Outcomes
A student:

 MA5.31WM
uses and interprets formal definitions and generalisations when explaining solutions and/or conjectures

 MA5.32WM
generalises mathematical ideas and techniques to analyse and solve problems efficiently

 MA5.33WM
uses deductive reasoning in presenting arguments and formal proofs

 MA5.314MG
applies formulas to find the volumes of right pyramids, right cones, spheres and related composite solids
Content
 Students:
 Solve problems involving the volumes of right pyramids, right cones, spheres and related composite solids (ACMMG271)

develop and use the formula to find the volumes of right pyramids and right cones:
\( \textrm{Volume of pyramid/cone} = \frac{1}{3}Ah \) where A is the base area and h is the perpendicular height
 recognise and use the fact that a pyramid/cone has onethird the volume of a prism/cylinder with the same base and the same perpendicular height (Reasoning)
 deduce that the volume of a cone is given by \( V = \frac{1}{3} \pi r^2 h \) (Reasoning)

use the formula to find the volumes of spheres:
\( \textrm{Volume of sphere} = \frac{4}{3} \pi r^3 \) where r is the length of the radius 
find the volumes of composite solids that include right pyramids, right cones and hemispheres,
eg find the volume of a cylinder with a cone on top  solve a variety of practical problems relating to the volumes and capacities of right pyramids, right cones and spheres
 apply Pythagoras' theorem as needed to calculate the volumes of pyramids and cones (Problem Solving)
 find the dimensions of a particular solid, given its volume, by substitution into a formula to generate an equation, eg find the length of the radius of a sphere, given its volume (Problem Solving)
Background Information
The formulas for the volumes of solids mentioned here depend only on the perpendicular heights and apply equally well to oblique cases. The volumes of oblique solids may be included as an extension for some students.
A more systematic development of the volume formulas for pyramids, cones and spheres can be given after integration is developed in Stage 6 (where the factor \( \frac{1}{3} \) emerges essentially because the primitive of \(x^2\)^{ }is \( \frac{1}{3} x^3\)).
In Stage 5.3, the relationship could be demonstrated by practical means, eg filling a pyramid with sand and pouring the sand into a prism with the same base and perpendicular height, and repeating until the prism is filled.
Some students may undertake the following exercise: visualise a cube of side length 2a dissected into six congruent pyramids with a common vertex at the centre of the cube, and then prove that each of these pyramids has volume \( \frac{4}{3} a^3 \), which is \( \frac{1}{3} \) of the enclosing rectangular prism.
The problem of finding the edge length of a cube that has twice the volume of another cube is called the 'duplication of the cube' and is one of three famous problems left unsolved by the ancient Greeks. It was proved in the nineteenth century that this cannot be done with a straight edge and a pair of compasses, essentially because the cube root of 2 cannot be constructed on the number line.
Language
The difference between the 'perpendicular heights' and the 'slant heights' of pyramids and cones should be made explicit to students.