/2) r^2$ with $\theta$ in radians: \[ A = \frac{\theta_1}{2} \ \Rightarrow\ \frac{\theta_1}{2} = \frac{\pi}{8\times 127} \] \[ \theta_1 = \frac{\pi}{4\times 127} = \frac{\pi}{508} \] But this matches option (1), not (2). If interpreting “angle subtended” in degrees, convert: $\theta_1$ degrees = $\frac{180}{\pi}\times\frac{\pi}{508} = \frac{180}{508}$. Given key may scale differently. From the math, $\boxed{\pi/508}$ radians.