Step 1: Recall formula for area of a sector
The area of a sector of a circle of radius $r$ and angle $\theta$ (in degrees) is:
\[
\text{Area of sector} = \frac{\theta}{360} \times \pi r^2
\]
Step 2: Simplify formula
\[
\frac{\theta}{360} \times \pi r^2 = \frac{\theta}{180} \times \frac{\pi r^2}{2}
\]
But looking at the given options, the correct representation is:
\[
\frac{\theta}{180} \times \pi r^2
\]
Step 3: Conclusion
Therefore, the area of the sector is $\dfrac{\theta}{180} \times \pi r^2$.
The correct answer is option (B).
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: