Question

From a cubical piece of wood of side 21 cm, a hemisphere is carved out in such a way that the diameter of the hemisphere is equal to the side of the cubical piece. Find the surface area and volume of the remaining piece. [CBSE 2014]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{the}\mathrm{edge}\mathrm{of}\mathrm{the}\mathrm{cubical}\mathrm{piece},a=21\mathrm{cm}\mathrm{and} \\ \mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{hemisphere},r=\frac{a}{2}=\frac{21}{2}\mathrm{cm} \\ \mathrm{The}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{piece}=\mathrm{TSA}\mathrm{of}\mathrm{cube}+\mathrm{CSA}\mathrm{of}\mathrm{hemisphere}-\mathrm{Area}\mathrm{of}\mathrm{circle} \\ =6a^2+2\mathrm{\pi }{r}^2-\mathrm{\pi }{r}^2 \\ =6a^2+\mathrm{\pi }{r}^2 \\ =6\times 21\times 21+\frac{22}{7}\times \frac{21}{2}\times \frac{21}{2} \\ =21\times 21\left(6+\frac{22}{7\times 4}\right) \\ =21\times 21\left(6+\frac{11}{14}\right) \\ =21\times 21\left(\frac{84+11}{14}\right) \\ =21\times 3\left(\frac{95}{2}\right) \\ =2992.5{\mathrm{cm}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Also}, \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{piece}=\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cube}-\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{hemisphere} \\ =a^3-\frac{2}{3}\mathrm{\pi }{r}^3 \\ =21\times 21\times 21-\frac{2}{3}\times \frac{22}{7}\times \left(\frac{21}{2}\right)\times \left(\frac{21}{2}\right)\times \left(\frac{21}{2}\right) \\ =21\times 21\times 21\times \left(1-\frac{2}{3}\times \frac{22}{7}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}\right) \\ =21\times 21\times 21\times \left(\frac{1}{1}-\frac{11}{42}\right) \\ =21\times 21\times 21\times \left(\frac{42-11}{42}\right) \\ =21\times 21\times \left(\frac{31}{2}\right) \\ =6835.5{\mathrm{cm}}^3 \end{gathered}$$