A figure formed from a polygonal base and a set of triangular faces. The triangular faces connect to one side of the base and all join together at a common point called the apex.

A pair of three dimensional solids. One has a rectangular base and triangular sides which meet at a point, the other has a triangular base and triangular sides which meet at a point.

Regular pyramid

A pyramid formed from a regular polygonal base. The triangular faces of a regular pyramid are congruent.

Rectangular pyramid

A pyramid with a rectangular base.

Right pyramid

A pyramid in which the apex lies directly above the centre of the base.

Triangular pyramid

A pyramid with a triangular base.

Cone

A solid with a circular base and an apex that is connected to the base by a collection of line segments that form a curved surface.

A three dimensional figure with a circular face, and a curved surface connected to the circle which meets at a point.

Right cone

A cone in which the apex lies directly above the centre of the base.

A cone with a dashed line segment drawn from the vertex of the cone to the center of the base. The angle the segment makes with the base is shown to be a right angle.

Oblique cone

A cone in which the apex does not lie directly above the centre of the base.

A cone with a dashed line segment drawn from the vertex of the cone to the center of the base. The angle the segment makes with the base is not a right angle.

Whereas prisms are formed by a pair of congruent polygons joined by rectangles, pyramids are instead formed from a single polygon as a base, with triangular sides that meet at a point (the apex).

Similarly, while a cylinder was formed by a pair of congruent circles joined by a curved surface, a cone is formed from a single circle with a curved surface that meets itself at a point (again called the apex).

The volume of a pyramid is exactly one third the volume of a prism formed from the same base polygon with the same perpendicular height. That is, the volume of a pyramid is given by

\displaystyle V = \frac{1}{3}Bh

\bm{B}

area of the base

\bm{h}

perpendicular height of the apex from the base

The volume of a cone can be calculated identically, and is exactly one third the volume of a cylinder with the same base circle and perpendicular height.

The surface area of a pyramid is the sum of the area of the base and the area of each triangle. For a regular pyramid, the triangles are congruent, and so the surface area of a regular pyramid is given by

\displaystyle SA = B + nA

\bm{B}

area of the base polygon

\bm{n}

number of sides of the base polygon

\bm{A}

area of one triangular face

The surface area of a cone is somewhat different. For a right cone, the surface area can be calculated using the formula

\displaystyle SA = B + \pi rh_s

\bm{B}

area of the circular base

\bm{r}

radius of the base

\bm{h_s}

slant height of the cone

A diagram of a right cone showing the height of the cone as a dashed segment from the apex to the center of the circular base. From the center of the circle to a point on the circumference is labelled r which is perpendicular to the height of the cone. The distance of the segment from the apex to a point on the circle is labelled h sub s.

The slant height of a right cone is the distance from any point on the circumference of the base to the apex.

Worked examples

Example 1

The Pyramid of Giza is a square pyramid, that is 280 Egyptian Royal cubits high and has a base length of 440 Egyptian Royal cubits. What is the volume of the Pyramid of Giza?

Approach

First we find the area of the base, then we can use that to find the volume. Since this solid is a square pyramid, the base is a square.

Solution

Finding the area of the base, we have:

\displaystyle B	\displaystyle =	\displaystyle \text{side length}^2	Area of a square
	\displaystyle =	\displaystyle 440^2	Substitute the side length
	\displaystyle =	\displaystyle 193\,600	Simplify

We can now use this to calculate the volume:

\displaystyle V	\displaystyle =	\displaystyle \dfrac{1}{3}Bh	Volume of a pyramid
	\displaystyle =	\displaystyle \dfrac{1}{3} \cdot 193\,600 \cdot 280	Substitute known values
	\displaystyle =	\displaystyle 18\,069\,333 \frac{1}{3}	Simplify

So the volume of the Pyramid of Giza is 18 \, 069 \, 333 \dfrac{1}{3} cubic Egyptian Royal cubits.

Example 2

A carrot has grown such that it can be approximated by a cone with radius 1 \text{ in} and length 12 \text{ in}. If the carrot weighs 4 \text{ lb}, find the approximate density of the carrot.

Approach

To find the density, we first want to calculate the volume of the cone. Before we can do that, we want to start by calculating the area of the circular base.

Solution

We can calculate the area of the base as follows:

\displaystyle B	\displaystyle =	\displaystyle \pi r^2	Area of a circle
	\displaystyle =	\displaystyle \pi \cdot 1^2	Substitute the radius
	\displaystyle =	\displaystyle \pi	Simplify

So the area of the circular base is \pi \text{ in}^2.

We can now use this to find the approximate volume of the carrot:

\displaystyle V	\displaystyle =	\displaystyle \dfrac{1}{3}Bh	Volume of a cone
\displaystyle {}	\displaystyle =	\displaystyle \dfrac{1}{3}\cdot\pi \cdot12	Substitute known values
\displaystyle {}	\displaystyle =	\displaystyle 4\pi	Simplify

So the approximate volume of the carrot is 4\pi \text{ in}^3.

Finally, we can approximate the density:

\displaystyle \text{Density}	\displaystyle =	\displaystyle \frac{\text{Mass}}{\text{Volume}}	Density formula
	\displaystyle =	\displaystyle \dfrac{4}{4\pi}	Substitute known values
	\displaystyle =	\displaystyle \frac{1}{\pi}	Simplify

So the density of the carrot is approximately \dfrac{1}{\pi} \, \text{lb}/\text{in}^3.

Reflection

Since this is only an approximation of the density we could also express it as a decimal value, rounded to some number of decimal places. In this case, the density is approximately 0.318 \, \text{lb}/\text{in}^3, to three decimal places.

Note that even when approximating, it is generally good practice to leave values in exact form throughout the work and only round at the final expression.

Outcomes

MA.912.GR.4.5

Solve mathematical and real-world problems involving the volume of three-dimensional figures limited to cylinders, pyramids, prisms, cones and spheres.

MA.912.GR.4.6

Solve mathematical and real-world problems involving the surface area of three-dimensional figures limited to cylinders, pyramids, prisms, cones and spheres.

11.05 Pyramids and cones

Concept summary

Worked examples

Example 1

Approach

Solution

Example 2

Approach

Solution

Reflection

Outcomes

MA.912.GR.4.5

MA.912.GR.4.6

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