Chapter 19 · Higher tier

GCSE Maths Volume, Surface Area and 3D Shapes (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

Name 3D solids and count their faces, edges and vertices
Recognise and draw nets of cuboids, prisms, pyramids and cylinders
Draw and use plans and front and side elevations
Work out the volume of cuboids, prisms and cylinders, and use units of volume and capacity
Use density = mass ÷ volume
Work out the surface area of cuboids, prisms and cylinders
Higher: use the formulae for pyramids, cones, spheres, hemispheres, composite solids and frustums, work backwards, and link to similar solids

You should already know

Area of rectangles, triangles, trapezia and circles (Chapter 18)
Metric units of length and capacity
Powers and square roots (Chapters 1 and 2)

Where this leads

Pythagoras and trigonometry in 3D (Chapter 20)
Similar shapes and scale factors for area and volume (Chapter 21)
Density, pressure and compound measures (Chapter 16)

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier. Content references: G12, G13, G16, G17.

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

19.1 Properties of 3D solids: Worksheet (PDF) · Answers (PDF) · lesson page
19.2 Nets of solids: Worksheet (PDF) · Answers (PDF) · lesson page
19.3 Plans and elevations: Worksheet (PDF) · Answers (PDF) · lesson page
19.4 Volume of cuboids, and units of volume and capacity: Worksheet (PDF) · Answers (PDF) · lesson page
19.5 Volume of prisms: Worksheet (PDF) · Answers (PDF) · lesson page
19.6 Volume of cylinders, and density: Worksheet (PDF) · Answers (PDF) · lesson page
19.7 Surface area of cuboids, prisms and cylinders: Worksheet (PDF) · Answers (PDF) · lesson page
19.8 Pyramids and cones: Worksheet (PDF) · Answers (PDF) · lesson page
19.9 Spheres and hemispheres: Worksheet (PDF) · Answers (PDF) · lesson page
19.10 Composite solids, frustums and similar solids: Worksheet (PDF) · Answers (PDF) · lesson page
Check-up and exam-style questions: Worksheet (PDF) · Answers (PDF) · lesson page

Exercise P Do you remember?

Non-calculator

FWork out the area of a rectangle that is 9 cm long and 4 cm wide.

Done

FWork out the area of a triangle with base 10 cm and perpendicular height 6 cm.

Done

FWork out the area of a circle of radius 5 cm. Give your answer in terms of π.

Done

FWork out the circumference of a circle of diameter 12 cm. Give your answer in terms of π.

Done

FChange 3.5 litres into millilitres.

Done

19.1

Properties of 3D solids

Edexcel 1MA1: G12
Grade 1-4

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Properties of 3D solidsComing soon

A flat shape such as a rectangle or a circle is 2D: it has length and width only. A 3D shape, also called a solid, takes up space, so it also has depth or height. A cereal box, a can of beans, a football and a tent are all solids. In Chapter 18 you worked with the flat shapes; in this chapter you meet the solids built from them.

To describe a solid we use three words. A face is a flat surface (the top of a box is a face). An edge is a line where two faces meet. A vertex is a corner where edges meet; the plural is vertices. Some solids also have a curved surface, which is not a flat face: a tin has one curved surface round its side.

cuboid

triangular prism

cylinder

cone

sphere

square-based pyramid

Six common solids. A cube is a cuboid whose faces are all squares.

Two families of solid are very important. A prism has the same shape (its cross-section) all the way along its length. A cuboid is a prism with a rectangle as its cross-section; a triangular prism has a triangle. A pyramid has a flat base and triangular faces that rise to a single point called the apex. We name them by the shape of the base: a square-based pyramid has a square base, a hexagonal pyramid has a hexagon.

Three solids have curved surfaces. A cylinder has two flat circular faces and one curved surface. A cone has one flat circular face, one curved surface and one vertex (its point). A sphere is a perfect ball: one curved surface and no edges or vertices.

Counting faces, edges and vertices gives a pattern. Look at the last column of this table:

Solid	Faces (F)	Edges (E)	Vertices (V)	V − E + F
Cuboid or cube (square or rectangular prism)	6	12	8	8 − 12 + 6 = 2
Triangular prism	5	9	6	6 − 9 + 5 = 2
Pentagonal prism	7	15	10	10 − 15 + 7 = 2
Hexagonal prism	8	18	12	12 − 18 + 8 = 2
Triangular-based pyramid (tetrahedron)	4	6	4	4 − 6 + 4 = 2
Square-based pyramid	5	8	5	5 − 8 + 5 = 2
Pentagonal-based pyramid	6	10	6	6 − 10 + 6 = 2
Hexagonal-based pyramid	7	12	7	7 − 12 + 7 = 2

For every solid with flat faces and no holes, vertices − edges + faces = 2. This is called Euler’s formula (say “Oiler”). It is useful as a check on your counting, and for finding a missing number.

Key ideas

Face: a flat surface. Edge: where two faces meet. Vertex: a corner.
A prism has a constant cross-section; a pyramid has a base and triangular faces meeting at the apex.
Cylinder: 2 flat faces, 1 curved surface. Cone: 1 flat face, 1 curved surface, 1 vertex. Sphere: 1 curved surface only.
An n-sided prism has n + 2 faces, 3n edges and 2n vertices. An n-sided pyramid has n + 1 faces, 2n edges and n + 1 vertices.
Euler’s formula: V − E + F = 2.

Worked example 19.1.1

Work out the number of faces, edges and vertices of this triangular prism.

Key words. A face is a flat surface. An edge is where two faces meet. A vertex (plural vertices) is a corner. A prism has the same cross-section all along its length. A pyramid has a base and triangular faces meeting at a point (the apex).

Method. Count faces, edges and vertices in turn. Split the solid into its base(s) and sides to count without missing any.

Name the solid. It helps to picture or sketch it.
The solid is a triangular prism.
Count the faces. A prism has 2 identical end faces and one rectangular face for each side of an end. The ends have 3 sides.
2 ends + 3 rectangles = 5 faces
Count the edges. Each end has 3 edges (that is 3 + 3) and 3 more edges join the two ends together.
3 + 3 + 3 = 9 edges
Count the vertices. Each end has 3 corners and there are two ends.
3 + 3 = 6 vertices

Check. Euler’s formula: vertices − edges + faces = 6 − 9 + 5 = 2. ✓

Watch out. Count each edge once only, and remember the base (or both ends) when you count faces.

Worked example 19.1.2

A solid has one square face and four triangular faces. Name the solid and say how many vertices it has.

Method. Use the faces to identify the family of solid, then count the corners.

Look at the faces. One square face and four triangles that meet at a point means a pyramid with a square base.
Solid: square-based pyramid
Count the vertices. There are 4 corners on the square base and 1 more at the top (the apex).
4 + 1 = 5
State the answer.
A square-based pyramid has 5 vertices.

Check. Euler’s formula: 5 − 8 + 5 = 2 (a square-based pyramid has 8 edges). ✓

Watch out. Do not count the apex twice, and do not forget it.

Worked example 19.1.3

A solid has 9 faces and 16 edges. Use Euler’s formula to work out its number of vertices.

Method. Use Euler’s formula: V − E + F = 2, and solve for the missing number.

Write Euler’s formula. For a solid with flat faces and no holes: vertices − edges + faces = 2.
V − E + F = 2
Put in the numbers you know.
V − 16 + 9 = 2
Simplify the known part.
V − 7 = 2
Solve for V. Add the same to both sides.
V = 2 + 7 = 9

Check. 9 − 16 + 9 = 2. ✓

Watch out. Be careful with the minus sign in front of E.

Worked example 19.1.4

Show that a prism whose ends have n sides has n + 2 faces, 3n edges and 2n vertices, and that Euler’s formula works for every prism.

Method. Count the faces, edges and vertices using n, then substitute into V − E + F.

Faces. There are two ends, and one rectangular face for each of the n sides of an end.
F = n + 2
Edges. Each end has n edges, and n more edges join the two ends together.
E = n + n + n = 3n
Vertices. Each end has n corners.
V = 2n
Substitute into V − E + F. Collect the terms in n and the numbers separately.
V − E + F = 2n − 3n + (n + 2) = 2n − 3n + n + 2 = 0n + 2 = 2
Conclude. The n terms cancel, so the answer is 2 whatever n is.
So V − E + F = 2 for every prism. ✓

Check. Try n = 3 (a triangular prism): 6 − 9 + 5 = 2. ✓

Watch out. In a proof, write each step and finish with a sentence that states what you have shown.

Watch out

Count the faces of a prism by splitting it up: the two ends plus the sides. A common slip is to forget the ends, or to count the same edge twice. A cylinder, cone and sphere have curved surfaces, so they do not have faces and edges in the same way as a cube.

19.2

Nets of solids

Edexcel 1MA1: G12, G13
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Nets of solidsComing soon

A net is a flat pattern that folds up to make a solid. Imagine cutting along some edges of a cardboard box and unfolding it flat: the flat shape you get is a net. Nets show you every face of a solid at once, which makes them very useful for finding surface areas (Section 19.7).

Every net is made from the faces of the solid. A cuboid has six rectangular faces, so its net has six rectangles. A triangular prism has two triangles and three rectangles. A square-based pyramid has one square and four triangles. A cylinder has two circles and one rectangle. Counting the faces helps you decide which solid a net makes.

Left: a net of a 5 cm by 3 cm by 2 cm cuboid. Right: a net of a triangular prism (the dashed line is the height of the triangle).

Nets of a cube. A cube has 6 identical square faces, so its net is 6 squares joined edge to edge. But not every arrangement of 6 squares folds into a cube. There are exactly 11 different nets of a cube. To test a net, imagine folding it: if two squares land on the same face, it fails. Squares in a straight row of four wrap round the cube like a belt, so a row of 5 or a 2 × 3 block never works.

works

fails

Four of the eleven cube nets (top row) and three arrangements of six squares that do not fold to a cube (bottom row).

Left: net of a square-based pyramid (the slant height of a triangle is dashed). Right: net of a cylinder. The rectangle wraps round the curved side, so its width is the circumference 2πr of the circle.

Nets of cylinders. If you peel the label off a tin of soup and lay it flat, you get a rectangle. Its height is the height of the tin. Its width is the distance once round the tin, which is the circumference of the circular end, 2πr. So a cylinder’s net is a rectangle (2πr by h) with a circle of radius r touching the top edge and another touching the bottom edge.

Net of a cone. The curved surface of a cone opens out into a sector (a slice of a circle) whose radius is the cone’s slant height l. The arc of the sector is exactly as long as the circumference of the base circle. Its angle is 360 × r ÷ l degrees, where r is the base radius.

Net of a cone with base radius 3 cm and slant height 5 cm: a circle and a sector of angle 216°.

Key ideas

A net is a flat pattern that folds to make a solid; it shows every face.
There are 11 different nets of a cube. If two squares fold onto the same face, the net fails.
Cuboid: 6 rectangles. Triangular prism: 2 triangles + 3 rectangles. Square-based pyramid: 1 square + 4 triangles. Cylinder: 2 circles + 1 rectangle.
The rectangle in the net of a cylinder has width 2πr (the circumference) and height h.
Higher: the sector in the net of a cone has radius l and angle 360 × r ÷ l.

Worked example 19.2.1

Does this net fold to make a cube?

Method. Mentally fold the net up. A net makes a cube when all six squares become six different faces.

