Question

A heap of wheat is in the form of a cone, whose diameter is 10.5 m and height 7 m. Find the volume of wheat in the heap. The heap is to be covered by canvas to protect it from rain, find the area of the canvas required.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{Base}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{conical}\mathrm{heap},r=\frac{10.5}{2}=\frac{21}{4}\mathrm{m}\mathrm{and} \\ \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{conical}\mathrm{heap},h=7\mathrm{m} \\ \mathrm{Now}, \\ \mathrm{Volume}\mathrm{of}\mathrm{wheat}\mathrm{in}\mathrm{the}\mathrm{heap}=\frac{1}{3}\mathrm{\pi }{r}^{2}h \\ =\frac{1}{3}\times \frac{22}{7}\times \frac{21}{4}\times \frac{21}{4}\times 7 \\ =\frac{1617}{8}{\mathrm{m}}^3 \\ =202\frac{1}{8}{\mathrm{m}}^3 \\ \mathrm{Also}, \\ \mathrm{the}\mathrm{slant}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{heap},l=\sqrt{h^2+r^2} \\ =\sqrt{{\left(\frac{21}{4}\right)}^2+7^2} \\ =\sqrt{\frac{441}{16}+49} \\ =\sqrt{\frac{1225}{16}} \\ =\frac{35}{4}\mathrm{m} \\ \mathrm{So},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{canvas}\mathrm{required}\mathrm{to}\mathrm{protect}\mathrm{the}\mathrm{heap}\mathrm{of}\mathrm{wheat}\mathrm{from}\mathrm{rain}=\mathrm{CSA}\mathrm{of}\mathrm{the}\mathrm{heap} \\ =\mathrm{\pi }rl \\ =\frac{22}{7}\times \frac{21}{4}\times \frac{35}{4} \\ =\frac{1155}{8}{\mathrm{m}}^2 \\ =144\frac{3}{8}{\mathrm{m}}^2 \end{gathered}$$