Find the area of the shaded region in figure, $\overline{APD}$, $\overline{AQB}$, $\overline{BRC}$ and $\overline{CSD}$, are
semi-circles of diameter 14 𝑐𝑚, 3. 5 𝑐𝑚, 7 𝑐𝑚 and 3. 5 𝑐𝑚 respectively.(Use π = 22/7 )

We are given the diameters of the semicircles, and we have to find the area of the shaded region.
So, the radius of $\overline{APD}$,$\overline{AQB}$ , $\overline{BRC}$ and $\overline{CSD}$ is 7 𝑐𝑚, 1. 75 𝑐𝑚, 3. 5 𝑐𝑚 and 1. 75 𝑐𝑚, respectively.
The Area of the shaded region can be found by subtracting the areas of the two semicircles of radius 1. 75 𝑐𝑚 from the Area of the semicircle of radius 7 𝑐𝑚 and then adding this to the Area of the semicircle of radius 3. 5 𝑐𝑚.
This can be written as
Area of the region = Area of the semicircle of radius 7 𝑐𝑚 − 2 × Area of the semicircles of radius 1. 75 𝑐𝑚 + Area of the semicircle of radius 3. 5 𝑐𝑚.
Now, the area of the semicircle of radius 7 𝑐𝑚 = $\frac{\mathrm{\pi }}{2}\times 7^2$

= $$\begin{gathered} \frac{22}{7\times 2}\times 7 \times 7 \\ 77 c{m}^2 \end{gathered}$$

The area of the semicircle of radius 1. 75 𝑐𝑚 = $\frac{\mathrm{\pi }}{2}\times 1.{75}^2$

= $$\begin{gathered} \frac{22}{7\times 2}\times 1.75 \times 1.75 \\ 4.81 c{m}^2 \end{gathered}$$

The area of the semicircle of radius 3. 5 𝑐𝑚 = $\frac{\mathrm{\pi }}{2}\times 3.5^2$

= $$\begin{gathered} \frac{22}{7\times 2}\times 3.5 \times 3.5 \\ 19.25 c{m}^2 \end{gathered}$$

So, area of shaded region = 77 − 2 × 4. 81 + 19. 25
⇒ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑎𝑑𝑒𝑑 𝑟𝑒𝑔𝑖𝑜𝑛 = 57. 75 − 9. 62
⇒ 𝑇ℎ𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠ℎ𝑎𝑑𝑒𝑑 𝑟𝑒𝑔𝑖𝑜𝑛 = 48. 13 𝑐𝑚²
Hence, the area of the shaded region is 48. 13 𝑐𝑚²