Chapter 18 · Foundation tier

GCSE Maths Perimeter, Area and Circles (Foundation): worked examples and questions with answers

By the end of this chapter you will be able to

Find the perimeter and area of rectangles, squares, triangles, parallelograms and trapezia, and explain where each formula comes from
Split compound shapes into simple pieces (or subtract a piece) to find their area and perimeter
Convert between units of area such as cm² and m², and solve area problems about paint, turf, fencing and cost
Name the parts of a circle and calculate circumference and area, with a calculator and in terms of π
Work with semicircles, quarter circles and shapes made from circles, and work backwards from a given area or circumference
Calculate arc lengths and areas of sectors; (Higher) segments, algebraic forms and exact answers with π and surds

You should already know

Multiplying and dividing whole numbers and decimals (Chapter 1)
Squares and square roots (Chapter 2)
Rounding to decimal places and significant figures (Chapter 4)
Substituting into formulae and solving simple equations (Chapters 8 and 9)

Where this leads

Volume and surface area (Chapter 19)
Pythagoras and trigonometry (Chapter 20)
Similar shapes: how area scales (Chapter 21)
Circle theorems (Chapter 26)

Edexcel GCSE (9–1) Mathematics 1MA1, Foundation tier. Content references: G9, G14, G16, G17, G18.

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

18.1 Perimeter and area of rectangles and squares: Worksheet (PDF) · Answers (PDF) · lesson page
18.2 Area of triangles and parallelograms: Worksheet (PDF) · Answers (PDF) · lesson page
18.3 Area of trapezia: Worksheet (PDF) · Answers (PDF) · lesson page
18.4 Compound shapes: Worksheet (PDF) · Answers (PDF) · lesson page
18.5 Units of area and area in context: Worksheet (PDF) · Answers (PDF) · lesson page
18.6 Circles: vocabulary and circumference: Worksheet (PDF) · Answers (PDF) · lesson page
18.7 Area of a circle, semicircles and composite shapes: Worksheet (PDF) · Answers (PDF) · lesson page
18.8 Arc lengths and areas of sectors: 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 3.5 × 4.

Done

FRound 3.14159 to 2 decimal places.

Done

FWork out (a) 6² (b) √49.

(a)6²

(b)√49

Done

FSolve 2x + 6 = 20.

Done

FChange 250 cm into metres.

Done

18.1

Perimeter and area of rectangles and squares

Edexcel 1MA1: G14, G16, G17
Grade 1-3

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

Video lesson: Perimeter and area of rectangles and squaresComing soon

Two measurements describe how big a flat shape is, and they are easy to mix up. The perimeter is the distance all the way round the outside edge. Imagine walking round the edge of a football pitch: the distance you walk is the perimeter. The area is the amount of flat surface inside the shape. Imagine painting the pitch: the amount of paint you need depends on the area.

Because perimeter is a distance, we measure it in ordinary units of length: millimetres (mm), centimetres (cm), metres (m) or kilometres (km). Area is measured by counting squares, so it uses square units: cm², m², and so on. A cm² (“one square centimetre”) is a square whose sides are each 1 cm long.

The shape covers 6 squares, so its area is 6 cm². The thick outline is its perimeter: it is made of 14 unit edges, so the perimeter is 14 cm.

Rectangles

Look at a rectangle 5 cm long and 3 cm wide. We can fill it with 1 cm squares: 3 rows, each with 5 squares. Instead of counting every square we use multiplication: area = length × width. That is 5 × 3 = 15 cm².

To go round the rectangle we add up its four sides. Opposite sides of a rectangle are equal, so the sides are length, width, length, width. The perimeter is 2 × (length + width), or simply add all four sides. Here that is 5 + 3 + 5 + 3 = 16 cm.

Area = 5 × 3 = 15 cm². Perimeter = 5 + 3 + 5 + 3 = 16 cm.

Squares

A square is a rectangle whose sides are all equal, so the area is side × side (which is why we say a number is “squared”) and the perimeter is 4 × side.

Working backwards

Sometimes you are given the area or the perimeter and have to find a missing side. Use the same formulas, but undo them. If the area is 54 cm² and one side is 9 cm, then 9 × (missing side) = 54, so the missing side is 54 ÷ 9 = 6 cm. If you know the perimeter, take away the sides you know and share the rest between the sides that are left.

