6.5.4 Volume of Three Dimensional Shapes, PT3 Practice

Question 11:
Diagram shows a composite formed by joining a quarter cylinder and a right prism at the rectangular plane EFGH. The trapezium ABGF and the quarter circle FGK are the uniform cross sections of the solid.


Using   $\pi =\frac{22}{7}$ , calculate the volume, in  cm³, of the composite solid.


Solution:
$\begin{array}{l}\text{Volume of }FGK\\ =\frac{1}{4}\left(\pi r^{2}h\right)\\ =\frac{1}{4}\left(\frac{22}{7}\times 5^2\times 7\right)\\ =\frac{1}{4}\left(\frac{22}{7}\times 25\times 7\right)\\ =137\frac{1}{2}{\text{ cm}}^3\\ \\ \text{Volume of }ABGF\\ =\left[\frac{1}{2}\left(10+5\right)5\right]\times 7\\ =\frac{5}{2}\left(15\right)\times 7\\ =\frac{525}{2}\\ =262\frac{1}{2}{\text{ cm}}^3\\ \\ \text{Total volume of solid}\\ =137\frac{1}{2}{\text{ cm}}^3+262\frac{1}{2}{\text{ cm}}^3\\ =400{\text{ cm}}^3\end{array}$

Question 12:
Diagram shows a cylindrical tank in a residential area which have 125 houses. Each house received equal volume of water.


The diameter of the cylindrical tank is 4 m. It is given that each house has a cuboid tank with a base area of 0.8 m².
Using  $\pi =\frac{22}{7}$ , calculate the height of the water level, in m, of each tank in the house.


Solution:
$\begin{array}{l}\text{Volume of water in cylindrical tank}\\ =\pi r^{2}h\\ =\frac{22}{7}\times 2^2\times 3.5\\ =44{\text{ m}}^3\\ \\ \text{Volume of water in each cuboid}\\ \text{tank}=\frac{44}{125}\\ 0.8\times h=\frac{44}{125}\\ h=0.44\text{ m}\\ \\ \therefore \text{Height of water level in each }\\ \text{cuboid tank}=0.44\text{ m}.\end{array}$