# 9.07 Surface area of pyramids and cones

Lesson

## Lateral area of pyramids

The lateral area of a pyramid is the total area of all the non-base faces. Consider the following regular right-pentagonal pyramid.

 Regular right-pentagonal pyramid

It's lateral area can be found by adding up the area of each triangular face, which we call the lateral faces. As shown in the net, we can see that there are five congruent triangles, each with the same area.

 The net of a regular right-pentagonal pyramid

So to find the lateral area, we find the area of one triangular face, and multiply this by five.

#### Exploration

In some instances, we can immediately find the area of one triangular face given the side length of the base and the slant height.

 Base side length of $10$10 units and slant height of $23$23 units

In the above pyramid, the slant height is $23$23 units, and the side length of the base is $10$10 units. We can find the area $A$A of one triangular face using the formula $A=\frac{1}{2}bh$A=12bh where $b$b is the base of the triangle and $h$h is the perpendicular height.

 One triangular face

The area of one triangular face is calculated below.

 $A$A $=$= $\frac{1}{2}bh$12​bh (Writing down the area formula) $A$A $=$= $\frac{1}{2}\times10\times23$12​×10×23 (Substitution) $A$A $=$= $115$115 units2 (Simplifying)

To find the lateral area of the pentagonal pyramid, we multiply the area of the triangular face by $5$5. So the lateral area is given by:

$115\times5=575$115×5=575 units2

In other cases, we might have to determine the slant height given the height of the pyramid and some length along the base. For instance, consider the following regular right-pentagonal pyramid.

 Regular right-pentagonal pyramid with unknown slant height $s$s

We can use the Pythagorean theorem to determine the slant height $s$s. A right triangle is constructed by the slant height $s$s, the perpendicular height of the pyramid, and the center of the base to its edge as shown.

 A right triangle

Using the Pythagorean theorem, we calculate $s$s below.

 $s^2$s2 $=$= $15^2+8^2$152+82 (Using the Pythagorean theorem) $s^2$s2 $=$= $225+64$225+64 (Simplifying the squares) $s^2$s2 $=$= $289$289 (Simplifying addition) $s$s $=$= $17$17 (Taking the square root)

From here we can find the lateral area by first finding the area of one triangular face like before, and then multiplying by the number of faces.

Lateral area

To find the lateral area of a regular right pyramid, find the area of one lateral face, and then multiply by the number of faces.

Sometimes we must find the slant height first by using the Pythagorean theorem.

#### Practice questions

##### question 1

Consider this regular pentagonal pyramid and its net.

1. How many identical triangular faces does it have?

$\editable{}$ triangular faces.

2. Find the area of one triangular face.

3. Find the lateral area of the pyramid.

##### question 2

Consider this regular square pyramid.

1. Use Pythagoras' theorem to find the slant height $p$p.

2. Find the area of one triangular face.

3. Find the lateral surface area of the pyramid.

##### question 3

Consider this regular square pyramid.

1. Find the side length of the square $x$x.

2. Find the area of one triangular face.

3. Find the lateral surface area of the pyramid.

## Surface area of pyramids and cones

### Pyramids

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 $2$2D triangle, which we will use in calculating surface area.

As we found with other $3$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

#### Practice questions

##### question 4

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

##### question 5

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.

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

2. Now find the surface area of the solid formed.

Make sure to include all faces in your calculations.

### Cones

A cone is made by connecting a circular base to an apex.  If the apex is directly perpendicular to the center 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.

#### Exploration

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.

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 $\pi s^2$πs2

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

The pink arc, 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$2πr

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.

 $\frac{\text{area of sector }}{\text{area of whole circle }}$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:

Surface area of right cone

Where $r$r is the cone's base radius and $l$l 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 $=$= $\pi r^2+\pi rl$πr2+πrl

#### Practice questions

##### Question 6

Find the surface area of the cone shown.

##### Question 7

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.

### Outcomes

#### G.13

Use surface area and volume of three-dimensional objects to solve practical problems