Pick one square to be the bottom of the cube. Imagine holding the first square flat on a table and folding the other squares up around it, like a paper box.
Fold each neighbouring square up, one at a time. Every square must become a different face of the cube: bottom, top, front, back, left, right.
Check that no two squares land on the same face. If all six squares go to six different faces, the net works.
Pairs of opposite faces: A is opposite D; B is opposite E; C is opposite F

Check. There are exactly 11 different nets of a cube. This one is among them. ✓

Watch out. Rows of 4 squares with one square above and one below always work; a row of 5 or a 2 × 3 block never does.

Worked example 19.2.2

The diagram shows a net of a cuboid. Work out the total area of the net.

Key words. The surface area of a solid is the total area of all its faces added together. It is measured in square units (cm²), because it is an area.

Method. Surface area is the total area of all the faces. Work out the area of each face (or matching pair of faces) and add.

Picture the six faces. A cuboid has 6 rectangular faces that come in 3 matching pairs: top and bottom, front and back, left and right.
Top and bottom. Each is length × width.
5 × 3 = 15 (and there are 2 of them, so 30)
Front and back. Each is length × height.
5 × 2 = 10 (2 of them: 20)
Left and right. Each is width × height.
3 × 2 = 6 (2 of them: 12)
Add the three pairs. The surface area is every face added together.
30 + 20 + 12 = 62
Write the unit. Area, so square units.
Surface area = 62 cm²

Check. Use the formula 2(lw + lh + wh) = 2 × (15 + 10 + 6) = 2 × 31 = 62. ✓

Watch out. Count every face: forgetting one pair of faces is the most common error. Give the answer in cm², not cm³.

Worked example 19.2.3

The net of a cylinder has circles of radius 3 cm. Work out the exact width of the rectangle in the net.

Method. The rectangle in the net of a cylinder is as wide as the circumference of the circular end.

Think about how the rectangle wraps round. When the net is folded up, the long edge of the rectangle goes right round the circular end once. So it is as long as the circumference of the circle.
Use the circumference formula. Circumference = 2πr (or πd).
C = 2 × π × 3 = 6π
Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
Width = 6π cm

Check. Circumference is a bit more than 3 × the diameter: 3 × 6 = 18. ✓

Watch out. Use the circumference (2πr) for the width of the rectangle, not the area of the circle.

Worked example 19.2.4

Work out the angle of the sector in the net of a cone with base radius 3 cm and slant height 5 cm.

Method. The arc of the sector is as long as the circumference of the base. Find what fraction of the big circle (radius l) the arc is, then find that fraction of 360°.

Circumference of the base. The base circle has radius 3.
2 × π × 3 = 6π cm
Circumference of the full circle of radius l = 5. If the sector were a whole circle, its arc would be this long.
2 × π × 5 = 10π cm
Find the fraction of the full circle. The arc is 6π out of 10π.
6π ÷ 10π = 6 ÷ 10 = 3 ÷ 5
Find that fraction of 360°.
3 ÷ 5 × 360 = 216°

Check. Formula: 360 × r ÷ l = 360 × 3 ÷ 5 = 216. ✓

Watch out. The radius of the sector is the slant height l, not the base radius r.

Watch out

Squares that touch in a net do not always touch on the cube, and squares that look far apart can be opposite. Always fold the net in your head one square at a time. For a cylinder the width of the rectangle is the circumference (2πr), not the radius and not the area of the circle.

19.3

Plans and elevations

Edexcel 1MA1: G13
Grade 2-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Plans and elevationsComing soon

Imagine walking round a model made of building blocks. From the front it looks one way, from the side another way, and from above another. Architects and engineers draw exactly these views to describe a solid on flat paper.

The view from above is the plan. The view from the front is the front elevation. The view from the side (we use the right-hand side) is the side elevation. Each view is drawn straight on, so you see flat shapes, never a 3D picture. For a solid made of cubes we draw each view on squared paper, one square for each cube face you can see.

A solid made from 7 cubes, with its front elevation, side elevation and plan. In the side elevation the front of the solid is on the left.

How to find an elevation of a stack of cubes. For the front elevation, look at each column from the left to the right and find the tallest stack: a tall stack hides any shorter stack behind it. For the side elevation, look at each row from the front to the back and find its tallest stack. For the plan, just mark every stack as one square (you can write the number of cubes in it).

Other solids. The views of cylinders, cones, spheres, pyramids and prisms are shapes you already know. A cylinder standing on its circular face looks like a rectangle from the front and the side, and like a circle from above. A cone looks like a triangle from the front and a circle (with a dot at the centre) from above.

Left: front elevation, side elevation and plan of a cone with radius 3 and height 5. Right: the same three views of a triangular prism (triangle 4 by 3, length 6). One square of the grid is 1 cm.

Key ideas

Plan: the view from above. Front elevation: the view from the front. Side elevation: the view from the side.
Draw only what you can see, straight on, with the right sizes on squared paper.
For cubes: use the tallest stack in each column (front) or row (side). In the side elevation the front is on the left.
Cylinder: rectangle, rectangle, circle. Cone: triangle, triangle, circle. Sphere: circle, circle, circle. Square-based pyramid: triangle, triangle, square with diagonals.

Worked example 19.3.1

The solid is made from 1 cm cubes. Draw the front elevation.

Key words. A plan is the view of a solid from directly above. An elevation is a view from the side (front elevation: from the front; side elevation: from the side). Pictures of the solid are drawn on a square grid.

Method. An elevation is a view from the side; the plan is the view from above. Only what can be seen is drawn, and a taller stack hides a shorter one behind it.

Stand in front of the solid, looking at it straight on. The front elevation is the view from the front. Each column of the plan lines up from left to right.
For each column, find the tallest stack. The tallest stack hides any shorter stack behind it, so only the tallest shows.
Column 1 (from the left): stacks 1, 3 → tallest = 3
Column 2 (from the left): stacks 2, 1 → tallest = 2
Draw the elevation on squared paper. In each column, shade as many squares as the tallest stack has cubes.
Heights from left to right: 3, 2
Here is the finished drawing. Check that your drawing matches it, square for square.

Check. The elevation has 5 squares: 3 + 2 = 5. ✓

Watch out. In the side elevation the front row is on the left when you look from the right. Do not add the numbers up: use the tallest in each column or row.

Worked example 19.3.2

Draw the side elevation of the same solid, looking from the right.

Method. An elevation is a view from the side; the plan is the view from above. Only what can be seen is drawn, and a taller stack hides a shorter one behind it.

Stand to the right of the solid, looking at its right-hand side. The side elevation shows the rows. Looking from the right, the front row is on the left of your view and the back row is on the right.
For each row, find the tallest stack. The tallest stack in a row hides the shorter ones behind it.
Front row (view position 1): stacks 3, 1 → tallest = 3
Back row (view position 2): stacks 1, 2 → tallest = 2
Draw the elevation on squared paper. Start with the front row on the left. Shade squares up to each row’s tallest stack.
Heights from left to right: 3, 2
Here is the finished drawing. Check that your drawing matches it, square for square.

Check. The elevation has 5 squares: 3 + 2 = 5. ✓

Watch out. In the side elevation the front row is on the left when you look from the right. Do not add the numbers up: use the tallest in each column or row.

Worked example 19.3.3

A cone stands on its circular base. The base has radius 3 cm and the cone is 5 cm tall. Draw the plan and the front elevation on a 1 cm grid.

Method. Think about what you would see from above and from the front, then draw it to the correct size.

Plan: look from above. From above you see the circular base, with the point of the cone in the middle.
A circle of radius 3 cm with a dot at its centre
Front elevation: look from the front. From the side-on view the cone looks like a triangle. Its base is the width of the circle and its height is the height of the cone.
Base = 2 × 3 = 6 cm, height = 5 cm: an isosceles triangle
Draw on the grid. Use one square for 1 cm so the drawings are the right size.

Check. The widest part of both drawings is 6 cm, as it should be. ✓

Watch out. From above a cone is a circle, not a triangle.

Worked example 19.3.4

A triangular prism has a triangle of base 4 cm and height 3 cm as its cross-section, and is 6 cm long. It lies on a rectangular face with the triangle facing you. Draw the front elevation, the side elevation and the plan.

Method. Draw each view of the triangular prism straight on, using one grid square for 1 cm.

Front elevation. Looking at the triangular end, we see the triangle: base 4 cm, height 3 cm.
Triangle with base 4 and height 3
Side elevation. From the side we see a rectangular face: as long as the prism (6 cm) and as tall as the triangle (3 cm).
Rectangle 6 cm by 3 cm
Plan. From above we see the two sloping rectangles side by side, each 2 cm wide (half of the base) and 6 cm long. The line between them is the top edge.
Rectangle 4 cm by 6 cm with a line down the middle
Draw on the grid. One square for 1 cm.

Check. The width is 4 in the front elevation and the plan, the height is 3 in both elevations, and the length is 6 in the side elevation and the plan. ✓

Worked example 19.3.5

A solid is made from cubes. Its front elevation shows columns of height 2, 1 and 3 (left to right). Its side elevation shows columns of height 3 and 2 (front to back). Work out the largest and the smallest number of cubes the solid could have.

Method. Each stack must not be taller than the front column and not taller than the side row it sits in. Work out the tallest each stack could be (largest total) and the fewest cubes that still match both views.

Set out the grid of stacks. There are 3 columns (from the front elevation) and 2 rows (from the side elevation), so 6 stacks.
Columns: 2, 1, 3. Rows (front to back): 3, 2.
Largest: make every stack as tall as allowed. A stack can be no taller than its column limit and no taller than its row limit, so use the smaller of the two.
Front row (limit 3): min(2,3), min(1,3), min(3,3) = 2, 1, 3 = 6 cubes
Back row (limit 2): min(2,2), min(1,2), min(3,2) = 2, 1, 2 = 5 cubes
Largest total = 6 + 5 = 11
Smallest: use as few cubes as possible. Every column must still reach its height and every row must still reach its height. Let one stack do two jobs at once: a stack that is the tallest in its column and the tallest in its row saves cubes.
One arrangement with the fewest cubes (front row, then back row): [0, 0, 3], [2, 1, 0]
Smallest total = 6

Check. The smallest arrangement has front elevation columns [2, 1, 3] and side elevation rows [3, 2], as required. ✓

Watch out. Check both views for every arrangement you try. A total that matches only one view is not enough.

Watch out

For the side elevation, the front of the solid is on the left when you look from the right. Do not add the numbers in a column: use the tallest stack, because the shorter stacks are hidden behind it. A plan shows one square for each stack, whatever its height.

19.4

Volume of cuboids, and units of volume and capacity

Edexcel 1MA1: G14, G16
Grade 1-5

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Volume of cuboids, and units of volume and capacityComing soon

Volume is the amount of space inside a solid. We measure it by counting how many unit cubes fit inside. A cubic centimetre (cm³) is the volume of a cube whose edges are all 1 cm long. A cubic metre (m³) is a cube with 1 m edges, and a cubic millimetre (mm³) a tiny cube with 1 mm edges.

Look at the cuboid below, which is 4 cm long, 3 cm wide and 2 cm high. The bottom layer holds 4 × 3 = 12 cubes. There are 2 layers, so the whole cuboid holds 12 × 2 = 24 cubes: its volume is 24 cm³.

Left: 24 unit cubes in 2 layers of 12. Right: the same cuboid with its lengths marked.

This always works, so the volume of a cuboid is length × width × height. A cube has all edges equal, so its volume is side × side × side. The three lengths must be in the same unit before you multiply.

Capacity. Liquids are measured in litres (l) and millilitres (ml). The capacity of a container is how much it can hold. The links to volume are worth learning:

Volume	Capacity
1 cm³	1 ml
1000 cm³	1 litre (1000 ml)
1 m³ = 1,000,000 cm³	1000 litres
1 cm³	1000 mm³

Why is 1 m³ equal to 1,000,000 cm³? Because 1 m = 100 cm, so a cube of side 1 m has volume 100 × 100 × 100 = 1,000,000 cm³. In the same way 1 cm = 10 mm, so 1 cm³ = 10 × 10 × 10 = 1000 mm³. Volume units are not like length units: you cube the conversion factor.

