Ch 3 · 5.4–5.5 Surface Area, Volume and Compound Shapes · O Level Mathematics 4024
Topic 3 of 3 in Mensuration — O Level Mathematics 4024 — 53 lessons.
Introduction
A can of beans is a cylinder, an ice-cream cone is a cone topped by a hemisphere, a lampshade is a frustum (a cone with its top cut off). This lesson finds how much material covers a solid (surface area) and how much it holds (volume), and how to handle shapes built from simpler pieces. (Syllabus sections 5.4 and 5.5.)
Key ideas — Prisms and cuboids
A prism has the same cross-section all along its length (including a cylinder, or a cylindrical sector):
🧑🏫 Teacher's explanation
Imagine a loaf of bread that has been sliced perfectly straight. Every single slice you pull out looks exactly the same. This is the essence of a prism. Because every slice is identical, you can find the total amount of bread (the volume) by simply calculating the area of one slice and multiplying it by how long the loaf is.
📖 Definitions
🔍 Worked example — Easy
A cuboid has dimensions 5 cm by 4 cm by 10 cm. Calculate its volume.
🔍 Worked example — Standard
A right prism has a cross-section in the shape of a triangle with base 6 cm and height 4 cm. The length of the prism is 12 cm. Calculate the total surface area.
🔍 Worked example — Exam-style
A compound solid is made by placing a small cuboid of 2 cm × 2 cm × 2 cm on top of a larger cuboid of 5 cm × 5 cm × 3 cm. Calculate the total surface area of the compound solid.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — The formulas given in the exam
These are given in the List of formulas. Total surface area adds any flat faces: a closed cylinder has 2π rh + 2π r^2. Use Pythagoras for a cone's slant height: l = √(r^2 + h^2).
🧑🏫 Teacher's explanation
Think of volume as the amount of water needed to fill a container, while surface area is the amount of wrapping paper needed to cover it. If you have a cylinder, the volume is just the area of the circular base, π r^2, stacked up to a height h. The curved surface area is like taking a label off a soup can and flattening it into a rectangle with length 2π r and height h.
📖 Definitions
🔍 Worked example — Easy
A cylinder has a radius of 3 cm and a height of 10 cm. Calculate its total surface area in terms of π.
🔍 Worked example — Standard
A cone has a base radius of 5 cm and a vertical height of 12 cm. Find its volume in terms of π.
🔍 Worked example — Exam-style
A solid is made by placing a hemisphere of radius 4 cm on top of a cylinder of radius 4 cm and height 10 cm. Find the total volume of the composite solid in terms of π.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Compound shapes and parts of shapes
For the surface area of a compound solid, include only the outside surfaces: the join between two pieces is hidden.
🧑🏫 Teacher's explanation
Imagine you are building something with LEGO bricks. If you stick a small cube onto a large cube, the total volume is simply the sum of the two cubes. However, if you were painting them, you wouldn't paint the faces that are touching each other because they are no longer on the 'outside'. This is the golden rule for compound surface area: only calculate the visible, external faces.
📖 Definitions
🔍 Worked example — Easy
A solid consists of a cylinder with a hemisphere attached to one end. The cylinder has a radius of 3 cm and a height of 10 cm. Calculate the total volume of the solid. Leave your answer in terms of π.
🔍 Worked example — Standard
A cone has a height of 12 cm and a base radius of 5 cm. A smaller cone of height 4 cm is removed from the top to create a frustum. Calculate the volume of the frustum. Leave your answer in terms of π.
🔍 Worked example — Exam-style
A wooden block is in the shape of a cuboid with dimensions 10 cm × 8 cm × 5 cm. A cylindrical hole of radius 2 cm is drilled through the entire 5 cm thickness. Calculate the total surface area of the remaining block. Leave your answer in terms of π.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked examples — Example 1: cylinder and cone
A cylinder with r = 3 cm, h = 10 cm: V = 90π = 283 cm³; total surface area = 60π + 18π = 78π = 245 cm². A cone with r = 6 cm, h = 8 cm: l = 10 cm; V = tfrac13π(36)(8) = 96π = 302 cm³; curved surface area = 60π cm².
🔍 Worked example — Easy
A cylinder has a radius of 3 cm and a height of 10 cm. Calculate its volume and total surface area. Give answers in terms of π.
🔍 Worked example — Standard
A cone has a radius of 6 cm and a vertical height of 8 cm. Calculate its volume and curved surface area. Give answers in terms of π.
🔍 Worked example — Exam-style
A solid is made by placing a cone of radius 3 cm and height 4 cm on top of a cylinder of radius 3 cm and height 5 cm. Find the total volume of the solid in terms of π.
⚠️ Common errors
Worked examples — Example 2: a sphere
Radius 4.5 cm: V = tfrac43π(4.5)^3 = 121.5π = 382 cm³; surface area = 4π(4.5)^2 = 81π = 254 cm².
🔍 Worked example — Easy
A sphere has a radius of 3 cm. Calculate its volume in terms of π.
🔍 Worked example — Standard
A sphere has a diameter of 9 cm. Calculate its surface area in terms of π.
🔍 Worked example — Exam-style
A hemisphere (half a sphere) has a radius of 6 cm. Calculate its total surface area (including the flat circular base) in terms of π.
⚠️ Common errors
Worked examples — Example 3: a compound solid
A solid is a cylinder (radius 3 cm, height 8 cm) with a hemisphere of radius 3 cm on top. Volume = 72π + tfrac23π(27) = 90π = 283 cm³. Surface area = 48π (curved cylinder) + 9π (base) + 18π (hemisphere) = 75π = 236 cm².
🔍 Worked example — Easy
A solid consists of a cylinder with radius r=2 cm and height h=5 cm, topped by a hemisphere of radius 2 cm. Calculate the total volume in terms of π.
🔍 Worked example — Standard
A solid is formed by a cylinder (radius 3 cm, height 8 cm) with a hemisphere (radius 3 cm) on top. Find the total surface area in terms of π.
🔍 Worked example — Exam-style
A composite object consists of a cone with radius 4 cm and slant height 5 cm placed on top of a cylinder with radius 4 cm and height 10 cm. Find the total volume of the object. (Use π ≈ 3.14)
⚠️ Common errors
Worked examples — Example 4: a frustum
A cone of radius 6 cm and height 12 cm has a cone of radius 2 cm and height 4 cm cut from its top. Frustum volume = tfrac13π(36 × 12 - 4 × 4) = 416π/3 = 436 cm³.
🔍 Worked example — Easy
A cone has a radius of 3 cm and a height of 6 cm. A smaller cone with a radius of 1 cm and a height of 2 cm is removed from the top. Find the volume of the remaining frustum in terms of π.
🔍 Worked example — Standard
A cone of radius 6 cm and height 12 cm has a cone of radius 2 cm and height 4 cm cut from its top. Find the volume of the frustum.
🔍 Worked example — Exam-style
A frustum is created by cutting a cone of height 15 cm and base radius 5 cm with a plane parallel to the base. The height of the removed top cone is 5 cm. Calculate the volume of the frustum in terms of π.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) V = 24 × 15 = 360 cm³; faces 2 × 24 + 15(6 + 8 + 10) = 408 cm². (2) tfrac23π(216) = 144π cm³. (3) tfrac13 × 36 × 4 = 48 cm³. (4) h = 12 cm; V = tfrac13π(25)(12) = 100π = 314 cm³.