# 8.2.1 Solid Geometry III, PT3 Practice 1

8.2.1 Solid Geometry III, PT3 Practice 1

Question 1:
The diagram below shows a cone with diameter 14 cm and height 6 cm.

Find the volume of the cone, in cm3.

Solution:
$\begin{array}{l}\text{Volume of a cone}\\ =\frac{1}{3}\pi {r}^{2}h\\ =\frac{1}{\overline{)3}}×\frac{22}{\overline{)7}}×{7}^{\overline{)2}}×\overline{)6}2\\ =308{\text{cm}}^{3}\end{array}$

Question 2:
In the pyramid shown, the base is a rectangle.
If the volume is 20 cm2, find the height of the pyramid, in cm.

Solution:
$\begin{array}{l}\text{Volume of a pyramid}=\frac{1}{3}×\text{base area}×h\\ \frac{1}{3}×\text{base area}×h=20\\ \frac{1}{3}×5×4×h=20\\ \text{}\frac{20}{3}×h=20\\ \text{}h=\overline{)20}×\frac{3}{\overline{)20}}\\ \text{}h=3\text{cm}\end{array}$

Question 3:
Diagram below shows a composite solid consisting of a right circular cone, a right circular cylinder and a hemisphere.

The volume of the cylinder is 1650 cm3. Calculate the height, in cm, if the cone.
$\left[\text{Use}\pi =\frac{22}{7}\right]$

Solution:
$\begin{array}{l}\text{Volume of a cylinder}=\pi {r}^{2}h\\ \frac{22}{7}×{r}^{2}×21=1650\\ \text{}{r}^{2}=\frac{1650×7}{22×21}\\ \text{}=25\\ \text{}r=5cm\\ \text{Thus the height of the cone}\\ =39-5-21\\ =13cm\end{array}$

Question 4:
The cross section of the prism shown is an isosceles triangle.

The volume of the prism, in cm3, is

Solution:
$\begin{array}{l}\text{Height of the}△=\sqrt{{10}^{2}-{6}^{2}}\\ \text{}=\sqrt{64}\\ \text{}=8cm\\ \\ \text{Volume of prism}\\ =\text{Area of cross section}×\text{Length}\\ =\left(\frac{1}{\overline{)2}}×{\overline{)12}}^{6}×8\right)×16\\ =768c{m}^{3}\end{array}$

Question 5:
A right circular cone has a volume of 77 cm3 and a circular base of radius 3.5 cm. Calculate its height.

Solution:
$\begin{array}{l}V=\frac{1}{3}\pi {r}^{2}h\\ h=\frac{3V}{\pi {r}^{2}}\\ \text{}=\frac{3×77}{\frac{22}{7}×3.5×3.5}\\ \text{}=6cm\end{array}$