Q.1 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 )

 

(OR)

Q.2 The internal and external diameters of a hollow hemispherical vessel are 16 𝑐𝑚 and 12 𝑐𝑚 respectively. If the cost of painting 1 cm² of the surface area is rupees 5. 00, find the total cost of painting the vessel all over. (Use π = 3. 14)

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
 

(OR)

We are given the internal and external diameters of a hollow hemispherical bowl, and we have to find how much it will cost to paint the bowl's surface area.
While painting, the inner area, the outer area and the upper thickness of the bowl will be painted.
So, we have to find the curved surface area of the internal and external parts of the hemisphere and the upper part.
Here, the diameter of the external part, 𝐷 = 16 𝑐𝑚
So, the radius of the external part, 𝑅 = 𝐷/2 
⇒ 𝑅 = 16/2  
⇒ 𝑅 = 8 𝑐𝑚
And the diameter of the internal part, 𝑑 = 12 𝑐𝑚
So, the radius of the external part, 𝑟 = 𝑑/2
⇒ 𝑟 = 12/2
⇒ 𝑟 = 6 𝑐𝑚
The surface area of the hemispherical bowl,

A = $2{\mathrm{\pi R}}^2 + 2{\mathrm{\pi r}}^2 + \mathrm{\pi }({\mathrm{R}}^2 - {\mathrm{r}}^2)$                                                              
⇒ 𝐴 = 2 × 3. 14 × 8 × 8 + 2 × 3. 14 × 6 × 6 + 3. 14(8 × 8 − 6 × 6)
⇒ 𝐴 = 401. 92 + 226. 08 + 3. 14(64 − 36)
⇒ 𝐴 = 628 + 3. 14 × 28
⇒ 𝐴 = 628 + 87. 92
⇒ 𝐴 = 715. 94 𝑐𝑚²
The cost of painting the surface area per 𝑐𝑚2 is 5. 00 rupees. So, cost of painting 715. 94 𝑐𝑚2 = 715. 94 × 5
= 3579. 6
Hence, the cost of painting the hemispherical bowl is rupees. 3579. 6