Key ideas

Volume of a cuboid = length × width × height. Volume of a cube = side³.
Volume is measured in cubic units: mm³, cm³, m³.
1 ml = 1 cm³. 1 litre = 1000 ml = 1000 cm³. 1 m³ = 1000 litres = 1,000,000 cm³.
To find a missing length, divide the volume by the other two lengths.
Put all lengths in the same unit before you multiply.

Worked example 19.4.1

Work out the volume of this cuboid.

Key words. Volume is the space inside a 3D shape. A cubic centimetre (cm³) is the volume of a cube with sides of 1 cm.

Method. Volume is the amount of space inside a solid. For a cuboid, multiply length × width × height.

Write down the formula. A cuboid is built from layers of 1-unit cubes. One layer holds length × width cubes and the height tells us how many layers there are.
Volume = length × width × height
Substitute the lengths. All three lengths are in cm, so the units match.
V = 8 × 5 × 4
Multiply the length and the width first. This is the area of the base: the number of cubes in one layer.
8 × 5 = 40
Multiply by the height. This counts all the layers.
40 × 4 = 160
Write the answer with its unit. Three lengths multiplied together give a cubic unit.
V = 160 cm³

Check. Multiply in a different order: 5 × 4 = 20, then 20 × 8 = 160. ✓

Watch out. Volume is in cm³, not cm or cm². Do not add the lengths.

Worked example 19.4.2

A cuboid has volume 240 cm³. Its length is 8 cm and its width is 5 cm. Work out its height.

Method. Use Volume = length × width × height with the unknown length as a letter, then divide to find it.

Write the volume formula with the numbers you know. Volume = length × width × height. We call the missing length h.
240 = 8 × 5 × h
Multiply the two known lengths. Join the known numbers into one number.
8 × 5 = 40, so 240 = 40 × h
Divide to undo the multiplication. Dividing is the inverse of multiplying.
h = 240 ÷ 40 = 6
Write the unit. This is a length, so it is in cm (not cm³).
h = 6 cm

Check. 8 × 5 × 6 = 240. ✓

Watch out. Divide the volume by the two known lengths, not by just one of them.

Worked example 19.4.3

A fish tank is a cuboid 80 cm long, 50 cm wide and 40 cm high. How many litres of water can it hold when it is full?

Method. Find the volume in cm³, then divide by 1000 to change cm³ into litres.

Work out the volume of the cuboid. Volume = length × width × height, all in cm.
80 × 50 = 4000
4000 × 40 = 160,000 cm³
Change cm³ to litres. 1 litre = 1000 cm³, so divide by 1000.
160,000 ÷ 1000 = 160
Write the answer.
Capacity = 160 litres

Check. 160 litres × 1000 = 160,000 cm³. ✓

Watch out. 1 litre is 1000 cm³. Do not divide by 100.

Worked example 19.4.4

A swimming pool holds 2.5 m³ of water in a small paddling section. Change 2.5 m³ into litres.

Key words. Capacity is how much a container can hold, measured in litres (l) and millilitres (ml). Volume is the space a solid takes up.

Method. Change the volume to cm³ first, using the fact that links the unit to cm³, then change to the unit you want.

Change m³ into cm³. 1 m = 100 cm, so 1 m³ = 100 × 100 × 100 = 1,000,000 cm³.
2.5 m³ = 2.5 × 1,000,000 = 2,500,000 cm³
Change cm³ into litres. 1 litre = 1000 ml = 1000 cm³. To go to the bigger unit, divide.
2,500,000 cm³ = 2,500,000 ÷ 1000 = 2500 litres
Write the answer. Check that the unit matches what the question asked for.
2.5 m³ = 2500 litres

Check. Go back the other way: 2500 litres gives 2.5 m³. ✓

Watch out. Volume units are not the same as length units: 1 m³ is 1,000,000 cm³, not 100 cm³.

Worked example 19.4.5

A cube has volume 343 cm³. Work out the length of one edge.

Method. A cube of side a has volume a × a × a. To find the side, find the number that multiplies by itself three times to give the volume (the cube root).

Write the volume as a cube. We are looking for a number a with a × a × a = the volume.
a × a × a = 343
Try whole numbers. Test cubes you know: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000.
7 × 7 × 7 = 49 × 7 = 343
State the side length. The side is the cube root of the volume.
a = 7 cm

Check. 7 × 7 × 7 = 343. ✓

Watch out. The cube root is not the volume divided by 3.

Worked example 19.4.6

A cuboid has edges l, w and h. Every edge is doubled. Show that the volume is multiplied by 8.

Method. Write the new volume using the doubled lengths, then compare it with the old volume.

Write the original volume. Volume = length × width × height.
V = l × w × h
Double each edge. The new edges are 2l, 2w and 2h.
New volume = 2l × 2w × 2h
Collect the numbers. Multiplication can be done in any order, so group the 2s.
= 2 × 2 × 2 × l × w × h = 8lwh
Compare. lwh is the original volume.
New volume = 8 × V. So the volume is multiplied by 8. ✓

Check. Try l = w = h = 1: volume 1. Doubled: 2 × 2 × 2 = 8. ✓

Watch out. Doubling every length multiplies the volume by 2³ = 8, not by 2. (See Chapter 21: similar solids.)

Watch out

Volume is in cubic units (cm³), so do not write cm or cm². Do not add the lengths: multiply them. And do not change m³ to cm³ by multiplying by 100: you must multiply by 100 × 100 × 100 = 1,000,000.

19.5

Volume of prisms

Edexcel 1MA1: G16
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Volume of prismsComing soon

A prism is a solid that has the same cross-section all the way along its length. Cut a loaf of bread into slices: every slice is identical. A cuboid, a triangular tent, a Toblerone box and a trough are all prisms. The two identical faces at the ends are the cross-section, and the distance between them is the length of the prism.

Here is why there is one simple rule. Imagine the cross-section as a thin slice. The volume of the slice is its area × its thickness. Stack up slices all along the length, and you get the whole prism. So:

Volume of a prism = area of cross-section × length

The skill is finding the area of the cross-section. Remember the area formulas from Chapter 18:

Shape	Area
rectangle	width × height
triangle	½ × base × perpendicular height
parallelogram	base × perpendicular height
trapezium	½ × (sum of the parallel sides) × the distance between them
L-shape or other compound shape	split into rectangles and add

A triangular prism, a trapezium prism and an L-shaped prism. The dashed line in the first two is the perpendicular height of the cross-section.

Which length is the “length”? A prism can lie on any face. Always find the cross-section first: it is the face that is repeated. The “length” is the distance between the two copies of it. If a triangular prism stands on one of its triangles, the “length” is its height!

Key ideas

Volume of a prism = area of cross-section × length.
Find the area of the cross-section first, using the perpendicular height.
The length is the distance between the two identical end faces.
To find the length, divide the volume by the area of the cross-section.
Volume is in cubic units: cm³, m³.

Worked example 19.5.1

Work out the volume of this triangular prism.

Key words. A prism is a 3D shape with the same cross-section along its whole length. The cross-section is the face you would see if you sliced straight through it.

Method. For a prism: Volume = area of the cross-section × length.

Identify the cross-section and the length. A prism has the same cross-section all the way along. The length of the prism is the distance between its two identical end faces.
Cross-section: triangle. Length = 10 cm
Work out the area of the triangular cross-section. Area of a triangle = ½ × base × perpendicular height.
A = ½ × 6 × 4 = 24 ÷ 2 = 12 cm²
Multiply the area by the length. Volume of a prism = area of cross-section × length. Think of the cross-section as a slice, and the prism as that slice repeated along the length.
V = 12 × 10 = 120
Write the answer with its unit. Cubic units, because the volume is area (cm²) × length (cm).
V = 120 cm³

Check. A triangular prism is half of a cuboid 6 × 4 × 10 = 240, and half of that is 120. ✓

Watch out. Use the perpendicular height of the triangle or trapezium, not a sloping side. The prism length is the distance between the two ends.

Worked example 19.5.2

Work out the volume of this prism, which has a trapezium as its cross-section.

Key words. A prism is a 3D shape with the same cross-section along its whole length. The cross-section is the face you would see if you sliced straight through it.

Method. For a prism: Volume = area of the cross-section × length.

Identify the cross-section and the length. A prism has the same cross-section all the way along. The length of the prism is the distance between its two identical end faces.
Cross-section: trapezium. Length = 12 cm
Work out the area of the trapezium cross-section. Area of a trapezium = ½ × (sum of the parallel sides) × the distance between them.
A = ½ × (4 + 8) × 3
A = ½ × 12 × 3 = 36 ÷ 2 = 18 cm²
Multiply the area by the length. Volume of a prism = area of cross-section × length. Think of the cross-section as a slice, and the prism as that slice repeated along the length.
V = 18 × 12 = 216
Write the answer with its unit. Cubic units, because the volume is area (cm²) × length (cm).
V = 216 cm³

Check. Estimate: the volume must be between 144 and 288 (the volumes of the prisms using only the shorter or only the longer parallel side), and 216 is in that range. ✓

Watch out. Use the perpendicular height of the triangle or trapezium, not a sloping side. The prism length is the distance between the two ends.

Worked example 19.5.3

Work out the volume of this prism, whose cross-section is an L-shape.

Key words. A prism is a 3D shape with the same cross-section along its whole length. The cross-section is the face you would see if you sliced straight through it.

Method. For a prism: Volume = area of the cross-section × length.

Identify the cross-section and the length. A prism has the same cross-section all the way along. The length of the prism is the distance between its two identical end faces.
Cross-section: L-shape. Length = 10 cm
Split the L-shape into two rectangles. We can only use the area formula on simple shapes, so cut the L into two rectangles.
Bottom rectangle: 8 by 2
Top rectangle: 3 by (6 − 2) = 3 by 4
Find the area of each rectangle. Width × height each time.
8 × 2 = 16 cm²
3 × 4 = 12 cm²
Add the two areas. The whole cross-section is the two parts together.
A = 16 + 12 = 28 cm²
Multiply the area by the length. Volume of a prism = area of cross-section × length. Think of the cross-section as a slice, and the prism as that slice repeated along the length.
V = 28 × 10 = 280
Write the answer with its unit. Cubic units, because the volume is area (cm²) × length (cm).
V = 280 cm³

Check. Split the L the other way: left rectangle 3 × 6 = 18, right rectangle 5 × 2 = 10; total 28 = 28. ✓

Watch out. Use the perpendicular height of the triangle or trapezium, not a sloping side. The prism length is the distance between the two ends.

Worked example 19.5.4

A triangular prism has a cross-section with base 6 cm. It is 15 cm long and its volume is 180 cm³. Work out the height of the triangle.

Method. Work backwards: the volume divided by the length gives the area of the cross-section. Then use the triangle area formula to find its height.

Use V = area × length to find the cross-section area. Divide both sides by the length.
180 = A × 15
A = 180 ÷ 15 = 12 cm²
Write the triangle area formula. Area of a triangle = ½ × base × height.
12 = ½ × 6 × h
Simplify the right-hand side. ½ × 6 = 3.
12 = 3 × h
Divide to find h. Undo the multiplication.
h = 12 ÷ 3 = 4 cm

Check. ½ × 6 × 4 = 12, and 12 × 15 = 180. ✓

Watch out. The volume is not the area of the triangle. Divide by the length first to get the area.

