Lesson

Much like how we can calculate other surface areas, the surface areas of cones and pyramids are the sums of the areas of their faces. We can do this by unfolding them to get their nets and then calculate the areas of all the faces.

When we unfold a pyramid to get its net, we can see that a pyramid is made up of one base face and a number of triangular faces.

We can calculate the area of the base using the appropriate formula for its shape, since its dimensions will match the dimensions of the pyramid.

Finding the area of the triangular faces is not quite as straightforward. The base length of each of the triangular faces is the length of the base side that they are joined to, which is easy to find. But since the triangular faces are tilted to meet at the vertex, the height of each triangle corresponds to the slope length (or slant height) of that face, not the height of the pyramid.

If we know the base side lengths and the height of a pyramid, we can find the slant heights of each triangular face using Pythagoras' theorem in 3D space.

Provided that the apex of the triangle is directly above the centre of the base then the two short sides of the right-angled triangle (as shown below) will always be (1) the height, and (2) half the base side length.

Aside from simply adding up the faces of a pyramid, we can also look for symmetries in the solid to make our calculations easier. For example, since the base sides of a square pyramid are all equal, each triangular face will also have the same area.

In fact, any triangular faces with the same base side length will have equal areas - provided that the apex of the pyramid is above the centre of the base.

When we unfold a cone to get its net, we find that the surface area of a cone will be the sum of the area of a circle and the area of a sector.

While finding the area of the circular base is not a problem, finding the area of the sector is a bit more complicated.

We can see that the radius of the sector is equal to the slope length $s$`s` of the cone, which can be calculated using Pythagoras' theorem in 3D as we did for the pyramid. However, to find the area we also need to know what fraction of the circle's area we have.

We can do this by comparing the arc length of the sector to the circumference of a full circle with radius $s$`s`.

Since the sector face must wrap around the circular base to form the cone, we know that the arc length of the sector must be equal to the circumference of the circular base. So the arc length of the sector will be equal to $2\pi r$2π`r`, where $r$`r` is the radius of the cone's base.

Since the circumference of a full circle with radius $s$`s` would be $2\pi s$2π`s`, we know that the sector is equal to $\frac{2\pi r}{2\pi s}$2π`r`2π`s` of the full circle. Simplifying this tells us that the sector's area will be equal to $\frac{r}{s}$`r``s` of the full circle's area.

The area of a full circle with radius $s$`s` is $\pi s^2$π`s`2. Taking $\frac{r}{s}$`r``s` of this area gives us the area of the sector:

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

We can then combine this area with the area of the circular base to get a formula for the surface area of a cone:

Surface area of a cone

$\text{Surface area of a cone}=\pi r^2+\pi rs$Surface area of a cone=π`r`2+π`r``s`

Where $r$`r` is the radius and $s$`s` is the slope length of the cone.

Consider the rectangular pyramid with rectangular base of dimensions $4$4cm by $10$10cm and perpendicular height $7$7cm.

Find the length of the slant height, $x$

`x`cm.Round your answer to two decimal places.

Find the length of the second slant height $y$

`y`cm.Round your answer to two decimal places.

Find the surface area of the pyramid. Make sure to include all faces in your calculation.

Round your answer to one decimal place.

Find the surface area of the cone.

Round your answer to two decimal places.

Find the surface area of a cone with a diameter of $10$10 cm and a perpendicular height of $12$12 cm.

Round your answer to two decimal places.

applies formulas to find the surface areas of right pyramids, right cones, spheres and related composite solids