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.
The product of $\sqrt{2}$ and $(2-\sqrt{2})$ will be:
If a tangent $PQ$ at a point $P$ of a circle of radius $5 \,\text{cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \,\text{cm}$, then length of $PQ$ will be:
In the figure $DE \parallel BC$. If $AD = 3\,\text{cm}$, $DE = 4\,\text{cm}$ and $DB = 1.5\,\text{cm}$, then the measure of $BC$ will be: