A right circular cone of radius 3 𝑐𝑚, has a curved surface area of 47.1cm²

Find the volume of the cone. (Use π = 3. 14)

                                                (OR)

In the given figure, ∆𝑃𝑄𝑅 is an equilateral triangle of side 8 𝑐𝑚 and 𝐷, 𝐸, 𝐹 are centres of circular arcs, each radius 4 𝑐𝑚. Find the area of shaded region. (Use π = 3. 14 and  $\sqrt{3}$     = 1. 732)

 

We need to find the volume of the cone of radius 3 𝑐𝑚 provided that its curved surface area of 47.1 cm².
It is known that the CSA of a cone is given as 𝐶𝑆𝐴 = π𝑟𝑙, where 𝑟 and 𝑙 are the radius and slant height respectively.
On substituting the values, we get the slant height as
⇒ 47. 1 = π(3)𝑙
⇒ 47. 1 = 9. 42𝑙
⇒ 𝑙 = 5 𝑐𝑚
Now, the height of the cone can be given as
ℎ = $\sqrt{5^2-3^2}$
⇒ ℎ = $\sqrt{16}$
⇒ ℎ = 4 𝑐𝑚
Also, the volume of the cone is given by 𝑉 = 1/3π𝑟² h . On substituting the values, we get
3 π𝑟 ℎ
⇒ 𝑉 = 1/3 π x 3 x 4
⇒ 𝑉 = 37.68 cm³
Hence, the volume of the cone is 37.68 cm³

                                                                           ^(OR)

We need to find the area of shaded region provided that ∆𝑃𝑄𝑅 is an equilateral triangle of side 8 𝑐𝑚 and 𝐷, 𝐸, 𝐹 are centres of circular arcs, each radius 4 𝑐𝑚.

It can be observed that the arcs form the sector in the triangle and hence, the area of shaded region is

𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑎𝑑𝑒𝑑 𝑟𝑒𝑔𝑖𝑜𝑛 = 𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑃𝑄𝑅 − 3(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑐𝑡𝑜𝑟)

⇒ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑎𝑑𝑒𝑑 𝑟𝑒𝑔𝑖𝑜𝑛 = $\frac{\sqrt{3}}{4}\times sid{e}^2 - 3 \times \left(\frac{\theta }{360°}\times {\mathrm{\pi r}}^2\right)$

⇒ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠ℎ𝑎𝑑𝑒𝑑 𝑟𝑒𝑔𝑖𝑜𝑛 = $$\begin{gathered} \frac{\sqrt{3}}{4}\times 8^2 - 3 \times \left(\frac{60°}{360°}\times \pi 4^2\right) \\ 16\sqrt{3} - 8\mathrm{\pi } \\ 16 \times 1.732 - 8 \mathrm{x} 3.14 \\ 27.712 - 25.12 \\ 2.592 {\mathrm{cm}}^2 \end{gathered}$$

Hence, the area of the shaded region is  2.592 cm².