A pyramid is a 3D shape that has a polygon as a base and sloping sides that meet at a point called the apex. If the apex is directly above the centre of the base, the pyramid is called a right pyramid. Pyramids are named according to the shape of their base.
A cone is made by connecting a circular base to an apex. If the apex is directly perpendicular to the centre of the base, it is called a right cone. Conical shapes appear everywhere in the real world.
Surface area of a pyramid
As we found with prisms, calculating the surface area of a pyramid is done by adding the area of all faces and so it is helpful to visualise the net:
 |
 |
| Regular right-pentagonal pyramid | The net of a regular right-pentagonal pyramid |
For pyramids, the net consists of the base and a number of triangular faces. If the pyramid is a right pyramid and if the base shape is regular (meaning its side lengths are all equal - like a square or equilateral triangle) then all the triangular faces are equivalent. On the other hand if the base is not regular, for example if it is rectangular, then the triangles will have different sizes.
$\text{Surface area of right pyramid }=\text{area of base }+\text{area of triangles }$Surface area of right pyramid =area of base +area of triangles
Practice Questions
question 1
Find the surface area of the square pyramid shown. Include all faces in your calculations.
question 2
Surface area of a cone
Using the interactive tool below you can see what happens when we unravel a right cone. This will help us to see the shapes we need to work out its surface area.
We observe that for a right cone the surface area will be the area of the circular base plus the area of a sector with radius of length $s$s (this length is called the slant height of the cone).
But how do we find the area of this sector without knowing the included angle?
Exploration: formula for the area of a cone
The base we can see is a circle, and will have area $\pi r^2$πr2 where $r$r is the radius of the base of the cone
Before we work out the area of the sector, let's first consider the entire circle the sector is a part of.
The area of this large circle with radius $s$s, would be $\pi s^2$πs2
The circumference of the large circle with radius $s$s, would be $2\pi s$2πs.
The green arc, arc $AB$AB, originally wrapped around the base of the cone, and so its length is the circumference of the base. So the length of arc$AB$AB is $2\pi r$2πr.
The ratio of the blue shaded sector to the area of the whole circle, is the same as the ratio of the green arc$AB$AB to circumference of the whole circle.
We can write this as an equation.
| $\frac{\text{area of sector }}{\text{area of whole circle }}$area of sector area of whole circle | $=$= | $\frac{\text{length of arc }}{\text{circumference of large circle }}$length of arc circumference of large circle |
By definition of ratios |
| $\frac{\text{area of sector }}{\pi s^2}$area of sector πs2 | $=$= | $\frac{2\pi r}{2\pi s}$2πr2πs |
Substituting in given information |
| $\text{area of sector }$area of sector | $=$= | $\frac{r}{s}\times\pi s^2$rs×πs2 |
Simplifying and multiplying both sides by $\pi s^2$πs2 |
| $\text{area of sector }$area of sector | $=$= | $\pi rs$πrs |
Simplifying |
Thus the total surface area of a right cone is:
Where $r$r is the cone's base radius and $s$s is the slant height:
| $\text{Surface Area of Right Cone}$Surface Area of Right Cone | $=$= | $\text{Area of Base }+\text{Area of Sector }$Area of Base +Area of Sector |
| $SA$SA | $=$= | $\pi r^2+\pi rs$πr2+πrs |
Practice question
Question 3
Find the surface area of the cone shown.
Round your answer to two decimal places.
Volume of a pyramid
The volume of a pyramid with a base area $A$A, and perpendicular height $h$h from base to apex is equal to $\frac{1}{3}$13 times the volume of the prism with the same base and height.
| Volume of a pyramid | $=$= | $\frac{1}{3}\times\text{area of base }\times\text{height }$13×area of base ×height |
| $V$V | $=$= | $\frac{1}{3}Ah$13Ah |
Note that the above formula is applicable for all pyramids. The base can be any polygon, and it is not necessary for the apex to be above the centre of the base.
Practice question
Question 4
Some very famous right pyramids are the Egyptian Pyramids.
Pictured here is the Great Pyramid. It has a square base of side length 230 metres and a vertical height of 146 metres. Find the volume of the Great Pyramid.
Write your answer correct to the nearest tenth.
Volume of a cone
In fact, it is not even necessary for the base to be a polygon when using the volume formula for a pyramid. By considering the base to be a circle, we get that the formula for the volume of a cone is essentially the same as the formula for a pyramid. Namely:
Volume = $\frac{1}{3}\times$13× area of base $\times$× height
where the height is the perpendicular height between the base of the cone and the apex. Since the base of a cone is a circle, we can substitute in the formula for the area of a circle:
| Volume of a cone | $=$= | $\frac{1}{3}\times\text{area of base }\times\text{height }$13×area of base ×height |
| $V$V | $=$= | $\frac{1}{3}\pi r^2h$13πr2h |
To reiterate, when we calculate the volume of a pyramid or cone we must use the perpendicular height, that is, the height perpendicular to the base that meets the apex. If we are given the height of one of the triangular faces (in the case of a pyramid) or the slant height (in the case of a cone), we can use Pythagoras' theorem with the radius to find the perpendicular height.
Practice question
Question 5
Consider the cone below.
Find the perpendicular height $h$h of the cone.
Enter each line of working as an equation.
What is the volume of the cone in cubic metres?
Round your answer to three significant figures.