Question

The diagram shows a section of a rocket firework. If this section can be completely filled with gunpowder what is the volume of gunpowder required?

Answer 1

$sin{60}^{\circ }=P/H=r/6=\sqrt{3}/2$

$\Rightarrow r=3\sqrt{3}cm$

In the cone $6^2=h^2+r^2$

$\Rightarrow 6^2=h^2+(3\sqrt{3}{)}^2=h^2+27$

$\Rightarrow h^2=36-27=9\therefore h=3cm$

Volume of gunpower = volume of cone + volume of cylinder

$=\frac{1}{3}\pi r^{2}h+\pi r^{2}h=\pi r^2(\frac{1}{3}h+h)$

$=\frac{22}{7}\times 27\times \frac{4h}{3}$

$=\frac{22}{7}\times 27\times \frac{4}{3}\times 3$

$=1782c{m}^3$