Key ideas

Perimeter = distance round the outside. Add all the sides. Unit: cm, m, ...
Area = space inside. Unit: cm², m², ...
Rectangle: area = length × width; perimeter = 2 × (length + width).
Square: area = side × side; perimeter = 4 × side.
To find a missing side from the area, divide the area by the side you know.

Watch out

Perimeter and area are different. Adding the sides gives the perimeter (in cm); multiplying gives the area (in cm²). A rectangle 7 cm by 5 cm has perimeter 24 cm but area 35 cm². Always write the units: they tell you which one you found.

Worked example 18.1.1

Work out the area of this rectangle.

Key words. Perimeter is the distance all the way round the outside of a shape (a length, measured in cm, m ...). Area is the amount of flat space inside a shape (measured in square units, cm², m² ...).

Method. Area counts the unit squares inside the shape. A rectangle has length squares in each row and width rows, so we multiply the two.

Write the formula. There are length squares in every row and width rows, so the number of squares is length × width.
Area = length × width
Substitute the numbers. Both lengths must be in the same unit before you multiply.
Area = 9 × 4
Multiply. Work out the product.
9 × 4 = 36
Write the units. Each little square is 1 cm by 1 cm, so the area is measured in square cm.
Area = 36 cm²

Check. Add the rows instead: 9 + 9 + 9 + 9 = 36. ✓

Watch out. Area is in square units (cm²), not cm. Do not add the sides: adding is for perimeter.

Worked example 18.1.2

A rectangular playground is 12.5 m long and 8 m wide. Work out its perimeter.

Method. Perimeter is the distance all the way round. A rectangle has two lengths and two widths, so add all four sides.

Mark the four sides. Opposite sides of a rectangle are equal, so the sides are length, width, length, width.
12.5, 8, 12.5, 8
Add the four sides. The perimeter is the total length of the boundary.
12.5 + 8 + 12.5 + 8 = 41
Write the units. Perimeter is a length, so we use m, not square units.
Perimeter = 41 m

Check. Shortcut: 2 × (length + width) = 2 × 20.5 = 41. ✓

Watch out. Perimeter is a length (m). Do not give it in square units.

Worked example 18.1.3

A rectangle has a perimeter of 38 cm. Its length is 11 cm. Work out its width.

Method. The perimeter is two lengths plus two widths. Take away the two lengths you know, then share what is left between the two widths.

Take away the two lengths. A rectangle has two sides of length 11 cm. Whatever is left of the perimeter is the two widths.
2 × 11 = 22
38 − 22 = 16
Share between the two widths. There are two equal widths, so divide by 2.
16 ÷ 2 = 8
Write the answer. Give the unit.
Width = 8 cm

Check. Add the four sides: 11 + 8 + 11 + 8 = 38 cm. ✓

Watch out. Half the perimeter is length + width, not just the width. Take away the length too.

Worked example 18.1.4

The length of a rectangle is 3 times its width. The perimeter is 48 cm. Work out the area of the rectangle.

Method. Find the width and length from the perimeter, then multiply them for the area.

Call the width w. The length is three times as long, so it is 3w.
width = w, length = 3w
Write the perimeter. Add the four sides: two widths and two lengths.
w + 3w + w + 3w = 8w
8w = 48
Solve for w. Divide both sides by 8.
w = 48 ÷ 8 = 6 cm
Find the length. Length = 3 × width.
3 × 6 = 18 cm
Find the area. Area = length × width.
18 × 6 = 108 cm²

Check. Perimeter check: 2 × (18 + 6) = 2 × 24 = 48 cm. ✓

Watch out. The question asks for the area, not for the width. Read the last line of the question before you finish.

18.2

Area of triangles and parallelograms

Edexcel 1MA1: G16, G14
Grade 2-4

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

Video lesson: Area of triangles and parallelogramsComing soon

Rectangles are easy because you can see the rows of squares. To find the area of a triangle or a parallelogram we turn it into a rectangle, so that we can see why the formula works.

Triangles

Choose one side of the triangle to be the base. The perpendicular height is the distance from the base straight up (at a right angle, 90°) to the opposite corner. The perpendicular height is usually shown as a dotted line with a small square at the bottom. It is not the same as a slanting side.

