Measurement

Lesson

A pyramid can be made in the following way: Use any polygon as a base. We can have square bases, triangular bases or even hexagonal bases. Then connect every vertex of the base to an apex point above the base, and you have a pyramid.

If the apex is directly above (ie perpendicular to) the centre of the base it is a right pyramid. These are all right pyramids.

We can identify a pyramid by the shape of the base. Square and rectangular based pyramids, are the most common you will come across in mathematics, but also in the real world.

As you can see from these pictures, the triangular shaped sides slope towards the apex. This introduces a new term we use in calculations with pyramids called the **Slant Height** (or *Slope Height*).

In the interactive below, look at the triangular faces of the pyramid. You can see that the slope height corresponds to the height of the 2D triangle, which we will use in calculating surface area.

As we found with other 3D shapes, calculating the surface area of a solid is done by adding the area of all faces. For right pyramids, we have the base and a number of triangular faces.

This results in:

Surface Area of Right Pyramid

$\text{Surface Area of Right Pyramid }=\text{Area of Base }+\text{Area of triangles }$Surface Area of Right Pyramid =Area of Base +Area of triangles

Find the surface area of the square pyramid shown. Include all faces in your calculations.

A small square pyramid of height $5$5 cm was removed from the top of a large square pyramid of height $10$10 cm to form the solid shown.

Find the length of the slant height of the sides of the new solid.

Round your answer to two decimal places.

Now find the surface area of the solid formed.

Round your answer to one decimal place.

Make sure to include all faces in your calculations.

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. Cone shapes appear everywhere in the real world.

Using the interactive below you can see what happens when we unravel a cone. This will help us to see the shapes we need to work out its surface area.

As we found with other 3D shapes, calculating surface areas is done by finding the total of the area of all faces. For right cones, we have the base and a circle sector.

The base we can see is a circle, and will have area $\pi r^2$π`r`2 where $r$`r` is the radius of the base of the cone.

The other piece is the sector. It is part of a larger circle that has a radius of $s$`s`, the slant height of the cone.

Before we work out the area of the sector, lets first consider the entire circle the sector is a part of.

The area of this large circle with radius s, would be $\pi s^2$π`s`2

The circumference of the large circle with radius s would be $2\pi s$2π`s`.

The pink arc 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 is $2\pi r$2π`r`

The ratio of the blue shaded sector to the area of the whole circle, is the same as the ratio of the pink arc AB to circumference of the whole circle.

We can write this as an equation.

$\frac{\text{area of sector }}{\text{area of whole circle }}=\frac{\text{length of arc }}{\text{circumference of large circle }}$area of sector area of whole circle =length of arc circumference of large circle

$\frac{\text{area of sector }}{\pi s^2}=\frac{2\pi r}{2\pi s}$area of sector π`s`2=2π`r`2π`s`

$\text{area of sector }=\frac{r}{s}\times\pi s^2$area of sector =`r``s`×π`s`2

$\text{area of sector }=\pi rs$area of sector =π`r``s`

(if you want to refresh our work on sectors, go here)

Thus the total surface area of a right cone is:

Surface Area of Right Cone

$\text{Surface Area of Right Cone}=\text{Area of Base }+\text{Area of Sector }$Surface Area of Right Cone=Area of Base +Area of Sector

$SA=\pi r^2+\pi rs$`S``A`=π`r`2+π`r``s`

Find the surface area of the cone shown.

Round your answer to two decimal places.

The top of a solid cone was sawed off to form the solid attached. Find the surface area of the solid formed correct to 2 decimal places.