Calculate the volume of the remaining wood correct to the nearest cubic centimetre. Use formulae: $\mathrm{V}=\pi \mathrm{r}^{2} \mathrm{~h}$ and $\mathrm{V}=\frac{1}{2} \times$ Area of base $\mathrm{x} \perp$ height.

Question

In the diagram below, $\Delta \mathrm{ECQ}$ is equilateral with sides $20 \mathrm{~cm}$. Vertices $\mathrm{E}, \mathrm{C}$ and $Q$ lie on the circumference of the circle at the top of a cylinder. M is the centre of the base of the cylinder. CEQM forms a triangular pyramid and is cut out of the wooden cylinder. The volume of the triangular pyramid is $3000 \mathrm{~cm}^{3}$.

Calculate the volume of the remaining wood correct to the nearest cubic centimetre. Use formulae:

$\mathrm{V}=\pi \mathrm{r}^{2} \mathrm{~h}$ and $\mathrm{V}=\frac{1}{2} \times$ Area of base $\mathrm{x} \perp$ height.

Answer 1

\begin{array}{l}
\mathrm{V}_{\text {cylinder }}=\pi \mathrm{r}^{2} \mathrm{~h} \\
\mathrm{~V}_{\text {pyramid }}=\frac{1}{3} \times \text { Area of base } \times \perp \text { height } \\
\therefore 3000=\frac{1}{3} \times\left(\frac{1}{2} \times 20 \times 20 \times \sin 60^{\circ}\right) \times \perp \mathrm{h} \\
\therefore \perp \mathrm{h}=51,961524 \ldots
\end{array}

\begin{array}{l}
\dfrac{\mathrm{r}}{\sin 30^{\circ}}=\dfrac{20}{\sin 120^{\circ}}  \\
\therefore \mathrm{r}=11,54700 \ldots  \\
\mathrm{V}_{\text {cylinder }}=\pi \mathrm{r}^{2} \mathrm{~h}=\pi \times 11,547 \ldots^{2} \times 51,9615 \ldots=21765,592 \ldots \\
\mathrm{V}_{\text {remaining }}=18766 \mathrm{~cm}^{3}
\end{array}