Now draw a rectangle round the triangle, with the same base and the same height. The triangle always fills exactly half of that rectangle: the two leftover pieces at the sides together make a copy of the triangle. So

area of a triangle = ½ × base × perpendicular height

The shaded triangle is half of the 10 cm by 6 cm rectangle: ½ × 10 × 6 = 30 cm².

For some triangles the height line falls outside the triangle. This happens when one angle is bigger than 90°. You extend the base with a dotted line and measure the height down to it. The formula still works.

The height can be outside the triangle. The base is 10 cm and the perpendicular height is 6 cm, so the area is ½ × 10 × 6 = 30 cm².

Parallelograms

A parallelogram has two pairs of parallel sides. Cut a right-angled triangle off one end and slide it to the other end: the parallelogram becomes a rectangle with the same base and the same perpendicular height. So the area of a parallelogram is

area = base × perpendicular height

(no half this time, because nothing was lost). The slanting sides are used for the perimeter, but never for the area.

Cutting off the orange triangle and moving it to the other end turns the parallelogram into a rectangle of the same area.

Key ideas

The perpendicular height is measured at 90° to the base. It is not a slanting side.
Triangle: area = ½ × base × perpendicular height.
Parallelogram: area = base × perpendicular height.
Perimeter = add all the outside sides (the slanting sides count; the dotted height does not).
To find a missing base or height, work backwards: double the area first for a triangle, then divide.

Watch out

Never use a slanting side as the height, and never forget the half for a triangle. In a triangle with sides 13 cm, 14 cm and 15 cm and a height of 12 cm, the area is ½ × 14 × 12 = 84 cm², not ½ × 14 × 13.

Worked example 18.2.1

Work out the area of this triangle.

Key words. The base is any side you choose. The perpendicular height is the distance from the base to the opposite corner, measured at 90° to the base.

Method. The area of a triangle is half of base × perpendicular height. The height must be at right angles to the base.

Pick the base and the perpendicular height. The height is the straight-up distance from the base to the top corner (the dotted line at a right angle). A slanting side is not the height. The slanting sides are not used for the area.
base = 14 cm, height = 12 cm
Write the formula. A triangle is half of the rectangle with the same base and height.
Area = ½ × base × height
Multiply base by height. This is the area of the rectangle round the triangle.
14 × 12 = 168
Halve it. The triangle is half of that rectangle.
168 ÷ 2 = 84
Write the units. Area is in square cm.
Area = 84 cm²

Check. Use half the base instead: 7 × 12 = 84. ✓

Watch out. Use the perpendicular (dotted) height, never the slanting side, and do not forget to halve.

Worked example 18.2.2

Work out the perimeter of the same triangle.

Method. Perimeter is the distance all the way round, so add the three sides. Use the sloping sides, not the dotted height.

Identify the three sides of the triangle. The dotted perpendicular height is not part of the boundary, so it is left out.
13 cm, 14 cm, 15 cm
Add them. Perimeter is the total length of the boundary.
13 + 14 + 15 = 42
Write the units. A perimeter is a length.
Perimeter = 42 cm

Check. Add in a different order: 15 + 14 + 13 = 42. ✓

Watch out. The dotted height is a helper line. Do not include it in the perimeter.

Worked example 18.2.3

Work out the area of this parallelogram.

Method. A parallelogram can be cut and re-arranged into a rectangle with the same base and perpendicular height, so area = base × height.

Pick the base and the perpendicular height. The height is the distance between the two parallel sides, measured at a right angle. The slanting side is not the height. Ignore the slant length.
base = 12 cm, height = 6 cm
Write the formula. Cut off the triangle at one end and slide it to the other end: you get a rectangle.
Area = base × perpendicular height
Multiply. Work out the product.
12 × 6 = 72
Write the units. Area is in square cm.
Area = 72 cm²

Check. Swap the order: 6 × 12 = 72. ✓

Watch out. Use the perpendicular height (dotted line), not the slanting side. There is no half in the parallelogram formula.

Worked example 18.2.4

Work out the perimeter of the same parallelogram.

Method. A parallelogram has two pairs of equal sides. Add all four sides; the slanting sides count, the dotted height does not.

List the four sides. Opposite sides are equal.
12, 10, 12, 10
Add. The perimeter is the total of the four sides.
12 + 10 + 12 + 10 = 44
Write the units. Perimeter is a length.
Perimeter = 44 cm

Check. 2 × (12 + 10) = 2 × 22 = 44. ✓

Watch out. The perimeter uses the slanting sides, not the perpendicular height.

Worked example 18.2.5

A triangle has an area of 60 cm² and a base of 15 cm. Work out its perpendicular height.

Method. Use Area = ½ × base × height. Put in the area and the base, then undo the operations to find the height.

Write the formula and substitute. Replace the letters with the numbers you know.
60 = ½ × 15 × ?
Undo the half by doubling. Multiplying by ½ halves the number; the opposite is to multiply by 2.
60 × 2 = 120
so 15 × ? = 120
Divide by the base. The opposite of multiplying by 15 is dividing by 15.
? = 120 ÷ 15 = 8
Write the answer. Give the unit.
The height is 8 cm

Check. ½ × 15 × 8 = 60 cm². ✓

Watch out. Remember to double the area first. A common mistake is to divide the area by the other length without doubling.

18.3

Area of trapezia

Edexcel 1MA1: G16, G14
Grade 3-5

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

Video lesson: Area of trapeziaComing soon

A trapezium is a four-sided shape with exactly one pair of parallel sides (sides that always stay the same distance apart). Think of the side view of a bucket, or a slice of a ramp. The two parallel sides have different lengths: call them a and b. The distance between them, measured at a right angle, is the perpendicular height h. The other two sides slant, and we do not need them for the area.

Where the formula comes from

Take two identical trapeziums and turn one upside down. Fitted together they make a parallelogram. Its base is a + b and its height is h, so the parallelogram has area (a + b) × h. Our trapezium is only half of that, so

area of a trapezium = ½ × (a + b) × h

Two copies of the trapezium (blue and orange) make a parallelogram with base a + b.

Another way to remember it: (a + b) ÷ 2 is the average of the two parallel sides, so the area is the average width times the height. Always do the bracket (the addition) first.

The perimeter of a trapezium is the sum of all four sides. This time you do need the two slanting sides, but still not the dotted height.

Key ideas

A trapezium has one pair of parallel sides, a and b. The height h is at 90° to them.
Area = ½ × (a + b) × h. Add the parallel sides first.
Do not use a slanting side as the height.
Perimeter = sum of all four sides (including the two slanting sides).
To find a missing measurement, put the numbers into the formula and undo it step by step.

Watch out

The two slanting sides are not the parallel sides. Parallel sides are the ones that look like a pair of railway lines. In a trapezium whose top is 8 cm and bottom is 22 cm, the 8 and 22 go in the bracket, whatever the slanting sides measure.

Worked example 18.3.1

Work out the area of this trapezium.

Key words. A trapezium is a four-sided shape with exactly one pair of parallel sides (sides that stay the same distance apart).

Method. A trapezium has one pair of parallel sides. Its area is the average of the two parallel sides times the height: ½ × (a + b) × h.

Pick out the two parallel sides and the height. The parallel sides are the two sides that never meet (a and b). The height is the perpendicular distance between them (the dotted line). The slanting sides are not used. Ignore the slant lengths.
a = 8 cm, b = 22 cm, h = 12 cm
Write the formula. Two copies of the trapezium fit together into a parallelogram of base (a + b) and height h; the trapezium is half of it.
Area = ½ × (a + b) × h
Add the parallel sides first. Do the bracket first (BIDMAS).
a + b = 8 + 22 = 30
Multiply by the height. This gives the area of the double (parallelogram).
30 × 12 = 360
Halve it. The trapezium is half of the parallelogram.
360 ÷ 2 = 180
Write the units. Area is in square cm.
Area = 180 cm²

Check. The average of the parallel sides is 30 ÷ 2 = 15; 15 × 12 = 180. ✓

Watch out. Add the two parallel sides before multiplying. Do not use a slanting side as the height.

Worked example 18.3.2

Work out the area of this trapezium.

Key words. A trapezium is a four-sided shape with exactly one pair of parallel sides (sides that stay the same distance apart).

Method. A trapezium has one pair of parallel sides. Its area is the average of the two parallel sides times the height: ½ × (a + b) × h.

Pick out the two parallel sides and the height. The parallel sides are the two sides that never meet (a and b). The height is the perpendicular distance between them (the dotted line). The slanting sides are not used.
a = 7 cm, b = 10 cm, h = 4 cm
Write the formula. Two copies of the trapezium fit together into a parallelogram of base (a + b) and height h; the trapezium is half of it.
Area = ½ × (a + b) × h
Add the parallel sides first. Do the bracket first (BIDMAS).
a + b = 7 + 10 = 17
Multiply by the height. This gives the area of the double (parallelogram).
17 × 4 = 68
Halve it. The trapezium is half of the parallelogram.
68 ÷ 2 = 34
Write the units. Area is in square cm.
Area = 34 cm²

Check. The average of the parallel sides is 17 ÷ 2 = 8.5; 8.5 × 4 = 34. ✓

Watch out. Add the two parallel sides before multiplying. Do not use a slanting side as the height.

Worked example 18.3.3

Work out the perimeter of the trapezium in the last example.

Method. Perimeter is the distance all the way round: add the four sides. The dotted height is not on the edge, so leave it out.

List the four sides. Two parallel sides and two slanting sides.
7, 10, 4, 5
Add them. The perimeter is the total of the four sides.
7 + 10 + 4 + 5 = 26
Write the units. Perimeter is a length.
Perimeter = 26 cm

Check. Add in pairs: (7 + 10) + (4 + 5) = 17 + 9 = 26. ✓

Watch out. Include both slanting sides. The dotted height is not part of the boundary.

Worked example 18.3.4

A trapezium has parallel sides of 7 cm and 11 cm and an area of 90 cm². Work out its height.

Method. Put the area and the parallel sides into the trapezium formula, then undo each operation to find the height.

Write the formula and substitute. Replace the letters with the numbers you know.
90 = ½ × (7 + 11) × h
90 = ½ × 18 × h
Double the area. Undo the ½.
90 × 2 = 180, so 18 × h = 180
Divide by the sum of the parallel sides. Undo the multiplication by 18.
h = 180 ÷ 18 = 10
Write the answer. Give the unit.
Height = 10 cm

Check. ½ × 18 × 10 = 90 cm². ✓

Watch out. Double the area first, then divide by (a + b), not by one of the sides.

Worked example 18.3.5

A trapezium has area 76 cm². One parallel side is 5 cm and the height is 8 cm. Work out the other parallel side.

Method. Put the area, one parallel side and the height into the formula. First find a + b, then take away the side you know.

Write the formula and substitute. Replace the letters with the numbers you know.
76 = ½ × (5 + b) × 8
Double the area, then divide by the height. This undoes the ½ and the × height, and leaves the bracket (a + b).
76 × 2 = 152
152 ÷ 8 = 19
so 5 + b = 19
Subtract the known side. Undo the “+ a”.
b = 19 − 5 = 14
Write the answer. Give the unit.
b = 14 cm

Check. ½ × (5 + 14) × 8 = 76. ✓

Watch out. After doubling and dividing you have a + b, not b. Do not forget to subtract a.

18.4

Compound shapes

Edexcel 1MA1: G16, G17, G14
Grade 3-5

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

Video lesson: Compound shapesComing soon

A compound shape (also called a composite shape) is made by joining simple shapes such as rectangles, triangles and trapezia. You already know how to find the area of each simple shape, so the trick is to break the big shape up. There are two ways.

Method 1: split. Draw a line (or lines) to cut the shape into simple pieces, work out the area of each piece, then add.

Method 2: subtract. Imagine the shape is a bigger simple shape with a piece missing. Work out the area of the big shape, work out the area of the missing piece, then subtract.

An L-shape. Split it into two rectangles, or take the 4 cm by 3 cm corner away from a 10 cm by 8 cm rectangle.

Finding missing lengths

Diagrams often leave some sides unlabelled. In a shape with only right angles, opposite directions must balance: the horizontal sides along the top add up to the same as the horizontal sides along the bottom, and the same for the vertical sides. So a missing side is a bigger length minus the lengths you already know.

Perimeter

For the perimeter, work out every missing side, then add up all the sides of the outline. Do not include any dotted lines you drew inside the shape. For slanting sides you must be told the length (or find it with Pythagoras, Chapter 20).

Triangles on top

A “house” shape is a rectangle with a triangle on top. The triangle needs the perpendicular height, which is shown by a dotted line up the middle, not the slanting roof sides.

A rectangle 12 cm by 5 cm with a triangle of base 12 cm and perpendicular height 8 cm on top.

Key ideas

Split into simple shapes and add the areas, or complete a bigger shape and subtract the missing part.
Work out any missing lengths first (big total minus the known parts).
Perimeter = add every side of the outline. Ignore dotted lines inside the shape.
For a shape with a border, area of border = outer area − inner area. The border is on both sides, so the inner lengths are 2 × (border width) shorter.
Always write the units: cm² for area, cm for perimeter.

Watch out

A common error is to use the wrong lengths for the small rectangles. If an L-shape is 10 cm wide and its notch is 4 cm wide, the part left on that edge is 10 − 4 = 6 cm, not 10 cm. Write the missing lengths on the diagram before you multiply.

Worked example 18.4.1

Work out the area of this L-shape by splitting it into two rectangles.

Method. Split the shape into two rectangles, find each area, then add.

Draw a line to split the shape. A vertical cut makes a tall rectangle on the left and a shorter rectangle on the right. Mark the sides of each.
Left rectangle: width 10 − 4 = 6 cm, height 8 cm
Right rectangle: width 4 cm, height 8 − 3 = 5 cm
Area of the left rectangle. Area = length × width.
6 × 8 = 48 cm²
Area of the right rectangle. Do the same for the second piece.
4 × 5 = 20 cm²
Add the two areas. The whole shape is made of the two pieces.
48 + 20 = 68 cm²

Check. Subtract method: whole rectangle 10 × 8 = 80, take away the notch 4 × 3 = 12: 80 − 12 = 68. ✓

Watch out. Work out the length of each missing side before multiplying. Do not use the labelled lengths of the whole shape for the small pieces.

Worked example 18.4.2

Work out the area of the same L-shape by subtracting a rectangle from a bigger rectangle.

Method. Imagine the shape is a full rectangle with a corner missing. Find the area of the big rectangle, then subtract the area of the missing piece.

Area of the big rectangle. Pretend the notch is filled in. The rectangle is W by H.
10 × 8 = 80 cm²
Area of the missing piece. The notch is a small rectangle.
4 × 3 = 12 cm²
Subtract. The shape is the big rectangle without the missing piece.
80 − 12 = 68 cm²

Check. Split method: (6 × 8) + (4 × 5) = 48 + 20 = 68. ✓

Watch out. Subtract the missing piece. Adding it would count empty space.

Worked example 18.4.3

Work out the perimeter of this shape.

Method. To find the perimeter, first work out every missing side, then add all six sides.

Find the missing horizontal side. The bottom side W is as long as the top side plus the horizontal part of the notch. So the notch part is the bottom minus the top.
10 − 6 = 4 cm
Find the missing vertical side. The left side H is as long as the lower right side plus the vertical part of the notch. So the notch part is the left minus the lower right.
8 − 5 = 3 cm
Add all six sides. Go round the shape once, ticking each side.
10 + 5 + 4 + 3 + 6 + 8 = 36 cm

Check. The steps of the notch add up to the same as the sides they replace, so the perimeter equals the perimeter of the big rectangle: 2 × (10 + 8) = 36. ✓

Watch out. Do not forget the missing sides: an L-shape has six sides, not four.

Worked example 18.4.4

Work out the area of this shape.

Method. Split the shape into a rectangle and a triangle. Find the area of each, then add.

Split the shape. The dotted line splits it into a rectangle at the bottom and a triangle on top.
Rectangle: 12 cm by 5 cm
Triangle: base 12 cm, perpendicular height 8 cm
Area of the rectangle. length × width.
12 × 5 = 60 cm²
Area of the triangle. half × base × perpendicular height.
½ × 12 × 8 = 96 ÷ 2 = 48 cm²
Add. The whole shape is the rectangle plus the triangle.
60 + 48 = 108 cm²

Check. Check the triangle by using half the base: 6 × 8 = 48. ✓

Watch out. The triangle needs the perpendicular height (the dotted line up the middle), not the slanting side.

Worked example 18.4.5

The picture frame is 30 cm by 22 cm. The border is 4 cm wide. Work out the area of the border.

Method. The border is the big outer rectangle with the inner rectangle taken away. Find both areas and subtract.

Area of the outer rectangle. length × width.
30 × 22 = 660 cm²
Find the sides of the inner rectangle. The border is 4 cm wide on both sides, so each inside length is 2 × 4 = 8 cm shorter.
30 − 8 = 22 cm
22 − 8 = 14 cm
Area of the inner rectangle. length × width.
22 × 14 = 308 cm²
Subtract. The border is what is left.
660 − 308 = 352 cm²

Check. Four strips: two of 30 × 4 = 120 and two of 14 × 4 = 56. 240 + 112 = 352. ✓

Watch out. Take 2 × 4 off each length, not just 4: the border is on both sides.

