Question

In the given figure, APB and CQD are semicircles of diameter 7 cm each, while ARC an BSD are semicircles of diameter 14 cm each. Find the
(i) perimeter, (ii) area of the shaded region. [CBSE 2011]

Answer 1

(i) Perimeter of shaded region = Length of the arc APB + Length of the arc ARC + Length of the arc BSD + Length of the arc CQD
$$\begin{gathered} =\frac{1}{2}\times 2\mathrm{\pi }{r}_1+\frac{1}{2}\times 2\mathrm{\pi }{r}_2+\frac{1}{2}\times 2\mathrm{\pi }{r}_3+\frac{1}{2}\times 2\mathrm{\pi }{r}_4 \\ =\frac{1}{2}\times 2\mathrm{\pi }\left(\frac{7}{2}\right)+\frac{1}{2}\times 2\mathrm{\pi }\left(7\right)+\frac{1}{2}\times 2\mathrm{\pi }\left(7\right)+\frac{1}{2}\times 2\mathrm{\pi }\left(\frac{7}{2}\right) \\ =2\mathrm{\pi }\left(\frac{7}{2}\right)+2\mathrm{\pi }\left(7\right) \\ =2\mathrm{\pi }\left(\frac{7}{2}+7\right) \\ =2\times \frac{22}{7}\times \frac{21}{2} \\ =66\mathrm{cm} \end{gathered}$$

(ii) Area of shaded region = Area of the arc ARC + Area of the arc BSD − (Area of the arc APB + Area of the arc CQD)
$$\begin{gathered} =\frac{1}{2}\times \mathrm{\pi }{\left(r_1\right)}^2+\frac{1}{2}\times \mathrm{\pi }{\left(r_2\right)}^2-\left[\frac{1}{2}\times \mathrm{\pi }{\left({\mathrm{r}}_3\right)}^2+\frac{1}{2}\times \mathrm{\pi }{\left({\mathrm{r}}_4\right)}^2\right] \\ =\frac{1}{2}\times \mathrm{\pi }{\left(7\right)}^2+\frac{1}{2}\times \mathrm{\pi }{\left(7\right)}^2-\left[\frac{1}{2}\times \mathrm{\pi }{\left(\frac{7}{2}\right)}^2+\frac{1}{2}\times \mathrm{\pi }{\left(\frac{7}{2}\right)}^2\right] \\ =\mathrm{\pi }{\left(7\right)}^2-\mathrm{\pi }{\left(\frac{7}{2}\right)}^2 \\ =\mathrm{\pi }\left(49-\frac{49}{4}\right) \\ =\frac{22}{7}\times \frac{147}{4} \\ =115.5{\mathrm{cm}}^2 \end{gathered}$$