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.
(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
$=\frac{1}{2} \times 2 \pi r_{1}+\frac{1}{2} \times 2 \pi r_{2}+\frac{1}{2} \times 2 \pi r_{3}+\frac{1}{2} \times 2 \pi r_{4}$
$=\frac{1}{2} \times 2 \pi\left(\frac{7}{2}\right)+\frac{1}{2} \times 2 \pi(7)+\frac{1}{2} \times 2 \pi(7)+\frac{1}{2} \times 2 \pi\left(\frac{7}{2}\right)$
$=2 \pi\left(\frac{7}{2}\right)+2 \pi(7)$
$=2 \pi\left(\frac{7}{2}+7\right)$
$=2 \times \frac{22}{7} \times \frac{21}{2}$
$=66 \mathrm{~cm}$
(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)
$=\frac{1}{2} \times \pi\left(r_{1}\right)^{2}+\frac{1}{2} \times \pi\left(r_{2}\right)^{2}-\left[\frac{1}{2} \times \pi\left(\mathrm{r}_{3}\right)^{2}+\frac{1}{2} \times \pi\left(\mathrm{r}_{4}\right)^{2}\right]$
$=\frac{1}{2} \times \pi(7)^{2}+\frac{1}{2} \times \pi(7)^{2}-\left[\frac{1}{2} \times \pi\left(\frac{7}{2}\right)^{2}+\frac{1}{2} \times \pi\left(\frac{7}{2}\right)^{2}\right]$
$=\pi(7)^{2}-\pi\left(\frac{7}{2}\right)^{2}$
$=\pi\left(49-\frac{49}{4}\right)$
$=\frac{22}{7} \times \frac{147}{4}$
$=115.5 \mathrm{~cm}^{2}$