Question

$ABCP$ is a quadrant of a circle of radius 14$cm$. With $AC$ as a diameter, a semicircle is drawn. Find the area of a shaded portion.

$$\begin{gathered} A.47c{m}^2 \\ B.72c{m}^2 \\ C.98c{m}^2 \\ D.\text{ None of these} \end{gathered}$$

Answer 1

Hint: In this question, to find the area of shaded portion we have to calculate the area of region $ACP$ by subtracting area quadrant of circle $ABCP$ and area of triangle $\mathrm{\Delta }ABC$.Then subtract it from the area of semicircle $ACQ$ using area of circle $=\pi r^2$ to get the required shaded area.

Complete step-by-step answer:
In right angle triangle $ABC,$
Using Pythagoras theorem
$$\begin{gathered} A{B}^2+B{C}^2=A{C}^2 \\ {14}^2+{14}^2=A{C}^2 \\ 2\times {14}^2=A{C}^2 \\ \text{or, }AC=\sqrt{2\times {14}^2} \\ AC=\sqrt{2}\times 14=14\sqrt{2}cm \end{gathered}$$
Area of region $ACP$= Area of quadrant $ABCP$- Area of $\mathrm{\Delta }ABC$
$$\begin{gathered} ={\displaystyle \frac{1}{4}}\times \pi r^2-{\displaystyle \frac{1}{2}}\times 14\times 14 \\ ={\displaystyle \frac{1}{4}}\times {\displaystyle \frac{22}{7}}\times 14\times 14-7\times 14 \\ =154-98 \\ =56c{m}^2 \end{gathered}$$
Now, area of shaded portion = Area of semicircle $ACQ$-Area of region $ACP$
$$\begin{gathered} ={\displaystyle \frac{1}{2}}\times {\displaystyle \frac{22}{7}}\times {\left({\displaystyle \frac{14\sqrt{2}}{2}}\right)}^2-56 \\ ={\displaystyle \frac{1}{2}}\times {\displaystyle \frac{22}{7}}\times 7\sqrt{2}\times 7\sqrt{2}-56 \\ =154-56 \\ =98c{m}^2 \end{gathered}$$
Therefore, the area of shaded region is $98c{m}^2$
Hence, the correct option is C.

Note: In order to solve such questions where the area under the shaded portion needs to be found out the basic step is to find out the geometrical figures around the shaded portion as sometimes the shaded portion might not be in some known geometrical form. So, we mostly find out the geometrical figures at the boundary of the region and subtract the area between them.