1.	$\begin{align}\text{Amount of spheres that can be made =n }\\[5pt] \end{align}$ $\begin{align}\text{Volume of the cuboid -(1)} \\[5pt] & 12\times 12\times 22\\[5pt] \end{align}$ $\begin{align}\text{Volume of the sphere} -(2) {\frac{4} {3}\pi r^3 }\\[5pt] &=n\times (2)=(1)\\[5pt] &=n\times\frac {4}{3}\times\frac {22}{7}\times 3^3=12\times12\times12\\[5pt] & n=\frac{12\times12\times22\times3\times7}{4\times22\times 3\times3\times3}\\[5pt] &=\underline{\underline{28}}\\[5pt] \end{align}$ $\begin{align}\text{Amount of spheres that can be made =28 }\\[5pt] \end{align}$
2.	$\begin{align}\text{Volume of the sphere} -(1) {\frac{4} {3}\pi r^3 }\\[5pt] \end{align}$ $\begin{align}\text{Volume of the Cone} -(2) {\frac{1} {3}\pi r^2h }\\[5pt] & (1)=(2)\\[5pt] &{\frac{1} {3}\pi r^2h }= {\frac{4} {3}\pi r^3 }\\[5pt] & h=\frac{4r^3}{r^2}\\[5pt] & h=4r\\[5pt] & h=4\times3.5\\[5pt] &=\underline{\underline{14cm}}\\[5pt] \end{align}$ $\begin{align}\text{Height of cone =14cm }\\[5pt] \end{align}$
3.	$\begin{align}\text{Circumference} -(1) {\frac{216} {360}\times2\pi r }={\frac{3} {5}\times2\pi r }\\[5pt] \end{align}$ $\begin{align}\text{Volume of the edge of Cone} -(2) {2\pi r_1}\\[5pt] & (1)=(2)\\[5pt] &{2\pi r_1}= {\frac{3} {5}\times2\pi r }\\[5pt] &{r_1}= {\frac{3} {5}r }\\[5pt] \end{align}$ $\begin{align}\text{Container height=h}\\[5pt] &h^2=r^2-r_1^2\\[5pt] &=r^2-(\frac{3}{5}r)^2\\[5pt] &=r^2-\frac{9}{25}r^2\\[5pt] &=\frac{25-9}{25}r^2\\[5pt] &=\frac{16}{25}r^2\\[5pt] &=\underline{\underline{h=\frac{4}{5}r}}\\[5pt] \end{align}$ $\begin{align}\text{n Ice cubes volume=Container volume}\\[5pt] &n\times{\frac{4} {3}\pi a^3 }= {\frac{1} {3}\pi r_1^2h }\\[5pt] &n\times4a^3= r_1^2h\\[5pt] &n\times4a^3= (\frac{3}{5}r)^2\times\frac{4}{5}r\\[5pt] &n\times4a^3= \frac{9}{25}r^2\times\frac{4}{5}r\\[5pt] &na^3= \frac{9r^3}{125}\\[5pt] &\underline{\underline{125na^3=9r^3}}\\[5pt] \end{align}$

By studying this lesson you will be able to

Compute the volume of a square based right pyramid, right circular cone and a sphere.