Step 1: Use arc length formula.
Length of arc is given by
\[
l=\frac{\theta}{360^\circ}\times 2\pi r.
\]
Step 2: Substitute values.
Here, $r=14$ cm and $\theta=30^\circ$, so
\[
l=\frac{30}{360}\times 2\pi \times 14
=\frac{1}{12}\times 28\pi
=\frac{28\pi}{12}
=\frac{7\pi}{3}.
\]
Step 3: Simplify and approximate.
\[
l=\frac{7\pi}{3}\ \text{cm}.
\]
Using $\pi\approx 3.14$,
\[
l=\frac{7\times 3.14}{3}=7.33\ \text{cm (approx)}.
\]
\boxed{\text{Arc length }=\frac{7\pi}{3}\ \text{cm } \approx 7.33\ \text{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: