Step 1: Formula for area of a sector
\[
\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2
\]
Step 2: Substitute values
Here, $\theta = 60^\circ$, $r = 6$ cm.
\[
\text{Area} = \frac{60}{360} \times \pi \times 6^2
\]
\[
= \frac{1}{6} \times \pi \times 36 = 6\pi\ \text{cm}^2
\]
Step 3: Verify carefully
Oops! Our result is $6\pi$ cm$^2$, not $12\pi$. Let's check options again:
Given options: $36\pi, 12\pi, 6\pi, 132$. Correct match is (C) $6\pi\ \text{cm}^2$.
\[
\boxed{\text{Area of sector} = 6\pi\ \text{cm}^2}
\]
Find the unknown frequency if 24 is the median of the following frequency distribution:
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency} & 5 & 25 & 25 & \text{$p$} & 7 \\ \hline \end{array}\]
Two concentric circles are of radii $8\ \text{cm}$ and $5\ \text{cm}$. Find the length of the chord of the larger circle which touches (is tangent to) the smaller circle.