Worked example 19.5.5

A water trough is a triangular prism. The triangle has base 40 cm and perpendicular height 30 cm, and the trough is 100 cm long. How many litres of water does it hold when full?

Method. Find the volume in cm³ using area of cross-section × length, then change to litres.

Area of the triangular cross-section. Area = ½ × base × perpendicular height.
½ × 40 × 30 = 600 cm²
Multiply by the length. Volume = cross-section × length.
600 × 100 = 60,000 cm³
Change to litres. 1 litre = 1000 cm³, so divide by 1000.
60,000 ÷ 1000 = 60 litres

Check. 60 litres × 1000 = 60,000 cm³. ✓

Watch out. Change units only at the end, once the volume is found in cm³.

Worked example 19.5.6

A prism has a trapezium cross-section with parallel sides x cm and (x + 4) cm, and the distance between them is 5 cm. The prism is 10 cm long and its volume is 350 cm³. Work out x.

Method. Write the area of the trapezium using x, form an equation for the volume, and solve it.

Area of the cross-section. Area of a trapezium = ½ × (sum of parallel sides) × distance between them.
Area = ½ × (x + x + 4) × 5 = ½ × (2x + 4) × 5 = (x + 2) × 5 = 5x + 10
Write the volume equation. Volume = area × length.
(5x + 10) × 10 = 350
Divide both sides by 10. This undoes the multiplication by 10.
5x + 10 = 35
Subtract 10, then divide by 5. Undo the + 10 first, then the × 5.
5x = 25, so x = 5

Check. Parallel sides 5 and 9: area = ½ × 14 × 5 = 35, and 35 × 10 = 350. ✓

Watch out. Write the area as an expression in x first. Do not substitute guesses.

Watch out

Use the perpendicular height of the triangle or trapezium, not the sloping side. And remember to multiply by the length at the end: finding only the area of the cross-section is not enough. Check the units: area is in cm², volume in cm³.

19.6

Volume of cylinders, and density

Edexcel 1MA1: G16, G17, R11
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Volume of cylinders, and densityComing soon

A cylinder is a prism with a circle as its cross-section. So the volume rule for prisms still works:

Volume of a cylinder = area of the circular end × height = πr²h

Here r is the radius of the circle (half the diameter) and h is the height (the distance between the two circular ends). You met the area of a circle, πr², in Chapter 18. For a cylinder with radius 4 cm and height 10 cm the area of the end is π × 4² = 16π cm², and so the volume is 16π × 10 = 160π cm³.

Left: a cylinder given by its radius. Right: a cylinder given by its diameter, so halve it first.

Exact answers and decimal answers. When a question says “give your answer in terms of π”, leave the π symbol in the answer: 160π cm³. This is the exact answer. When it says “give your answer to 3 significant figures” (or 1 decimal place), use the π button on a calculator: 160π = 502.65… = 503 cm³ to 3 s.f. Never use 3.14 unless the question tells you to.

Density. Lead is heavy and cork is light, even for the same size piece. Density measures how much mass is in each unit of volume: for example, steel has a density of about 7.8 g/cm³, so every cm³ of steel has a mass of 7.8 g. Density, mass and volume are linked by

density = mass ÷ volume, mass = density × volume, volume = mass ÷ density

Always keep the units together: a mass in grams and a volume in cm³ give a density in g/cm³; a mass in kg and a volume in m³ give a density in kg/m³. Compound measures are covered in more depth in Chapter 16.

Key ideas

Volume of a cylinder = πr²h, where r is the radius. If you are given the diameter, halve it.
“In terms of π” means leave π in the answer: for example 160π.
For a decimal answer use the π button, and only round at the end.
Density = mass ÷ volume. Mass = density × volume. Volume = mass ÷ density.
To find h from the volume, divide the volume by πr².

Worked example 19.6.1

Work out the volume of this cylinder. Give your answer in terms of π.

Key words. The radius (r) goes from the centre to the edge of a circle; the diameter is twice the radius. In terms of π means leave π as a symbol in the answer.

Method. Volume of a cylinder = area of the circular end × height = πr²h.

Write down the formula. A cylinder is a circular prism: the cross-section is a circle (area πr²) and we multiply by the height.
V = πr²h
Substitute the radius and the height.
V = π × 4² × 10
Square the radius first. Powers come before multiplying (BIDMAS). r² means r × r.
4² = 4 × 4 = 16
Multiply by the height.
16 × 10 = 160, so V = 160π
Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
V = 160π cm³

Check. Estimate with π ≈ 3: 3 × 16 × 10 = 480, which is close to 503. ✓

Watch out. Square only the radius, not π or the height. If you are given the diameter, halve it first.

Worked example 19.6.2

A cylinder has diameter 10 cm and height 6 cm. Work out its volume. Give your answer correct to 3 significant figures.

Key words. The radius (r) goes from the centre to the edge of a circle; the diameter is twice the radius. In terms of π means leave π as a symbol in the answer.

Method. Volume of a cylinder = area of the circular end × height = πr²h.

Find the radius. The formulas use the radius, which is half the diameter.
r = 10 ÷ 2 = 5
Write down the formula. A cylinder is a circular prism: the cross-section is a circle (area πr²) and we multiply by the height.
V = πr²h
Substitute the radius and the height.
V = π × 5² × 6
Square the radius first. Powers come before multiplying (BIDMAS). r² means r × r.
5² = 5 × 5 = 25
Multiply by the height.
25 × 6 = 150, so V = 150π
Use your calculator. Type the calculation in using the π button on your calculator (not 3.14). Write down the whole display before you round.
π × 5² × 6 = 471.2389…
Round the answer. The question wants 3 significant figures. Look at the digit just after the last one you are keeping: 5 or more rounds up, 4 or less leaves it.
471.2389… = 471 cm³

Check. Estimate with π ≈ 3: 3 × 25 × 6 = 450, which is close to 471. ✓

Watch out. Square only the radius, not π or the height. If you are given the diameter, halve it first.

Worked example 19.6.3

A cylinder has radius 6 cm and volume 180π cm³. Work out its height.

Method. Put the known numbers into V = πr²h, then cancel the π and divide to find h.

Write the volume formula with the numbers. We know V and r. The height h is unknown.
180π = π × 6² × h
Work out r².
6² = 36, so 180π = 36π × h
Divide both sides by π. This cancels the π on each side.
180 = 36 × h
Divide by the number in front of h. Undo the multiplication.
h = 180 ÷ 36 = 5 cm

Check. π × 6² × 5 = 180π. ✓

Watch out. Divide by π as well as by r². If the volume is given in terms of π, the π simply cancels.

Worked example 19.6.4

A can is a cylinder with diameter 8 cm and height 12 cm. How many litres of drink does it hold when full? Give your answer to 2 decimal places.

Method. Find the volume in cm³ using πr²h, then change cm³ to litres by dividing by 1000.

Find the radius. The formula uses the radius, which is half of the diameter.
r = 8 ÷ 2 = 4 cm
Write the volume formula and substitute.
V = π × 4² × 12 = π × 16 × 12 = 192π
Use your calculator. Use the π button and keep the whole display.
192π = 603.1858… cm³
Change to litres. 1 litre = 1000 cm³, so divide by 1000.
603.1858… ÷ 1000 = 0.60319… litres
Round to 2 decimal places. The third decimal place decides whether the second goes up.
0.60 litres

Check. Estimate: 3 × 16 × 12 = 576 cm³, so about 0.6 litres. ✓

Watch out. Divide by 1000 to change cm³ to litres. Do not round before the last step.

Worked example 19.6.5

A prism has a semicircle of radius 4 cm as its cross-section. The prism is 10 cm long. Work out its volume in terms of π.

Method. A semicircle is half a circle, so the prism is half of a cylinder with the same radius and length.

Volume of the full cylinder. V = πr²h with r = 4 and h = 10 (the length).
π × 4² × 10 = 160π
Halve it. The prism is half of the cylinder.
160π ÷ 2 = 80π
Write the unit.
80π cm³

Check. Area of a semicircle = ½πr² = 8π, and 8π × 10 = 80π. ✓

Watch out. Halve the volume of the cylinder; do not halve the radius.

Worked example 19.6.6

A metal block has mass 450 g and volume 50 cm³. Work out its density.

Key words. Density tells you how much mass is packed into each unit of volume (for example grams in every cm³). Dense materials like lead feel heavy for their size.

Method. Density = mass ÷ volume.

Write down the formula. Density is mass shared out over each unit of volume.
Density = mass ÷ volume
Substitute. Mass is in g and volume in cm³, so the answer will be in g/cm³.
Density = 450 ÷ 50
Divide.
450 ÷ 50 = 9
Write the unit.
Density = 9 g/cm³

Check. Mass = density × volume = 9 × 50 = 450 g. ✓

Watch out. Divide the mass by the volume, not the other way round.

Worked example 19.6.7

Oak has a density of 750 kg/m³. Work out the mass of 2 m³ of oak.

Key words. Density tells you how much mass is packed into each unit of volume (for example grams in every cm³). Dense materials like lead feel heavy for their size.

Method. Mass = density × volume.

Write down the formula. Rearrange Density = mass ÷ volume by multiplying both sides by the volume.
Mass = density × volume
Substitute. Density is 750 kg/m³: every m³ has a mass of 750 kg.
Mass = 750 × 2
Multiply.
750 × 2 = 1500
Write the unit.
Mass = 1500 kg

Check. 1500 ÷ 2 = 750, the density we started with. ✓

Watch out. Check that the volume is in the same unit as in the density.

Worked example 19.6.8

A solid cylinder has radius 2 cm and height 5 cm. It is made of a metal with density 8 g/cm³. Work out the mass of the cylinder. Give your answer to the nearest gram.

Method. First find the volume of the cylinder, then use mass = density × volume.

Volume of the cylinder. V = πr²h.
V = π × 2² × 5 = 20π cm³
Use the calculator for the volume. Keep the whole display.
20π = 62.8319… cm³
Use mass = density × volume. Every cm³ has a mass of 8 g.
8 × 62.8319… = 502.655… g
Round to the nearest gram.
503 g

Check. Exact: mass = 8 × 20π = 160π g, and 160π is about 502.65. ✓

Watch out. Find the volume first. Do not multiply the density by the radius or the height.

Worked example 19.6.9

The volume of a cylinder is 90π cm³ and its height is 10 cm. Work out the radius.

Method. Put the known numbers into V = πr²h, cancel π, then find r² and square root it.

Write the volume formula with the numbers. We know V and h. The radius r is unknown.
90π = π × r² × 10
Divide both sides by π. The π cancels.
90 = r² × 10
Divide by the height. This leaves r² on its own.
r² = 90 ÷ 10 = 9
Square root. The radius is the number that multiplies by itself to give r².
r = √9 = 3 cm

Check. π × 3² × 10 = 90π. ✓

Watch out. Finish with a square root. r² = 25 means r = 5, not 25.

Watch out

Use the radius, not the diameter, in πr²h. If the diameter is given, halve it first. Square only the radius. And for density, make sure the volume is in the same unit as in the density (cm³ with g/cm³, m³ with kg/m³).

19.7

Surface area of cuboids, prisms and cylinders

Edexcel 1MA1: G16, G17
Grade 3-6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Surface area of cuboids, prisms and cylindersComing soon

The surface area of a solid is the total area of all of its faces. It tells you how much wrapping paper you need for a present, or how much paint covers a tank. Because it is an area, it is measured in square units such as cm² or m². Do not mix it up with volume, which is the space inside (cubic units).

The best way to find a surface area is to think of the net (Section 19.2). The net shows every face flat, so you can work out each area and add.

Cuboid. There are six rectangles in three matching pairs. Surface area = 2lw + 2lh + 2wh. For a cube with edge a all six faces are the same: surface area = 6a².

A 5 cm by 4 cm by 3 cm cuboid and its net. The six rectangles are 5 × 4 (twice), 5 × 3 (twice) and 4 × 3 (twice).

Prisms. A prism has two identical end faces (the cross-section) and a rectangle along each side. Each rectangle is one edge of the cross-section multiplied by the length of the prism. So

surface area = 2 × (area of cross-section) + (perimeter of cross-section) × length

The second part works because all the rectangles have the same length, so together they make one big rectangle: the perimeter by the length.

Cylinders. The net of a cylinder is two circles and a rectangle. The rectangle wraps round the side, so its width is the circumference 2πr and its height is h.

surface area of a cylinder = 2πr² + 2πrh

The first part is the two circles, the second part is the curved surface. If the cylinder has no lid (an open can) there is only one circle, so the surface area is πr² + 2πrh.

The net of a cylinder: two circles of area πr² and a rectangle of area 2πr × h.

Key ideas

Surface area = total area of all the faces, in square units.
Cuboid: 2lw + 2lh + 2wh. Cube: 6a².
Prism: 2 × (cross-section area) + (perimeter of cross-section) × length.
Cylinder: 2πr² + 2πrh (curved surface 2πrh). An open cylinder has only one circle.
Use the net to check that you have included every face.

Worked example 19.7.1

Work out the surface area of this cuboid.

Key words. The surface area of a solid is the total area of all its faces added together. It is measured in square units (cm²), because it is an area.

Method. Surface area is the total area of all the faces. Work out the area of each face (or matching pair of faces) and add.

Picture the six faces. A cuboid has 6 rectangular faces that come in 3 matching pairs: top and bottom, front and back, left and right.
Top and bottom. Each is length × width.
5 × 4 = 20 (and there are 2 of them, so 40)
Front and back. Each is length × height.
5 × 3 = 15 (2 of them: 30)
Left and right. Each is width × height.
4 × 3 = 12 (2 of them: 24)
Add the three pairs. The surface area is every face added together.
40 + 30 + 24 = 94
Write the unit. Area, so square units.
Surface area = 94 cm²

Check. Use the formula 2(lw + lh + wh) = 2 × (20 + 15 + 12) = 2 × 47 = 94. ✓

Watch out. Count every face: forgetting one pair of faces is the most common error. Give the answer in cm², not cm³.

Worked example 19.7.2

Work out the total surface area of this triangular prism. The triangle is right-angled.

Key words. The surface area of a solid is the total area of all its faces added together. It is measured in square units (cm²), because it is an area.

Method. Surface area = the area of the two ends + the area of all the rectangular faces. Using the net helps you not to miss a face.

Picture the faces. A prism has two identical ends (triangles) and a rectangle for every edge of the end: each rectangle is the edge length × the length of the prism (10 cm).
Area of one triangular end. It is a right-angled triangle, so use the two shorter sides as base and height: ½ × base × height.
½ × 3 × 4 = 6 cm²
Double it for the two ends. The two end faces are identical.
2 × 6 = 12 cm²
Find each rectangular face. Each rectangle is an edge of the end multiplied by the length of the prism.
Rectangle on the side of length 3: 3 × 10 = 30
Rectangle on the side of length 4: 4 × 10 = 40
Rectangle on the side of length 5: 5 × 10 = 50
Add all the faces together. Surface area = the two ends + all the rectangles.
12 + 30 + 40 + 50 = 132
Write the unit. Area, so square units.
Surface area = 132 cm²

Check. The rectangles together make (perimeter of an end) × length = 12 × 10 = 120. Then 12 + 120 = 132. ✓

Watch out. Do not forget the two ends, and make sure each rectangle uses the edge length (not the height of the triangle).

Worked example 19.7.3

Work out the surface area of this prism, which has an L-shaped cross-section.

Key words. The surface area of a solid is the total area of all its faces added together. It is measured in square units (cm²), because it is an area.

Method. Surface area = the area of the two ends + the area of all the rectangular faces. Using the net helps you not to miss a face.

Picture the faces. A prism has two identical ends (L-shapes) and a rectangle for every edge of the end: each rectangle is the edge length × the length of the prism (10 cm).
Split the L-shape into two rectangles. We can only use the area formula on simple shapes, so cut the L into two rectangles.
Bottom rectangle: 8 by 2
Top rectangle: 3 by (6 − 2) = 3 by 4
Find the area of each rectangle. Width × height each time.
8 × 2 = 16 cm²
3 × 4 = 12 cm²
Add the two areas. The whole cross-section is the two parts together.
A = 16 + 12 = 28 cm²
Double it for the two ends. The two end faces are identical.
2 × 28 = 56 cm²
Find each rectangular face. Each rectangle is an edge of the end multiplied by the length of the prism.
Bottom edge: 8 × 10 = 80
Right edge: 2 × 10 = 20
Inner horizontal edge: 5 × 10 = 50
Inner vertical edge: 4 × 10 = 40
Top edge: 3 × 10 = 30
Left edge: 6 × 10 = 60
Add all the faces together. Surface area = the two ends + all the rectangles.
56 + 80 + 20 + 50 + 40 + 30 + 60 = 336
Write the unit. Area, so square units.
Surface area = 336 cm²

Check. The rectangles together make (perimeter of an end) × length = 28 × 10 = 280. Then 56 + 280 = 336. ✓

Watch out. Do not forget the two ends, and make sure each rectangle uses the edge length (not the height of the triangle).

Worked example 19.7.4

Work out the total surface area of this cylinder. Give your answer in terms of π.

Key words. The surface area of a solid is the total area of all its faces added together. It is measured in square units (cm²), because it is an area.

Method. Surface area of a cylinder = two circles + the curved surface (a rectangle).

Think of the net. A cylinder opens out into two circles (top and bottom) and a rectangle. The rectangle wraps round the curved side, so its width is the circumference of the circle, 2πr, and its height is h.
Area of the two circles. Area of a circle = πr². There are two of them.
2 × π × 3² = 18π
Area of the curved surface. This is the rectangle: circumference × height = 2πr × h.
2 × π × 3 × 10 = 60π
Add them together. Total surface area = circles + curved surface.
18π + 60π = 78π
Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
Surface area = 78π cm²

Check. Using the factorised formula: 2πr(r + h) = 2 × π × 3 × 13 = 78π. ✓

Watch out. The curved surface is 2πr × h. Do not use the diameter for the circles, and remember both circular ends (unless the question says one end is open).

Worked example 19.7.5

An open tank (no lid) is a cuboid 2 m long, 1.5 m wide and 1 m high. The outside faces are to be painted. Paint costs £4 for each square metre. How much does it cost to paint the five outside faces?

Method. Work out the area of the five faces (base and four sides, no lid) and multiply by the cost per square metre.

Area of the base. length × width.
2 × 1.5 = 3 m²
Area of the two long sides. Each is length × height.
2 × 1 = 2 m², so 2 of them: 4 m²
Area of the two short sides. Each is width × height.
1.5 × 1 = 1.5 m², so 2 of them: 3 m²
Add to get the area to paint. There is no lid so we do not count one.
3 + 4 + 3 = 10 m²
Multiply by the cost. £4 for each m².
10 × 4 = £40

Check. A closed tank would be 13 m²; removing the lid (3 m²) leaves 10 m². ✓

Watch out. Check whether the question includes the top. An open tank has 5 faces.

Worked example 19.7.6

A cylinder has radius r and height 2r. Show that its total surface area is 6πr².

Method. Substitute h = 2r into the surface area formula and simplify.

Write the formula. Two circles plus the curved surface.
SA = 2πr² + 2πrh
Replace h by 2r. The height is twice the radius.
SA = 2πr² + 2πr × 2r
Multiply out the second term. 2 × 2 = 4 and r × r = r².
SA = 2πr² + 4πr²
Add the like terms. 2 + 4 = 6 lots of πr².
SA = 6πr² ✓

Check. With r = 1: cylinder height 2: 2π + 4π = 6π. ✓

Watch out. Replace h first, then simplify.

Watch out

Do not forget faces! Use the net to count them. A cuboid has six faces, a triangular prism has five, and a closed cylinder has two circles and one curved surface. Give the answer in square units, and do not confuse surface area with volume.

Exercise 19I Surface area of cylinders (calculator)

Calculator allowed

FWork out the total surface area of each cylinder. Give your answers to 3 significant figures.

(a)radius 4 cm, height 10 cm

(b)radius 2.5 cm, height 7.2 cm

(c)diameter 12 cm, height 15 cm

Done

PSAn open cylinder (no lid) has radius 6 cm and height 14 cm. Work out the outside surface area to 1 decimal place.

Done

PSA tin of soup is a cylinder of diameter 7.4 cm and height 10.8 cm. The paper label covers the curved surface only. Work out the area of the label, to 3 significant figures.

Done

PSAn open cylindrical tank has radius 0.8 m and height 1.5 m. The outside (base and curved surface) is painted. One litre of paint covers 4 m². How many litres of paint are needed, to the nearest litre?

Done

RA cylinder has radius 3 cm and height 8 cm. Work out the ratio of the curved surface area to the total surface area, in its simplest form.

Done

RThe surface area of a closed cylinder is 400 cm² and its radius is 5 cm. Work out the height to 3 significant figures.

Done

19.8

Pyramids and cones

Edexcel 1MA1: G16, G17
Grade 5-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Pyramids and conesComing soon

Pyramids and cones both come to a point, called the apex. They are special: each takes up exactly one third of the prism or cylinder with the same base and the same height. You can see this by filling a hollow cone with sand and pouring it into a cylinder with the same base and height: it takes exactly three cones to fill the cylinder.

Volume of a pyramid = ⅓ × area of base × vertical height

Volume of a cone = ⅓πr²h

The height h is always the vertical height: straight down from the apex to the base, at right angles to it. It is different from the slant height l, which runs along the sloping surface. The slant height is the hypotenuse of a right-angled triangle, so you can use Pythagoras’ theorem (see Chapter 20): for a cone, l² = r² + h².

Left: a square-based pyramid with vertical height 4 cm (orange) and slant height 5 cm (green). Right: a cone with vertical height 12 cm and slant height 13 cm.

Surface areas. A square-based pyramid has a square base and four identical triangular faces. Each triangle has base b and height equal to the slant height l, so its area is ½bl. A cone has a flat circular base of area πr² and a curved surface whose area is

curved surface area of a cone = πrl, total surface area = πrl + πr²

The curved surface of a cone opens out into a sector (Section 19.2), and this formula comes from the area of that sector. On the Edexcel Higher paper the volume of a cone and the curved surface area of a cone are printed on the formula sheet, but you must be able to use them, and you should learn the pyramid volume rule.

Key ideas

Pyramid: V = ⅓ × base area × vertical height. Cone: V = ⅓πr²h.
Use the vertical height h for volume and the slant height l for the sloping surfaces.
Cone: curved surface area = πrl; total surface area = πrl + πr².
Square-based pyramid surface area = base + 4 triangles = b² + 2bl.
Slant height from Pythagoras: l² = h² + r² (cone) or l² = h² + (b ÷ 2)² (square pyramid).

Worked example 19.8.1

Work out the volume of this square-based pyramid.

Key words. The apex is the top point of a pyramid. The vertical height is measured straight down from the apex to the base. The slant height is the height of a sloping triangular face.

Method. Volume of a pyramid = one third × area of the base × vertical height.

Write down the formula. A pyramid takes up exactly one third of the prism with the same base and the same height.
Volume = ⅓ × base area × vertical height
Find the area of the base. The base is a square 6 cm by 6 cm.
6 × 6 = 36 cm²
Multiply by the vertical height. Use the vertical height, from the top point straight down to the base (not the slanting edge).
36 × 4 = 144
Divide by 3. Taking one third means dividing by 3.
144 ÷ 3 = 48
Write the unit.
V = 48 cm³

Check. The prism with the same base and height has volume 144. Three pyramids fill it: 3 × 48 = 144. ✓

Watch out. Do not forget the ⅓. Use the vertical (perpendicular) height, not the sloping edge or slant height.

Worked example 19.8.2

A pyramid has a rectangular base 8 cm by 5 cm and vertical height 9 cm. Work out its volume.

Key words. The apex is the top point of a pyramid. The vertical height is measured straight down from the apex to the base. The slant height is the height of a sloping triangular face.

Method. Volume of a pyramid = one third × area of the base × vertical height.

Write down the formula. A pyramid takes up exactly one third of the prism with the same base and the same height.
Volume = ⅓ × base area × vertical height
Find the area of the base. The base is a rectangle 8 cm by 5 cm.
8 × 5 = 40 cm²
Multiply by the vertical height. Use the vertical height, from the top point straight down to the base (not the slanting edge).
40 × 9 = 360
Divide by 3. Taking one third means dividing by 3.
360 ÷ 3 = 120
Write the unit.
V = 120 cm³

Check. The prism with the same base and height has volume 360. Three pyramids fill it: 3 × 120 = 360. ✓

Watch out. Do not forget the ⅓. Use the vertical (perpendicular) height, not the sloping edge or slant height.

Worked example 19.8.3

Work out the total surface area of this square-based pyramid. The vertical height is 4 cm.

Key words. The surface area of a solid is the total area of all its faces added together. It is measured in square units (cm²), because it is an area.

Method. Surface area = area of the square base + the area of the 4 triangular faces (each uses the slant height).

Find the slant height first. The sloping faces are triangles. Their height is the slant height, which is the hypotenuse of a right-angled triangle made from the vertical height and half of the base.
Draw the right-angled triangle. The slant height of a sloping face l is the longest side (the hypotenuse) of a right-angled triangle with shorter sides 4 cm and 3 cm.
Use Pythagoras: a² + b² = c². The square on the hypotenuse equals the sum of the squares on the other two sides.
l² = 4² + 3² = 16 + 9 = 25
Square root. Undo the squaring.
l = √25 = 5 cm
Area of the square base.
6 × 6 = 36
Area of one triangular face. Area of a triangle = ½ × base × height. Here the height of the triangle is the slant height.
½ × 6 × 5 = 15
Four triangular faces. A square-based pyramid has 4 identical sloping faces.
4 × 15 = 60
Add the base to the triangles. Surface area = base + 4 triangles.
36 + 60 = 96
Write the unit.
Surface area = 96 cm²

Check. Quick check: 4 triangles = 2 × base × slant = 2 × 6 × 5 = 60; adding the base 36 gives 96. ✓

Watch out. The triangles use the slant height, not the vertical height of the pyramid.

Worked example 19.8.4

Work out the volume of this cone. Give your answer in terms of π. The slant height is 13 cm.

Key words. The slant height (l) of a cone is the distance from the point (apex) down the sloping side to the edge of the base. The vertical height (h) goes straight down from the apex to the centre of the base. They are different lengths.

Method. Volume of a cone = one third of πr²h, where h is the vertical height.

Find the vertical height first. The volume formula needs the vertical height, but we were given the slant height.
Draw the right-angled triangle. The vertical height h is one of the shorter sides. The slant side 13 cm is the hypotenuse and the other short side is 5 cm.
Use Pythagoras with the unknown on a short side. a² + b² = c², so b² = c² − a².
h² = 13² − 5² = 169 − 25 = 144
Square root. Undo the squaring.
h = √144 = 12 cm
Write down the formula. A cone is one third of the cylinder with the same base and height.
V = ⅓πr²h
Substitute.
V = ⅓ × π × 5² × 12
Work out r² and multiply by h. Powers first, then multiply.
5² = 25
25 × 12 = 300
Take one third. Divide by 3.
300 ÷ 3 = 100, so V = 100π
Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
100π cm³

Check. The cylinder with the same r and h has volume 300π; the cone is one third of it: 100π. ✓

Watch out. Use the vertical height h in V = ⅓πr²h. If you are given the slant height, use Pythagoras to find h first.

Worked example 19.8.5

Work out the total surface area of the cone in the last example. Give your answer in terms of π.

Method. Total surface area of a cone = curved surface (πrl) + circular base (πr²).

Write down the formula. Total surface area = the curved surface + the circular base.
Total surface area = πrl + πr²
Curved surface.
π × 5 × 13 = 65π
Circular base.
π × 5² = 25π
Add them. Add the numbers in front of π.
65π + 25π = 90π
Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
90π cm²

Check. Factorise: πr(l + r) = π × 5 × 18 = 90π. ✓

Watch out. The curved surface area uses the slant height l. Add the base circle only if the question asks for the total surface area.

Worked example 19.8.6

A cone has radius 4 cm and vertical height 7 cm. Work out its curved surface area, correct to 3 significant figures.

Method. The area formula uses the slant height, so find it with Pythagoras first, then use πrl.

Find the slant height. It is the hypotenuse of the right-angled triangle with sides 4 and 7.
l² = 4² + 7² = 16 + 49 = 65
Square root. Use the calculator and keep the full display.
l = √65 = 8.06226… cm
Use the curved surface area formula. Curved surface area = πrl.
π × 4 × 8.06226… = 101.3133…
Round to 3 significant figures.
101 cm²

Check. Estimate: π × 4 × 8 = 100.5, which is close. ✓

Watch out. Do not use the vertical height 7 in πrl. The formula needs the slant height, which is longer.

Worked example 19.8.7

A square-based pyramid has volume 96 cm³ and a base of side 6 cm. Work out its vertical height.

Method. Write the volume formula with the numbers, then undo it step by step.

Area of the base. The base is a 6 cm square.
6 × 6 = 36 cm²
Write the formula with the numbers. V = ⅓ × base area × height.
96 = ⅓ × 36 × h = 12h
Divide to find h. ⅓ × 36 = 12, so 12 lots of h make 96.
h = 96 ÷ 12 = 8 cm

Check. ⅓ × 36 × 8 = 12 × 8 = 96. ✓

Watch out. You can multiply both sides by 3 first to remove the ⅓: 288 = 36h.

Watch out

The slant height and the vertical height are different lengths. Volume needs the vertical height h; curved and sloping surface areas need the slant height l. And do not forget the ⅓ in the volume of a cone or pyramid.

19.9

Spheres and hemispheres

Edexcel 1MA1: G17
Grade 5-8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Spheres and hemispheresComing soon

A sphere is a perfectly round solid, like a football or a marble. Every point on its surface is the same distance, the radius r, from the centre. A hemisphere is exactly half a sphere, for example a bowl or a dome. Two formulae describe a sphere, and both are printed on the Edexcel Higher formula sheet:

Volume of a sphere = &frac43;πr³ Surface area of a sphere = 4πr²

Notice that volume uses r³ (three lengths multiplied, cubic units) and surface area uses r² (two lengths, square units). The surface area is exactly four times the area of the widest circle of the sphere.

A sphere of radius 4 cm and a solid hemisphere of radius 6 cm. The flat face of the hemisphere is a circle.

Hemispheres. Halve everything for the volume and the curved surface:

Volume = ⅔πr³ Curved surface area = 2πr² Total surface area = 2πr² + πr² = 3πr²

The extra πr² in the total surface area is the flat circular face. It is only included if the hemisphere is solid (a dome sitting on the ground counts the flat face as part of the surface of the solid; an open bowl does not have it).

Working backwards. If you are given the volume or surface area and must find the radius, write the formula with the numbers, cancel π, then undo the operations in reverse: for volume you finish with a cube root, for surface area with a square root.

Key ideas

Sphere: V = &frac43;πr³, surface area = 4πr².
Hemisphere: V = ⅔πr³, curved surface = 2πr², total surface = 3πr² (including the flat face).
Use the radius: halve the diameter first.
“In terms of π”: leave π as a symbol.
To find r from a volume: cancel π, rearrange, then cube root. From a surface area: cancel π, divide by 4 (or 2 or 3), then square root.

Worked example 19.9.1

Work out the volume of a sphere with radius 3 cm. Give your answer in terms of π.

Key words. A sphere is a perfectly round solid (like a ball). A hemisphere is half of a sphere. The radius is the same distance from the centre to any point on the surface.

Method. Volume of a sphere = &frac43;πr³.

Write down the formula. This is the volume of a sphere (it is on the formula sheet).
V = &frac43;πr³
Substitute.
V = &frac43; × π × 3³
Cube the radius. r³ means r × r × r.
3³ = 3 × 3 × 3 = 27
Multiply by &frac43;. Multiply by 4, then divide by 3 (or divide by 3 first if it is easier).
27 × 4 = 108, then ÷ 3 = 36
Write with π.
V = 36π
Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
36π cm³

Check. Estimate: a sphere fills about half of the cube that just holds it. That cube has volume 216, half is 108, and our answer 113 is a little more than that. ✓

Watch out. r³ means r × r × r, not r × 3. Use the radius: half the diameter.

Worked example 19.9.2

A sphere has diameter 10 cm. Work out its surface area, correct to 3 significant figures.

Key words. A sphere is a perfectly round solid (like a ball). A hemisphere is half of a sphere. The radius is the same distance from the centre to any point on the surface.

Method. Surface area of a sphere = 4πr².

Find the radius. The formulas use the radius, which is half the diameter.
r = 10 ÷ 2 = 5
Write down the formula. The surface area of a sphere is exactly four times the area of its widest circle.
Surface area = 4πr²
Substitute and square the radius first. Powers before multiplying.
4 × π × 5² = 4 × π × 25
Multiply.
4 × 25 = 100, so the area is 100π
Use your calculator. Type the calculation in using the π button on your calculator (not 3.14). Write down the whole display before you round.
4 × π × 5² = 314.1593…
Round the answer. The question wants 3 significant figures. Look at the digit just after the last one you are keeping: 5 or more rounds up, 4 or less leaves it.
314.1593… = 314 cm²

Check. It is 4 times the area of a circle with the same radius: π × 5² = 25π, and 4 × 25π = 100π. ✓

Watch out. Square the radius first, then multiply by 4π. Do not square the 4.

Worked example 19.9.3

Work out the volume of a hemisphere with radius 6 cm. Give your answer in terms of π.

Key words. A sphere is a perfectly round solid (like a ball). A hemisphere is half of a sphere. The radius is the same distance from the centre to any point on the surface.

Method. Volume of a hemisphere = half of &frac43;πr³ = ⅔πr³.

Write the sphere formula. A hemisphere is half of a sphere, so its volume is half of the sphere volume.
Sphere: V = &frac43;πr³
Halve it. ½ of &frac43; is ⅔.
Hemisphere: V = ½ × &frac43;πr³ = ⅔πr³
Substitute and cube the radius.
V = ⅔ × π × 6³, and 6³ = 216
Multiply by ⅔. Multiply by 2, then divide by 3.
216 × 2 = 432, ÷ 3 = 144, so V = 144π
Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
144π cm³

Check. Half the sphere volume: 288π ÷ 2 = 144π. ✓

Watch out. A hemisphere has half the volume of the full sphere: halve the &frac43;, do not halve the radius.

Worked example 19.9.4

Work out the total surface area of a solid hemisphere with radius 3 cm. Give your answer in terms of π.

Key words. A sphere is a perfectly round solid (like a ball). A hemisphere is half of a sphere. The radius is the same distance from the centre to any point on the surface.

Method. Total surface area of a solid hemisphere = curved surface (2πr²) + flat circle (πr²) = 3πr².

Curved surface. Half of the sphere surface area 4πr².
2πr² = 2 × π × 3² = 18π
Flat circular face. A solid hemisphere has a flat circle on the bottom. Area of a circle = πr².
π × 3² = 9π
Add them. Add the numbers in front of π.
18π + 9π = 27π
Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
27π cm²

Check. 3πr² = 3 × π × 9 = 27π. ✓

Watch out. Do not forget the flat circular face of a solid hemisphere: 2πr² + πr² = 3πr².

Worked example 19.9.5

A sphere has volume 36π cm³. Work out its radius.

Method. Put the volume into V = &frac43;πr³, cancel π, get r³ on its own, then cube root.

Write the formula with the numbers. The volume is given as a multiple of π. The radius r is unknown.
36π = &frac43; × π × r³
Divide both sides by π. The π cancels.
36 = &frac43; × r³
Multiply by 3 and divide by 4. This undoes the &frac43;. (Multiply by the reciprocal ¾.)
r³ = 36 × 3 ÷ 4 = 108 ÷ 4 = 27
Cube root. Find the number that multiplies by itself three times to give r³.
r = ∛27 = 3 cm (because 3 × 3 × 3 = 27)

Check. &frac43; × π × 3³ = &frac43; × π × 27 = 36π. ✓

Watch out. Undo &frac43; by multiplying by ¾. Then take a cube root, not a square root.

Worked example 19.9.6

A tennis ball has diameter 6.7 cm. Work out its volume, correct to 3 significant figures.

Key words. A sphere is a perfectly round solid (like a ball). A hemisphere is half of a sphere. The radius is the same distance from the centre to any point on the surface.

Method. Volume of a sphere = &frac43;πr³.

Find the radius. The formulas use the radius, which is half the diameter.
r = 6.7 ÷ 2 = 3.35
Write down the formula. This is the volume of a sphere (it is on the formula sheet).
V = &frac43;πr³
Substitute.
V = &frac43; × π × 3.35³
Cube the radius. r³ means r × r × r.
3.35³ = 3.35 × 3.35 × 3.35 = 37.595375
Multiply by &frac43;. Multiply by 4, then divide by 3 (or divide by 3 first if it is easier).
37.595375 × 4 = 150.3815, then ÷ 3 = 507636000
Write with π.
V = 3007636000π
Use your calculator. Type the calculation in using the π button on your calculator (not 3.14). Write down the whole display before you round.
&frac43; × π × 3.35³ = 157.4791…
Round the answer. The question wants 3 significant figures. Look at the digit just after the last one you are keeping: 5 or more rounds up, 4 or less leaves it.
157.4791… = 157 cm³

Check. Estimate: a sphere fills about half of the cube that just holds it. That cube has volume 300.763, half is 150.3815, and our answer 157 is a little more than that. ✓

Watch out. r³ means r × r × r, not r × 3. Use the radius: half the diameter.

Watch out

Cube the radius for volume (r³) and square it for surface area (r²). Do not confuse the formulae. If you are given the diameter, halve it first. A solid hemisphere has a flat face, so its total surface area is 3πr², not 2πr².

19.10

Composite solids, frustums and similar solids

Edexcel 1MA1: G16, G17, R12
Grade 6-9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Composite solids, frustums and similar solidsComing soon

Many real objects are made by joining simple solids: an ice-cream is a cone with a hemisphere on top, a grain silo is a cylinder with a cone roof, a capsule is a cylinder with a hemisphere at each end. These are composite solids. To find their volume or surface area, split them into the standard solids you know, work out each part, and combine.

Volume. Add the volumes of the parts. If a part is removed (a hole), subtract its volume. Surface area. Add only the surfaces you can touch from the outside. Where two parts are joined, the faces that meet are hidden inside, so they are not counted. For a cylinder with a hemisphere on top, the top circle of the cylinder is hidden by the hemisphere, so the surface area is: base circle + curved surface of the cylinder + curved surface of the hemisphere.

Three composite solids: a cylinder with a hemisphere on top, a cylinder with a cone on top, and an ice-cream shape (a cone with a hemisphere on top).

Frustums. A frustum is what is left of a cone when a smaller cone is cut off its top by a plane parallel to the base (a flower pot and a bucket are frustums). To find its volume, imagine the whole cone, then subtract the small cone that was removed:

volume of a frustum = volume of large cone − volume of small cone

The small cone is a smaller copy (an enlargement) of the large cone, so its measurements are in the same ratio: radius ÷ height is the same for both. Use this to find the missing height.

Left: a frustum. Right: the small cone cut off the top, put back on, to complete the large cone.

Similar solids (see Chapter 21). Two solids are similar if one is an enlargement of the other. If the lengths are multiplied by the scale factor k, then every area is multiplied by k² and every volume (and the mass of a solid of the same material) by k³. A model that is half the size (k = ½) has ¼ of the surface area and ⅛ of the volume.

Key ideas

Composite solid: split into standard solids, then add volumes (subtract any hole).
For surface area count only the outside surfaces: faces that are joined are hidden and are not counted.
Frustum volume = large cone − small cone; find the missing height with similar triangles (radius ÷ height is the same for both cones).
Similar solids with length scale factor k: area scale factor k², volume scale factor k³.
Work backwards by writing the formula with the numbers, cancelling π, then undoing each step.

Worked example 19.10.1

The solid is a cylinder with a hemisphere on top. Work out its volume, in terms of π.

Method. Split the solid into a cylinder and a hemisphere. Work out the two volumes and add them.

Identify the parts. The solid is a cylinder (radius 3 cm, height 7 cm) with a hemisphere of radius 3 cm on top.
Volume of the cylinder. Volume of a cylinder = πr²h.
π × 3² × 7 = π × 9 × 7 = 63π
Volume of the hemisphere. Half a sphere: ½ × &frac43;πr³ = ⅔πr³.
⅔ × π × 3³ = ⅔ × 27π = 18π
Add the parts. The whole solid is made of these parts, so add them. Keep the π terms together.
63π + 18π = 81π
Write the final answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
81π cm³

Check. Estimate with π ≈ 3: the total is about 243, close to the answer 254. ✓

Worked example 19.10.2

Work out the total surface area of the solid in the last example. Give your answer in terms of π.

Method. Add only the outside surfaces: the base circle, the curved surface of the cylinder and the curved surface of the hemisphere. The circle where they join is inside the solid.

Decide which surfaces are on the outside. The top circle of the cylinder is covered by the hemisphere, so it is not counted. The bottom circle is outside.
Surfaces: base circle + cylinder curved surface + hemisphere curved surface
Area of the base circle. Area of a circle = πr².
π × 3² = 9π
Area of the curved surface of the cylinder. A rectangle of width 2πr and height h: area = 2πrh.
2 × π × 3 × 7 = 42π
Area of the curved surface of the hemisphere. Half of the sphere surface 4πr², which is 2πr².
2 × π × 3² = 18π
Add the parts. The whole solid is made of these parts, so add them. Keep the π terms together.
9π + 42π + 18π = 69π
Write the final answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
69π cm²

Check. Estimate with π ≈ 3: the total is about 207, close to the answer 217. ✓

Worked example 19.10.3

An ice-cream is a cone with a hemisphere on top. The cone has radius 3 cm and vertical height 4 cm. Work out the volume of the ice-cream, to 3 significant figures.

Method. Split the ice-cream into a cone and a hemisphere with the same radius. Add the volumes.

Identify the parts. A cone of radius 3 cm and vertical height 4 cm, with a hemisphere of radius 3 cm on top.
Volume of the cone. Volume of a cone = ⅓πr²h (h is the vertical height).
⅓ × π × 3² × 4 = ⅓ × 36π = 12π
Volume of the hemisphere. Half a sphere: ½ × &frac43;πr³ = ⅔πr³.
⅔ × π × 3³ = ⅔ × 27π = 18π
Add the parts. The whole solid is made of these parts, so add them. Keep the π terms together.
12π + 18π = 30π
Use your calculator. Type the calculation in using the π button on your calculator (not 3.14). Write down the whole display before you round.
12π + 18π = 94.2478…
Round the answer. The question wants 3 significant figures. Look at the digit just after the last one you are keeping: 5 or more rounds up, 4 or less leaves it.
94.2478… = 94.2 cm³

Check. Estimate with π ≈ 3: the total is about 90, close to the answer 94. ✓

Worked example 19.10.4

A cuboid 10 cm by 6 cm by 8 cm has a cylindrical hole of radius 2 cm drilled right through it, from the front to the back. Work out the volume of the remaining solid, to 3 significant figures.

Method. The solid is a cuboid with a cylinder removed. Work out both volumes and subtract.

Identify the parts. The hole goes right through the 6 cm depth of the cuboid, so it is a cylinder of radius 2 cm and height (length) 6 cm.
Volume of the cuboid. Volume = length × width × height.
10 × 6 × 8 = 480
Volume of the cylindrical hole. Volume of a cylinder = πr²h.
π × 2² × 6 = π × 4 × 6 = 24π
Subtract the parts. We take away the part that is missing, so subtract. Keep the π terms together.
480 − 24π = 480 − 24π
Use your calculator. Type the calculation in using the π button on your calculator (not 3.14). Write down the whole display before you round.
480 − 24π = 404.6018…
Round the answer. The question wants 3 significant figures. Look at the digit just after the last one you are keeping: 5 or more rounds up, 4 or less leaves it.
404.6018… = 405 cm³

Check. Estimate with π ≈ 3: the total is about 408, close to the answer 405. ✓

Worked example 19.10.5

Work out the volume of this frustum. Give your answer in terms of π.

Key words. A frustum is what is left of a cone when the top is cut off by a plane parallel to the base. Similar shapes have the same shape but different sizes: all their lengths are in the same ratio.

Method. Frustum volume = volume of the large cone − volume of the small cone that was cut off.

Imagine the missing top of the cone. A frustum is a cone with a smaller cone cut off its top. If we put the small cone back we get a whole large cone.
Large cone: radius 6, height 4 + x. Small cone: radius 3, height x.
Use similar triangles to find x. The two cones have the same shape, so (radius ÷ height) is the same for both.
3 ÷ x = 6 ÷ (4 + x)
Cross-multiply and solve. Multiply both sides to remove the fractions.
3(4 + x) = 6x
12 + 3x = 6x
12 = 3x
x = 4
Heights of the two cones. The large cone is the frustum height plus x.
Large cone height = 4 + 4 = 8 cm
Small cone height = 4 cm
Volume of the large cone. V = ⅓πr²h.
⅓ × π × 6² × 8 = 96π
Volume of the small cone. Same formula with the smaller measurements.
⅓ × π × 3² × 4 = 12π
Subtract. The frustum is what is left when the small cone is removed.
96π − 12π = 84π
Write the answer in terms of π. Leave the answer in terms of π: write the number in front of π and do not multiply by 3.14.
V = 84π cm³

Check. Formula check: ⅓πh(R² + Rr + r²) = ⅓ × π × 4 × (36 + 18 + 9) = ⅓ × 252π = 84π. ✓

Watch out. Subtract the volumes (not the heights or radii): large cone − small cone.

Worked example 19.10.6

A solid is a cylinder of radius 3 cm and height h cm with a hemisphere of radius 3 cm on top. Its total volume is 117π cm³. Work out h.

