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10&10a

9.04 Surface area of pyramids and cones

Lesson

Introduction

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.

Surface area of a pyramid

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.

The net of a square pyramid with a square base and four triangles.

Unfolding a square pyramid gives us a square base and four triangles.

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.

This image shows the slope height is the hypotenuse of a right-angled in the pyramid. Ask your teacher for more information.

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) will always be (1) the height, and (2) half the base side length.

The slope height is the hypotenuse of a right-angled triangle in the pyramid.

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.

Examples

Example 1

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

A rectangular pyramid with a height of 7 centimetres. Ask your teacher for more information.
a

Find the length of the slant height labelled x cm. Round your answer to two decimal places.

Worked Solution
Create a strategy

Use Pythagoras' theorem given by c^2=a^2+b^2.

Apply the idea

We are given c=x, a=7, and b=\dfrac{4}{2}=2.

\displaystyle c^2\displaystyle =\displaystyle a^2+b^2Write the formula
\displaystyle x^2\displaystyle =\displaystyle 7^2+2^2Substitute the values
\displaystyle =\displaystyle 53Evaluate
\displaystyle x\displaystyle =\displaystyle 7.28\text{ cm}Take the square root of both sides
b

Now find the length of the other slant height, labelled y cm. Round your answer to two decimal places.

Worked Solution
Create a strategy

Use Pythagoras' theorem given by c^2=a^2+b^2.

Apply the idea

We are given c=y, a=7, and b=\dfrac{10}{2}=5.

\displaystyle c^2\displaystyle =\displaystyle a^2+b^2Write the formula
\displaystyle y^2\displaystyle =\displaystyle 7^2+5^2Substitute the values
\displaystyle =\displaystyle 74Evaluate
\displaystyle y\displaystyle =\displaystyle 8.60Take the square root of both sides
c

Find the surface area of the pyramid. Make sure to include all faces in your calculation. Round your answer to one decimal place.

Worked Solution
Create a strategy

The surface area of the pyramid is equal to the sum of the areas of four triangles and a rectangle.

Apply the idea

Based from the diagram, two of the triangles have a base of 10 and a height of 7.28, and the other two have a base of 4 and a height of 8.60. The rectangle has a width of 4 and length of 10.

\displaystyle \text{Rectangle}\displaystyle =\displaystyle lwWrite the formula
\displaystyle =\displaystyle 4\times 10Subsitute the dimensions
\displaystyle =\displaystyle 40 \text{ cm}^2Evaluate
\displaystyle \text{Side triangles}\displaystyle =\displaystyle \dfrac{1}{2}bh \times 2Multiply the formula by 2
\displaystyle =\displaystyle \dfrac{1}{2}\times 4 \times 8.60 \times 2Substitute the dimensions
\displaystyle =\displaystyle 34.4 \text{ cm}^2Evaluate
\displaystyle \text{Front and back triangles}\displaystyle =\displaystyle \dfrac{1}{2}bh \times 2Multiply the formula by 2
\displaystyle =\displaystyle \dfrac{1}{2}\times 10 \times 7.28 \times 2Substitute the dimensions
\displaystyle =\displaystyle 72.8 \text{ cm}^2Evaluate
\displaystyle \text{Surface area}\displaystyle =\displaystyle 40+34.4+72.8Add all the areas
\displaystyle =\displaystyle 147.2 \text{ cm}^2Evaluate
Idea summary

To find the surface area of a pyramid we add the area of the base and the area of each triangular face.

To find the height of each triangular face, we will need to use Pythagoras' theorem.

Surface area of a cone

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.

A cone with radius R, height H, and slant height S, and its net made up of a circle with radius R and a sector with radius S.

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 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.

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, where r is the radius of the cone's base.

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

The area of a full circle with radius s is \pi s^2. Taking \dfrac{r}{s} of this area gives us the area of the sector:\text{Area of the sector}=\pi s^2\times \dfrac{r}{s}=\pi rs

We can then combine this area with the area of the circular base to get a formula for the surface area of a cone given by SA=\pi r^2+\pi rs where r is the radius and s is the slope length of the cone.

Examples

Example 2

Find the surface area of the cone. Round your answer to two decimal places.

A cone with a slant height of 8 centimetres and base radius of 3 centimetres.
Worked Solution
Create a strategy

Use the surface area of a cone given by: \text{Surface area}=\pi r^2+\pi rs.

Apply the idea

We are given r=3 and s=8.

\displaystyle \text{Surface area}\displaystyle =\displaystyle \pi r^2+\pi r sWrite the formula
\displaystyle =\displaystyle \pi \times 3^2+\pi \times 3\times 8Substitute the values
\displaystyle =\displaystyle 103.67 \text{ cm}^2Evaluate
Idea summary
\displaystyle \text{Surface area of a cone}=\pi r^2+\pi rs
\bm{r}
is the radius
\bm{s}
is the slant height of the cone

Outcomes

ACMMG271 (10a)

Solve problems involving surface area and volume of right pyramids, right cones, spheres and related composite solids

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