Question

In the adjoining figure 6.56, PR = 6 units and PQ = 8 units. Semicircles are drawn taking sides PR, RQ and PQ as diameter as shown in the figure. Find the area of the shaded portion. ($\mathrm{\pi }=3.14$)

Answer 1

Given:
RQ is the diameter.
Angle in the semicircle is 90$°$.
∴ $\angle \mathrm{RPQ}=90°$
ar($∆$PRQ) = $\frac{1}{2}$ × PQ × PR
=$\frac{1}{2}$ × 8 × 6
= 24 sq. units
Using the Pythagoras theorem, we have:

$$\begin{gathered} \mathrm{QR}=\sqrt{{\mathrm{PR}}^2+{\mathrm{PQ}}^2} \\ =\sqrt{6^2+8^2} \\ =\sqrt{36+64} \\ =\sqrt{100} \\ =10\mathrm{units} \end{gathered}$$

Area of the semicircle with diameter RQ = $\frac{\mathit{\pi }{\mathit{r}}^{\mathit{2}}}{\mathit{2}}$

= $\frac{3.14\times 10\times 10}{2}=157\mathrm{sq}.\mathrm{units}$
Area of the semicircle with diameter PR = $\frac{\mathit{\pi }{\mathit{r}}^{\mathit{2}}}{\mathit{2}}$

= $\frac{3.14\times 6\times 6}{2}=56.52\mathrm{sq}.\mathrm{units}$
Area of the semicircle with diameter PQ = $\frac{\mathit{\pi }{\mathit{r}}^{\mathit{2}}}{\mathit{2}}$

= $\frac{3.14\times 8\times 8}{2}=100.48\mathrm{sq}.\mathrm{units}$
Thus, we have:
Area of the shaded portion = Area of the semicircle with diameter PR + Area of the semicircle with diameter PQ $-$ Area of the semicircle with diameter RQ + ar($∆$PRQ)
= 56.52 + 100.48 $-$157 + 24
= 24 sq. units