8.2.1 Solid Geometry III, PT3 Practice

8.2.1 Solid Geometry III, PT3 Practice

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} Volume of a cone \\ = \frac{1}{3} π r^{2} h \\ = \frac{1}{3} \times \frac{22}{7} \times 7^{2} \times 62 \\ = 308 {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} Volume of a pyramid = \frac{1}{3} \times base area \times h \\ \frac{1}{3} \times base area \times h = 20 \\ \frac{1}{3} \times 5 \times 4 \times h = 20 \\ \frac{20}{3} \times h = 20 \\ h = 20 \times \frac{3}{20} \\ h = 3 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, of the cone.

[Use π = \frac{22}{7}]

Solution:

\begin{array}{l} Volume of a cylinder = π r^{2} h \\ \frac{22}{7} \times r^{2} \times 21 = 1650 \\ r^{2} = \frac{1650 \times 7}{22 \times 21} \\ = 25 \\ r = 5 c m \\ Thus the height of the cone \\ = 39 - 5 - 21 \\ = 13 c m \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} Height of the △ = \sqrt{10^{2} - 6^{2}} \\ = \sqrt{64} \\ = 8 c m \\ Volume of prism \\ = Area of cross section \times Length \\ = (\frac{1}{2} \times 12^{6} \times 8) \times 16 \\ = 768 c 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} π r^{2} h \\ h = \frac{3 V}{π r^{2}} \\ = \frac{3 \times 77}{\frac{22}{7} \times 3.5 \times 3.5} \\ = 6 cm \end{array}