18.5

Units of area and area in context

Edexcel 1MA1: G14, G16, G17
Grade 3-5

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

Video lesson: Units of area and area in contextComing soon

Area units are square units, so converting them is different from converting lengths. Here is the key picture. A square that is 1 m by 1 m has an area of 1 m². But 1 m is 100 cm, so the same square measures 100 cm by 100 cm. Its area is 100 × 100 = 10 000 cm². So

1 m² = 10 000 cm² (not 100!)

In the same way 1 cm = 10 mm, so 1 cm² = 10 × 10 = 100 mm², and 1 km = 1000 m, so 1 km² = 1 000 000 m². The rule: square the length conversion. Going from a big unit to a small unit gives a bigger number (multiply); going from small to big gives a smaller number (divide).

A square metre is 100 cm by 100 cm, which is 10 000 cm².

Area problems in context

Real problems use area to decide how much to buy and what it costs. Follow the same plan every time: (1) change all lengths to the same unit, (2) find the area, (3) divide by the amount one unit covers (one tile, one tin, one roll), (4) round up to a whole number of items, (5) multiply by the price.

Why round up? If you need 7.2 tins you cannot buy 7.2 tins, and 7 tins would not be enough. Use perimeter for things that go round the edge (fences, borders, edging strip) and area for things that cover the surface (paint, turf, tiles, carpet).

A 3.2 m by 2.4 m floor covered in 40 cm square tiles: 8 tiles along and 6 tiles up.

Key ideas

1 m² = 100 × 100 = 10 000 cm²; 1 cm² = 10 × 10 = 100 mm²; 1 km² = 1 000 000 m².
Area conversions: square the length conversion. Big unit to small unit: multiply. Small to big: divide.
Make all lengths the same unit before finding an area.
Items that cover a surface (paint, tiles, turf) use area. Items round the edge (fencing, edging) use perimeter.
If you need part of an item, round up to a whole item, then multiply by the price.

Watch out

To change m² to cm² multiply by 10 000, not by 100. A quick check: a 1 m by 1 m square is 100 cm by 100 cm, and 100 × 100 = 10 000.

Worked example 18.5.1

Change 3.5 m² into cm².

Method. To change units of area, work out how many small squares make one big square: it is the length conversion squared.

Compare the lengths. Area units come from squares, so first see how the lengths compare.
1 m = 100 cm
Square it to compare the areas. 1 m² is a square 1 m by 1 m. Measured in cm it is 100 cm by 100 cm, which is 100 × 100 small squares.
1 m² = 100 × 100 = 10,000 cm²
Convert: m² to cm² gives a bigger number. Each m² is 10,000 cm², so multiply.
3.5 × 10,000 = 35,000 cm²

Check. Go back: 35,000 ÷ 10,000 = 3.5 m². ✓

Watch out. 1 m = 100 cm, but 1 m² = 10,000 cm², not 100. Areas need the conversion squared.

Worked example 18.5.2

Change 450 000 cm² into m².

Method. To change units of area, work out how many small squares make one big square: it is the length conversion squared.

Compare the lengths. Area units come from squares, so first see how the lengths compare.
1 m = 100 cm
Square it to compare the areas. 1 m² is a square 1 m by 1 m. Measured in cm it is 100 cm by 100 cm, which is 100 × 100 small squares.
1 m² = 100 × 100 = 10,000 cm²
Convert: cm² to m² gives a smaller number. It takes 10,000 cm² to make one m², so divide.
450,000 ÷ 10,000 = 45 m²

Check. Go back: 45 × 10,000 = 450,000 cm². ✓

Watch out. 1 m = 100 cm, but 1 m² = 10,000 cm², not 100. Areas need the conversion squared.

Worked example 18.5.3

A rectangular rug is 2.5 m long and 80 cm wide. Work out its area in m².

Method. Change both lengths to the same unit first, then multiply. Here we change centimetres to metres.

Change 80 cm to metres. There are 100 cm in 1 m, so divide by 100.
80 ÷ 100 = 0.8 m
Multiply the lengths. