Comprehensive Guide to Volume, Capacity, and Three-Dimensional Solids

Units of Volume and Capacity Conversion

• Fundamental equivalence relationships between metric volume and capacity:

  • $1\,\text{cm}^3 = 1\,\text{mL}$

  • $1000\,\text{cm}^3 = 1\,\text{L}$

  • $1\,\text{m}^3 = 1000\,\text{L} = 1\,\text{kL}$

• Metric capacity conversion hierarchy and rules:

  • Megalitres ($\text{ML}$) to kilolitres ($\text{kL}$): multiply by $1000$

  • Kilolitres ($\text{kL}$) to litres ($\text{L}$): multiply by $1000$

  • Litres ($\text{L}$) to millilitres ($\text{mL}$): multiply by $1000$

  • Millilitres ($\text{mL}$) to litres ($\text{L}$): divide by $1000$

  • Litres ($\text{L}$) to kilolitres ($\text{kL}$): divide by $1000$

  • Kilolitres ($\text{kL}$) to megalitres ($\text{ML}$): divide by $1000$

• Worked Example 14: Converting units of volume and capacity:

  • Part a: Convert $13.2\,\text{L}$ into $\text{cm}^3$

  • Multiply litres by $1000$ to convert to millilitres:       $13.2 \times 1000 = 13\,200\,\text{mL}$

  • Apply the direct equivalent $1\,\text{mL} = 1\,\text{cm}^3$:       $13\,200\,\text{mL} = 13\,200\,\text{cm}^3$

  • Part b: Convert $3.1\,\text{m}^3$ into litres:

  • Multiply by $1000$ based on $1\,\text{m}^3 = 1000\,\text{L}$:       $3.1 \times 1000 = 3100\,\text{L}$

  • Part c: Convert $0.13\,\text{cm}^3$ into $\text{mm}^3$:

  • Multiply by $1000$ (since $1\,\text{cm}^3 = 10^3\,\text{mm}^3 = 1000\,\text{mm}^3$):       $0.13 \times 1000 = 130\,\text{mm}^3$

  • Part d: Convert $3.8\,\text{kL}$ into $\text{m}^3$:

  • Apply the direct equivalence $1\,\text{kL} = 1\,\text{m}^3$:       $3.8\,\text{kL} = 3.8\,\text{m}^3$

• Fluency Practice Problems:

  • Exercise 1: Convert into $\text{mL}$:

  • Part a: Convert $325\,\text{cm}^3$ into $\text{mL}$:       $325\,\text{cm}^3 = 325\,\text{mL}$

  • Part b: Convert $2.6\,\text{m}^3$ into $\text{mL}$:       $2.6\,\text{m}^3 = 2.6 \times 100^3\,\text{mL}$       $2.6 \times 1\,000\,000\,\text{mL} = 2\,600\,000\,\text{mL}$

  • Part c: Convert $5.1\,\text{L}$ into $\text{mL}$:       $5.1\,\text{L} = 5.1 \times 1000\,\text{mL} = 5100\,\text{mL}$

  • Part d: Convert $0.63\,\text{kL}$ into $\text{mL}$:       $0.63\,\text{kL} = 0.63 \times 1000\,\text{L} = 630\,\text{L}$       $630\,\text{L} = 630 \times 1000\,\text{mL} = 630\,000\,\text{mL}$

  • Exercise 2: Convert into $\text{cm}^3$:

  • Part a: Convert $5.8\,\text{mL}$ into $\text{cm}^3$:       $5.8\,\text{mL} = 5.8\,\text{cm}^3$

  • Part b: Convert $6.1\,\text{L}$ into $\text{cm}^3$:       $6.1\,\text{L} = 6.1 \times 1000\,\text{mL} = 6100\,\text{mL}$       $6100\,\text{mL} = 6100\,\text{cm}^3$

  • Part c: Convert $3.2\,\text{m}^3$ into $\text{cm}^3$:       $3.2\,\text{m}^3 = 3.2 \times 100^3\,\text{cm}^3$       $3.2 \times 1\,000\,000\,\text{cm}^3 = 3\,200\,000\,\text{cm}^3$

  • Part d: Convert $59.3\,\text{mm}^3$ into $\text{cm}^3$:       $59.3\,\text{mm}^3 = 59.3 \div 10^3\,\text{cm}^3$       $59.3 \div 1000\,\text{cm}^3 = 0.0593\,\text{cm}^3$

Geometric Definition and Properties of Prisms

• Definition of a prism:

  • A solid object with identical parallel ends, flat faces, and the same uniform cross-section along its length.

  • Prisms are named strictly according to the shape of their cross-section:

  • Triangular prism (triangular cross-section)

  • Rectangular prism (rectangular cross-section)

  • Hexagonal prism (hexagonal cross-section)

• Concrete cross-sectional visualization (the loaf of bread analogy):

  • Slicing through a loaf of bread parallel to its front face reveals slices with the exact same cross-sectional area and shape throughout the entire length.

  • The face being sliced represents the cross-section (base area $A$), and the distance along the slices represents the length or perpendicular height ($h$).

• Distinction between prisms and non-prisms:

  • Solids categorized as prisms:

  • Triangular prisms

  • Rectangular prisms

  • Pentagonal prisms

  • Hexagonal prisms

  • Octagonal prisms

  • Any polyhedron possessing flat faces and a uniform polygonal cross-section throughout

  • Solids categorized as not prisms:

  • Cylinders: possess curved surfaces rather than flat polygonal faces

  • Pyramids: cross-section scales continuously down to an apex rather than remaining uniform

  • Truncated pyramids (frustums): cross-section varies in size along its height

  • Cones: curved surfaces and non-uniform cross-sections tapering to an apex

Volume Formulas for Prisms, Cubes, and Cuboids

• General volume formula for any prism:   $V_{\text{prism}} = A \times h$

  • Where:

  • $A$ is the area of the base or cross-section of the prism

  • $h$ is the perpendicular height of the prism

  • Crucial geometric conventions:

  • The cross-section ($A$) is often referred to as the "base", even if the prism is oriented such that the cross-section is not physically at the bottom.

  • The height ($h$) must always be measured perpendicular to the plane of the base/cross-section, regardless of whether the solid is standing upright or lying on its side.

• Volume of a Cube:

  • Base area equals $l^2$

  • Height equals $l$

  • Formula derivation:     $\text{Volume} = \text{base area} \times \text{height}$     $\text{Volume} = l^2 \times l = l^3$

• Volume of a Cuboid (Rectangular Prism):

  • Base area equals $l \times w$

  • Height equals $h$

  • Formula derivation:     $\text{Volume} = \text{base area} \times \text{height}$     $\text{Volume} = lw \times h = lwh$

• Worked Example 15: Calculating the volume of a hexagonal prism:

  • Given: cross-sectional area $A = 40\,\text{cm}^2$, perpendicular length/height $h = 8\,\text{cm}$

  • Calculation steps:     $V = A \times h$     $V = 40 \times 8$     $V = 320\,\text{cm}^3$

• Worked Example 16: Calculating the volume of a triangular prism:

  • Given dimensions of the triangular base:

  • Base edge of triangle $b = 4\,\text{cm}$

  • Perpendicular height of triangle $h_{\text{tri}} = 3\,\text{cm}$

  • Length/height of prism $h = 8\,\text{cm}$

  • Step 1: Calculate the area of the triangular cross-section ($A$):     $A = \frac{1}{2}bh$     $A = \frac{1}{2} \times 4 \times 3$     $A = 6\,\text{cm}^2$

  • Step 2: Calculate the volume of the prism ($V$):     $V = A \times h$     $V = 6 \times 8$     $V = 48\,\text{cm}^3$

Volume and Capacity of Cylinders

• Geometric structure of a cylinder:

  • A solid with a circular base of radius $r$ and a uniform cross-section along its perpendicular height $h$.

• Derivation of cylinder volume:   $\text{Volume} = \text{base area} \times \text{height}$   $\text{Volume} = \text{area of circle} \times \text{height}$   $\text{Volume} = \pi r^2 \times h$   $\text{Volume} = \pi r^2 h$

• Worked Example 17: Calculating the capacity of a cylinder:

  • Problem statement: Calculate the capacity, in litres, of a cylindrical water tank that has a diameter of $5.4\,\text{m}$ and a height of $3\,\text{m}$. Give your answer correct to the nearest litre.

  • Step 1: Determine the radius ($r$):     $r = \frac{5.4}{2} = 2.7\,\text{m}$

  • Step 2: Compute volume in cubic metres ($\text{m}^3$):     $V = \pi r^2 h$     $V = \pi \times (2.7)^2 \times 3$     $V \approx 68.706\,631\,\text{m}^3$

  • Step 3: Convert cubic metres to litres ($1\,\text{m}^3 = 1000\,\text{L}$):     $V = 68.706\,631 \times 1000$     $V = 68\,706.631\,\text{L}$

  • Step 4: Round to the nearest whole litre:     $V \approx 68\,707\,\text{L}$

Volume of Compound Solids

• Composite solid methodology:

  • Dissect the complex shape into non-overlapping fundamental geometric components.

  • Calculate the individual volumes ($V_1, V_2, \dots$) using the appropriate shape formulas.

  • Sum the individual component volumes to obtain the total solid volume ($V = V_1 + V_2 + \dots$).

• Worked Example 18: Calculating the volume of a solid:

  • The solid consists of a rectangular prism (square base) with a vertical cylinder mounted centered on top.

  • Component 1: Rectangular base prism ($V_1$):

  • Base edge lengths $l = 16\,\text{cm}$, $w = 16\,\text{cm}$

  • Height $h_1 = 6\,\text{cm}$

  • Volume formula and execution:       $V_1 = A h_1$       $V_1 = 16^2 \times 6$       $V_1 = 256 \times 6 = 1536\,\text{cm}^3$

  • Component 2: Mounted cylinder ($V_2$):

  • Cylinder radius $r = 5\,\text{cm}$

  • Cylinder height $h_2 = 12\,\text{cm}$

  • Volume formula and execution:       $V_2 = \pi r^2 h_2$       $V_2 = \pi \times 5^2 \times 12$       $V_2 = \pi \times 25 \times 12 = 300\pi$       $V_2 \approx 942.478\,\text{cm}^3$

  • Total combined volume ($V$):     $V = V_1 + V_2$     $V = 1536 + 942.478$     $V = 2478.478\,\text{cm}^3$     $V \approx 2478\,\text{cm}^3$

Extension: Volume of Spheres, Pyramids, and Cones

• Scope notice:

  • This material is classified as advanced extension content (non-assessable within standard Year 9 syllabus parameters).

• Volume of a Sphere:

  • Geometric definition: A perfectly symmetrical three-dimensional solid consisting of all points equidistant from a central point at radius $r$.

  • Volume formula:     $V = \frac{4}{3}\pi r^3$

• Worked Example 10: Calculating the volume of a sphere:

  • Given: radius $r = 9\,\text{cm}$

  • Problem requirement: answer correct to $1$ decimal place.

  • Step-by-step substitution and solution:     $V = \frac{4}{3}\pi r^3$     $V = \frac{4}{3} \times \pi \times 9^3$     $V = \frac{4}{3} \times \pi \times 729$     $V = 972\pi$     $V \approx 3053.6\,\text{cm}^3$

• Volume of Pyramids:

  • Distinction from prisms: Pyramids are not prisms because their horizontal cross-sectional area steadily diminishes from the base plane upward to the apex.

  • Volumetric ratio principle: The volume of any pyramid is exactly one-third the volume of a prism sharing the identical base area and perpendicular height.

  • Formula:     $V_{\text{pyramid}} = \frac{1}{3}AH$

  • Where $A$ is the area of the polygonal base

  • $H$ is the perpendicular height from the base to the apex

• Volume of Cones:

  • Volumetric ratio principle: A cone is geometrically analogous to a pyramid with a circular base; its volume is exactly one-third that of a cylinder possessing the same base radius $r$ and perpendicular height $h$.

  • Formula:     $V_{\text{cone}} = \frac{1}{3}\pi r^2 h$

• Worked Example 11: Calculating the volume of pyramids and cones:

  • Part a: Calculate the volume of a cone with radius $r = 8\,\text{cm}$ and perpendicular height $h = 10\,\text{cm}$:     $V = \frac{1}{3}\pi r^2 h$     $V = \frac{1}{3} \times \pi \times 8^2 \times 10$     $V = \frac{1}{3} \times \pi \times 64 \times 10$     $V = \frac{640\pi}{3}$     $V \approx 670.21\,\text{cm}^3$

  • Part b: Calculate the volume of a right square-based pyramid with base edge length $l = 8\,\text{cm}$ and vertical height $H = 12\,\text{cm}$:

  • Step 1: Base area calculation:       $A = l^2$       $A = 8^2 = 64\,\text{cm}^2$

  • Step 2: Pyramid volume computation:       $V = \frac{1}{3}AH$       $V = \frac{1}{3} \times 64 \times 12$       $V = 64 \times 4 = 256\,\text{cm}^3$