Step 1: Formula for length of an arc.
\[ \text{Length of arc} = \dfrac{\theta}{360^\circ} \times 2\pi r \] Step 2: Substitute the given values.
\[ \theta = 30^\circ, \quad r = 6 \text{ cm} \] \[ \text{Arc length} = \dfrac{30}{360} \times 2\pi \times 6 \] Step 3: Simplify.
\[ = \dfrac{1}{12} \times 12\pi = \dfrac{\pi}{1} = \pi \] Wait — we simplify carefully: \[ \dfrac{30}{360} = \dfrac{1}{12}, \quad 2\pi \times 6 = 12\pi \] \[ \text{Arc length} = \dfrac{1}{12} \times 12\pi = \pi \text{ cm} \] Step 4: Correct the simplification (angle check).
Oops — on rechecking, angle \(30^\circ\) gives: \[ \text{Arc length} = \dfrac{30}{360} \times 2\pi \times 6 = \dfrac{1}{12} \times 12\pi = \pi \text{ cm} \] So the correct answer is actually (D) π cm.
Step 5: Conclusion.
The measure of the arc = $\pi$ cm.

In the following figure chord MN and chord RS intersect at point D. If RD = 15, DS = 4, MD = 8, find DN by completing the following activity: 
Activity :
\(\therefore\) MD \(\times\) DN = \(\boxed{\phantom{SD}}\) \(\times\) DS \(\dots\) (Theorem of internal division of chords)
\(\therefore\) \(\boxed{\phantom{8}}\) \(\times\) DN = 15 \(\times\) 4
\(\therefore\) DN = \(\frac{\boxed{\phantom{60}}}{8}\)
\(\therefore\) DN = \(\boxed{\phantom{7.5}}\)
Find the unknown frequency if 24 is the median of the following frequency distribution:
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency} & 5 & 25 & 25 & \text{$p$} & 7 \\ \hline \end{array}\]