Method. Write the volume of the solid as an expression in h, put it equal to the total, and solve.

Volume of the cylinder. V = πr²h with r = 3.
π × 3² × h = 9πh
Volume of the hemisphere. ⅔πr³ with r = 3.
⅔ × π × 27 = 18π
Add and put equal to the total. The total volume is 117π.
9πh + 18π = 117π
Divide every term by π. The π cancels.
9h + 18 = 117
Solve. Subtract 18, then divide by 9.
9h = 99, so h = 11 cm

Check. Cylinder 9π × 11 = 99π plus hemisphere 18π = 117π. ✓

Watch out. Write the unknown height in the cylinder part only: the hemisphere has fixed size.

Worked example 19.10.7

Two similar cones have heights 4 cm and 6 cm. The volume of the smaller cone is 40 cm³. Work out the volume of the larger cone.

Key words. Two solids are similar if one is an enlargement of the other: every length is multiplied by the same scale factor. See Chapter 21 for more on similar shapes.

Method. For similar solids with length scale factor k: areas scale by k² and volumes by k³.

Find the length scale factor. Divide a length on the solid we are going to by the matching length on the solid we are going from.
6 ÷ 4 = 1.5
Find the volume scale factor. Areas are multiplied by the scale factor squared, and volumes (and masses of the same material) by the scale factor cubed. So we raise 1.5 to the power 3.
1.5³ = 1.5 × 1.5 × 1.5 = 3.375
Multiply the volume. Multiply the known amount by the scale factor.
40 × 3.375 = 135
Write the unit.
135 cm³

Check. Reverse it: 135 ÷ 3.375 = 40. ✓

Watch out. Do not multiply the volume by just the length scale factor 1.5: you must use 1.5³.

Worked example 19.10.8

Two solids are similar. The volumes are 8 cm³ and 27 cm³. Work out the ratio of their lengths and the ratio of their surface areas.

Method. The ratio of volumes is the ratio of lengths cubed. So take the cube root of each part to get the ratio of lengths, then square it for the ratio of areas.

Simplify the volume ratio. Divide by the highest common factor, 1.
8 : 27 = 8 : 27
Cube root of each part. Undo the cubing to go back to lengths.
∛8 : ∛27 = 2 : 3
Square the length ratio for areas. Areas go with the length ratio squared.
2² : 3² = 4 : 9

Check. Cube the lengths again: 2³ : 3³ = 8 : 27. ✓

Watch out. For volumes use the cube root, not the square root.

Watch out

For a composite solid do not add the surface areas of the separate solids: the faces where they are joined are hidden. For similar solids, do not multiply an area or volume by the length scale factor alone: use k² for area and k³ for volume.

Exercise 19L Composite solids and working backwards

Non-calculator

FThe solid is a cylinder with a hemisphere on top. The cylinder has radius 3 cm and height 7 cm. Work out, in terms of π, (a) the volume and (b) the total surface area.

(a)Volume

(b)Total surface area

Done

FA solid is made from a cylinder and a cone. The cylinder has radius 3 cm and height 6 cm. The cone has the same radius and vertical height 4 cm. Work out, in terms of π, (a) the volume and (b) the total surface area.

(a)Volume

(b)Total surface area

Done

FAn ice-cream is a cone with a hemisphere on top. The cone has radius 3 cm and vertical height 4 cm. Work out, in terms of π, (a) the volume of the ice-cream, (b) the area of its curved surface (the area that is outside).

(a)Volume

(b)Outside surface area

Done

PSA cuboid 10 cm by 6 cm by 8 cm has a cylindrical hole of radius 2 cm drilled right through from front to back. Work out the volume of the remaining solid, (a) in terms of π and (b) correct to 3 significant figures.

(a)In terms of π

(b)To 3 significant figures

Done

PSA medicine capsule is a cylinder of radius 2 cm and length 6 cm with a hemisphere of radius 2 cm attached to each end. Work out the total volume, correct to 3 significant figures.

Done

PSA hollow hemispherical bowl has outer radius 6 cm and inner radius 3 cm. Work out the volume of material used to make the bowl, in terms of π.

Done

PSA solid is a cylinder of radius 3 cm and height h cm with a hemisphere of radius 3 cm on top. Its total volume is 117π cm³. Work out h.

Done

RA solid is a cone on top of a cylinder. Both have radius 6 cm. The cylinder has height 10 cm. The total volume is 396π cm³. Work out the vertical height of the cone.

Done

RA cylinder of radius 3 cm and height 7 cm has a hemisphere of radius 3 cm stuck on top. Without using the formula for the whole solid, explain why its surface area (69π cm²) is less than the sum of the surface areas of the separate solid cylinder and solid hemisphere.

Done

Exercise 19M Frustums and similar solids

Non-calculator

FWork out the volume of each frustum. Give your answers in terms of π.

(a)

(b)

Done

PSA bucket is the shape of a frustum. The base has radius 10 cm, the top has radius 15 cm and the height is 20 cm. Work out the capacity of the bucket, (a) in cm³ to 3 significant figures and (b) in litres to 3 significant figures.

(a)Capacity in cm³

(b)Capacity in litres

Done

PSTwo similar solids have lengths in the ratio shown. Work out each missing quantity.

(a)Two similar cones have heights 4 cm and 6 cm. The smaller has volume 40 cm³. Find the volume of the larger.

(b)Two similar bottles have surface areas 100 cm² (height 5 cm) and a height of 8 cm. Find the surface area of the larger bottle.

(c)Two similar statues, made of the same metal, are 10 cm and 15 cm tall. The smaller has mass 250 g. Find the mass of the larger.

(d)Two similar jugs have heights 20 cm and 30 cm. The smaller holds 400 ml. Find the capacity of the larger.

(e)Two similar solids have heights 6 cm and 9 cm. The larger has volume 540 cm³. Find the volume of the smaller.

Done

FThe lengths of two similar solids are in the ratio 6 : 15. Write down the ratio of their volumes.

Done

PSTwo similar solids have volumes 27 cm³ and 64 cm³. Work out (a) the ratio of their lengths and (b) the ratio of their surface areas.

(a)Ratio of lengths

(b)Ratio of surface areas

Done

RThe larger of two similar solids has 8 times the volume of the smaller. How many times larger is its surface area?

Done

RA cone is cut into two pieces by a plane parallel to its base, at one third of the height from the top (apex). The small cone that is cut off is similar to the whole cone. Show that the frustum is 26/27 of the volume of the whole cone.

Done

PSA scale model of a ship is made at a scale of 1 : 50. The model has a hull surface area of 0.4 m² and holds 0.02 m³ of water. Work out (a) the surface area and (b) the volume of the real ship, in m² and m³.

(a)Surface area of the real ship

(b)Volume of the real ship

Done

RA frustum is made by removing the top of a cone. The large cone has radius 12 cm and height 18 cm. The small cone removed has radius 4 cm. Work out the volume of the frustum, in terms of π.

Done

Chapter summary

Skill	Method
Properties of solids	Face = flat surface, edge = where two faces meet, vertex = corner. Euler: V − E + F = 2. n-sided prism: n + 2 faces, 3n edges, 2n vertices. n-sided pyramid: n + 1 faces, 2n edges, n + 1 vertices
Nets	A net folds to make the solid. Cube: 6 squares (11 different nets). Cylinder: rectangle 2πr by h and two circles. Cone (H): sector of radius l and angle 360r ÷ l
Plans and elevations	Plan = from above, front elevation = from the front, side elevation = from the side (front on the left). Use the tallest stack in each column or row
Volume of a cuboid	length × width × height. Cube: side³
Units of volume	1 ml = 1 cm³. 1 litre = 1000 cm³. 1 m³ = 1000 litres = 1,000,000 cm³
Volume of a prism	area of cross-section × length (use the perpendicular height)
Cylinder	V = πr²h. Total surface area = 2πr² + 2πrh. Curved surface = 2πrh
Density	density = mass ÷ volume; mass = density × volume; volume = mass ÷ density
Surface area of a cuboid or prism	Cuboid: 2lw + 2lh + 2wh. Prism: 2 × (cross-section area) + (perimeter of the cross-section) × length
Pyramid and cone (H)	Pyramid V = ⅓ × base area × vertical height. Cone V = ⅓πr²h, curved surface πrl, total πrl + πr². l² = r² + h²
Sphere and hemisphere (H)	Sphere V = &frac43;πr³, surface 4πr². Hemisphere V = ⅔πr³, curved 2πr², total 3πr²
Composite solids and frustums (H)	Split into standard solids; add volumes, subtract holes; count only outside surfaces. Frustum = large cone − small cone (similar triangles)
Similar solids (H)	Length scale factor k: area × k², volume (and mass) × k³

Can you…?

I can name solids and count their faces, edges and vertices, and use Euler’s formula.
I can recognise nets of cubes, cuboids, prisms, pyramids and cylinders.
I can draw the plan and elevations of a solid made of cubes and of standard solids.
I can work out the volume of a cuboid, prism and cylinder, and convert between cm³, litres and m³.
I can use density = mass ÷ volume.
I can work out the surface area of a cuboid, prism and cylinder.
Higher: I can work out the volume and surface area of pyramids, cones, spheres and hemispheres.
Higher: I can solve composite-solid, frustum, working-backwards and similar-solid problems.

Key terms

Solid (3D shape): A shape with length, width and height.
Face: A flat surface of a solid.
Edge: A line where two faces of a solid meet.
Vertex: A corner of a solid (plural: vertices).
Prism: A solid with the same cross-section all along its length.
Cross-section: The face you see when you slice straight through a prism.
Pyramid: A solid with a polygon base and triangular faces meeting at an apex.
Net: A flat pattern that folds up to make a solid.
Plan: The view of a solid from above.
Elevation: A view of a solid from the front or the side.
Volume: The amount of space inside a solid, in cubic units.
Capacity: The amount a container can hold, in litres or millilitres.
Surface area: The total area of all the faces of a solid.
Density: The mass of a material in each unit of volume.
Slant height: The length along a sloping surface of a cone or pyramid.
Frustum: A cone with the top cut off by a plane parallel to the base.
Similar solids: Solids that are enlargements of each other, with all lengths in the same ratio.

Total 56 marks. Allow about 56 minutes.

← Chapter 18: Perimeter, area and circles Chapter 20: Pythagoras and trigonometry →

GCSE Maths Volume, Surface Area and 3D Shapes (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

Properties of 3D solids

Key ideas

Watch out

Exercise 19A Properties of 3D solids

Nets of solids

Key ideas

Watch out

Exercise 19B Nets of solids

Plans and elevations

Key ideas

Watch out

Exercise 19C Plans and elevations

Volume of cuboids, and units of volume and capacity

Key ideas

Watch out

Exercise 19D Volume of cuboids and units of volume

Volume of prisms

Key ideas

Watch out

Exercise 19E Volume of prisms

Volume of cylinders, and density

Key ideas

Watch out

Exercise 19F Cylinders and density: exact answers (non-calculator)

Exercise 19G Cylinders and density (calculator)

Surface area of cuboids, prisms and cylinders

Key ideas

Watch out

Exercise 19H Surface area (non-calculator)

Exercise 19I Surface area of cylinders (calculator)

Pyramids and cones

Key ideas

Watch out

Exercise 19J Pyramids and cones

Spheres and hemispheres

Key ideas

Watch out

Exercise 19K Spheres and hemispheres

Composite solids, frustums and similar solids

Key ideas

Watch out

Exercise 19L Composite solids and working backwards

Exercise 19M Frustums and similar solids

Chapter summary

Can you…?

Key terms

Exercise 19Z Check-up (non-calculator)

Exercise 19X Exam-style questions