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.

Answer 1

We have, The edge of the cubical piece, a = 21 cm and

The radius of the hemisphere, $r=\frac{a}{2}=\frac{21}{2}$ cm

The surface area of the remaining piece = TSA of cube + CSA of the hemisphere - Area of circle

$=6a^2+2\pi r^2–\pi r^2=6a^2+\pi r^2=6\times 21\times 21+\frac{22}{7}\times \frac{21}{2}\times \frac{21}{2}=21\times 21(6+\frac{22}{7}\times 4)=21\times 21(6+\frac{11}{14})=21\times 21\left(\frac{84+11}{14}\right)=21\times 3\left(\frac{95}{2}\right)=2992.5c{m}^2$

Also

Volume of remaining piece = volume of cube - volume of the hemisphere

$=a^3-\frac{2}{3}\pi r^3=21\times 21\times 21–\frac{2}{3}\times \frac{22}{7}\times \left(\frac{21}{2}\right)\times \left(\frac{21}{2}\right)\times \left(\frac{21}{2}\right)=21\times 21\times 21\times (1–\frac{2}{3}\times \frac{22}{7}\times \frac{1}{2}\times \frac{1}{2}\times \frac{1}{2})=21\times 21\times 21\times (1–\frac{11}{42})=21\times 21\times 21\times \left(\frac{42–11}{42}\right)=21\times 21\times \left(\frac{31}{2}\right)=6835.5c{m}^3$