The angle (in degrees) made by a sector having area one-sixth of the area of a semicircle is
15°
30°
45°
60°
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CAPF – 2024
The area of the sector is given as one-sixth of the area of the semicircle.
Area of sector $= \frac{1}{6} \times \left(\frac{1}{2} \pi r^2\right) = \frac{1}{12} \pi r^2$.
The formula for the area of a sector with angle $\theta$ (in degrees) and radius $r$ is $\frac{\theta}{360°} \pi r^2$.
$\frac{\theta}{360°} \pi r^2 = \frac{1}{12} \pi r^2$.
Assuming $r > 0$, we can cancel $\pi r^2$ from both sides:
$\frac{\theta}{360°} = \frac{1}{12}$.
Now, solve for $\theta$:
$\theta = \frac{360°}{12} = 30°$.
The angle made by the sector is 30 degrees.