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9.07 Further composite solids

Lesson

Finding the volume and surface area of straight-edged composite solids is generally a case of adding together the area or volume of the basic shapes that make up the solid.

When our composite solids are also made from pyramids and curved surface solids like cones and spheres, there are a couple of new techniques that can be useful for calculating their surface area or volume.

 

Truncated pyramids and cones

To truncate a solid is to cut off some part of it in order to shorten it. We can do this to pyramids and cones by cutting off the tops to get composite solids that look like this:

A truncated pyramid is called a frustum and is made by subtracting a smaller pyramid from a larger pyramid.

Similarly, a truncated cone is made by subtracting a smaller cone from a larger cone.

It is worth noting that the smaller sections that are removed will always be similar to the larger, original solid.

 

Volume of truncated pyramids and cones

We can find the volume of a frustum or truncated cone by subtracting the volume of the removed section from the original solid.

Worked example

Find the volume of the frustum.

Think: We can see from the diagram that the frustum is formed by removing a small square pyramid from a larger, original square pyramid. The original pyramid had a base area of $81$81 and a height of $12$12. The removed section had a base area of $9$9 and a height of $4$4.

To find the volume of the frustum, we can subtract the volume of the removed section from the volume of the original pyramid.

Do: Using the formula for the volume of a pyramid, $V=\frac{1}{3}Ah$V=13Ah, we get:

$\text{Volume of original pyramid}=\frac{1}{3}\times81\times12$Volume of original pyramid=13×81×12 $=$= $324$324

$\text{Volume of removed section}=\frac{1}{3}\times9\times4$Volume of removed section=13×9×4 $=$= $12$12

Subtracting the volumes gives us:

$\text{Volume of the frustum}=312$Volume of the frustum=312

Reflect: We found the volume of the frustum by taking the difference between the original and removed volumes. We found each volume using the dimensions of each section.

We can use the same method to find the volumes of truncated cones.

 

Surface area of truncated pyramids and cones

When we unfold a frustum for its net we get two base faces, the top and the bottom, and some trapeziums.

This square-based frustum has a net consisting of two squares and four trapeziums.

To find the surface area of a frustum, we add the base areas to the areas of the various trapeziums.

So what about the surface area of a truncated cone?

Since there isn't a curved equivalent to a trapezium, we will need a different method to find the surface area of a truncated cone. This method involves taking the difference between the curved surface areas of the original and removed cone sections.

Exploration

Consider this truncated cone.

If we unfold it, we find that the net of a truncated cone looks like this:

Notice that we have two circular bases, the top and the bottom, and a surface formed by subtracting a small sector from a larger sector.

In other words, the surface area of a truncated cone is equal to the surface area of the original cone, minus the curved surface area of the removed cone, plus the base area of the removed cone.

For the truncated cone above, we can calculate each of those areas to be:

$\text{Surface area of the original cone}=\pi\times5^2+\pi\times5\times20$Surface area of the original cone=π×52+π×5×20 $=$= $125\pi$125π

$\text{Curved area of the removed cone}=\pi\times5^2+\pi\times5\times20$Curved area of the removed cone=π×52+π×5×20 $=$= $36\pi$36π

$\text{Base area of the removed cone}=\pi\times3^2$Base area of the removed cone=π×32 $=$= $9\pi$9π

As such, the surface area of this truncated cone is:

$\text{Surface area of the truncated cone}=125\pi-36\pi+9\pi$Surface area of the truncated cone=125π36π+9π $=$= $100\pi$100π

These types of calculations work particularly well with cones because the original and removed sections are always similar solids.

 

Other composite solids

In addition to truncated pyramids and cones, we can also make composite solids out of spheres, hemispheres, cylinders and other curved surface solids.

As we can see, while some dimensions will be common between the component solids, there is no guarantee that these parts will be similar in shape. As such, we can instead find the surface area and volume of these solids in the usual way- by adding and subtracting solids that we already know.

 

Practice questions

Question 1

A small square pyramid of height $6$6 cm was removed from the top of a large square pyramid of height $12$12 cm to form the solid shown.

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

    Round your answer to two decimal places.

  2. 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.

Question 2

The solid is a truncated cone, formed by cutting off a cone shaped section from the top of a larger, original, cone.

  1. What was the volume of the cone section that was cut off?

    Give your answer as an exact value.

  2. What was the volume of the original cone?

    Give your answer as an exact value.

  3. Find the volume of the truncated cone.

    Round your answer to two decimal places.

Question 3

A pyramid has been removed from a rectangular prism, as shown in the figure. Find the volume of this composite solid.

Question 4

Find the surface area of the composite figure shown, which consists of a cone and a hemisphere joined at their bases.

Round your answer to two decimal places.

Outcomes

VCMMG365 (10a)

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

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