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A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is $2$ cm and the diameter of the base is $4$ cm. Determine the volume of the toy. Also, find the surface area of the toy. (Take $\pi = 3\cdot14$)
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Sol. Height of the cone $(h) = 2$ cm
Radius of the hemisphere $=$ radius of the base of the cone $= r = 2$ cm
Volume of the toy $$\begin{aligned}& = \frac{1}{3} \times 3.14 \times (2)^2 \times 2 + \frac{2}{3} \times 3.14 \times (2)^3 \\ & = 25.12 \text{ cm}^3\end{aligned}$$ Slant height of the cone $(l) = \sqrt{2^2 + 2^2} = 2\sqrt{2}$ cm
Surface area of the toy $$\begin{aligned}& = 3.14 \times 2 \times 2\sqrt{2} + 2 \times 3.14 \times (2)^2 \\ & = 12.56 (2 + \sqrt{2}) \text{ cm}^2\end{aligned}$$
Radius of the hemisphere $=$ radius of the base of the cone $= r = 2$ cm
Volume of the toy $$\begin{aligned}& = \frac{1}{3} \times 3.14 \times (2)^2 \times 2 + \frac{2}{3} \times 3.14 \times (2)^3 \\ & = 25.12 \text{ cm}^3\end{aligned}$$ Slant height of the cone $(l) = \sqrt{2^2 + 2^2} = 2\sqrt{2}$ cm
Surface area of the toy $$\begin{aligned}& = 3.14 \times 2 \times 2\sqrt{2} + 2 \times 3.14 \times (2)^2 \\ & = 12.56 (2 + \sqrt{2}) \text{ cm}^2\end{aligned}$$