Mensuration: Area, Surface Area, and Volume

Shape Shifters: Mastering Area & Volume 📐

Introduction

1. Introduction

Ever wondered how game developers calculate the size of a map, or how much paint you'd actually need to redecorate your room? That's all mensuration! It’s the maths of measurement, and it's way more useful than you think. In this chapter, we're going to start with the basics: finding the perimeter and area of 2D shapes - think figuring out the perfect dimensions for your next Instagram post. Then, we'll level up to 3D solids, calculating their surface area and volume, so you'll know exactly how much wrapping paper you need for a gift or how much soda is in your can. Finally, we'll tackle the boss level: compound shapes. We'll learn how to break down complex objects, like a swimming pool or a custom-built speaker, into simple parts to find their measurements. Let's get this sorted.

A = l \times w

Worked example

The Weird Backyard Problem 🏡

Your part-time job is landscaping. A client has a backyard shaped like the diagram below, which consists of a rectangle and a trapezium. The rectangular part is 10m by 6m. The trapezium part shares the 6m side with the rectangle, has a top parallel side of 4m, and a perpendicular height of 5m. The two slanted sides of the trapezium are both 5.5m. Calculate: a) The total area of the backyard to lay down new grass. b) The perimeter of the backyard to build a new fence.

V = \text{Area of cross-section} \times \text{length}

A = 4\pi r^2

Worked example

The Grain Silo Challenge 🚜

r=5

1
$h_{cone}$
$h_{cone}^2 + r^2 = l^2 \\ h_{cone}^2 + 5^2 = 13^2 \\ h_{cone}^2 + 25 = 169 \\ h_{cone}^2 = 144 \\ h_{cone} = 12 \text{ m}$
2
$V_{cyl} = \pi r^2 h$
$V_{cylinder} = \pi (5)^2 (12) = 300\pi \text{ m}^3 \\ V_{cone} = \frac{1}{3}\pi (5)^2 (12) = 100\pi \text{ m}^3$
3
$For the total volume, we just add them together. Simple as that.$
$V_{total} = V_{cylinder} + V_{cone} = 300\pi + 100\pi = 400\pi \approx 1260 \text{ m}^3 \text{ (3 s.f.)}$
4
$For the surface area, think about what you would actually paint. You'd paint the circular floor, the curved wall of the cylinder, and the curved roof of the cone. You wouldn't paint the circle where they join because it's inside.$
$A_{total} = A_{base} + A_{cylinder\_curved} + A_{cone\_curved}$
5
$A = \pi r^2$
$A_{base} = \pi (5)^2 = 25\pi \\ A_{cylinder\_curved} = 2\pi (5)(12) = 120\pi \\ A_{cone\_curved} = \pi (5)(13) = 65\pi$
6
$Finally, add them all up to get the total area to be painted and round to 3 significant figures. Job done!$
$A_{total} = 25\pi + 120\pi + 65\pi = 210\pi \approx 660 \text{ m}^2 \text{ (3 s.f.)}$

Answer

A_{total} = 25\pi + 120\pi + 65\pi = 210\pi \approx 660 \text{ m}^2 \text{ (3 s.f.)}

Alright, let's talk about compound shapes. Think of them like a playlist mashup or a custom character build in a game. You're not dealing with one basic shape anymore; you're combining several simple ones (rectangles, circles, cylinders, cones) to create something more complex and interesting. The key to absolutely crushing these problems is to see the simple shapes hiding inside the complex one. It's like finding the secret level—once you see it, you can't unsee it. For 2D compound shapes, we have two missions: find the perimeter and the area. For perimeter, imagine you're walking around the absolute edge of the shape. Don't fall into the trap of adding up all the individual perimeters! If you join a rectangle and a semi-circle to make an ice-rink shape, the shared edge disappears from the perimeter calculation. You're only walking on the outside track.

For area, it's usually much simpler: you just calculate the area of each component shape and add them together. Sometimes, you might have to subtract, like finding the area of a donut by calculating the area of the big circle and subtracting the area of the hole. Now, let's go 3D. When we're calculating the volume of a compound solid (like a silo made from a cylinder and a hemisphere), it's just like the 2D area—you find the volume of the cylinder, the volume of the hemisphere, and add them up. Easy. But surface area is where you have to be slick. Just like with perimeter, the surfaces that are stuck together don't count. If you put a cone on top of a cylinder, the circular base where they join is now internal, not on the surface. So, you'd calculate the curved area of the cone and the curved area of the cylinder, plus the area of the bottom circular base.

\pi

Worked example

The Ultimate Ice Cream Cone Challenge 🍦

h=12

1
$First, let's break this down. We have two shapes: a cone and a hemisphere. For part (a), the total volume is simply the volume of the cone plus the volume of the hemisphere. Let's start with the cone's volume.$
$V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \pi \times 5^2 \times 12 = \frac{1}{3} \times \pi \times 25 \times 12 = 100\pi \text{ cm}^3$
2
$\frac{4}{3}\pi r^3$
$V_{hemisphere} = \frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi (5)^3 = \frac{2}{3}\pi \times 125 = \frac{250}{3}\pi \text{ cm}^3$
3
$Now we add them together for the total volume. We can leave it as a fraction or convert to a mixed number, but a single fraction is usually tidier.$
$V_{total} = V_{cone} + V_{hemisphere} = 100\pi + \frac{250}{3}\pi = \frac{300\pi}{3} + \frac{250}{3}\pi = \frac{550}{3}\pi \text{ cm}^3$
4
$A = \pi rl$
$l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \text{ cm}$
5
$4\pi r^2$
$A_{cone} = \pi rl = \pi \times 5 \times 13 = 65\pi \text{ cm}^2 <br> A_{hemisphere} = \frac{1}{2}(4\pi r^2) = 2\pi r^2 = 2\pi (5)^2 = 50\pi \text{ cm}^2$
6
$Finally, add the two surface areas together for the grand total. This is the total area you'd have to lick to eat all the ice cream off the outside. Done!$
$A_{total} = A_{cone} + A_{hemisphere} = 65\pi + 50\pi = 115\pi \text{ cm}^2$

Answer

A_{total} = A_{cone} + A_{hemisphere} = 65\pi + 50\pi = 115\pi \text{ cm}^2

10

3\text{ m}^2

Worked example

0.5

Worked example

20

Worked example

120°

Worked example

3.5

Worked example

r = 12

Worked example

4

Unlock the full chapter: practice questions, flashcards, mock papers and notes, free.

Continue revising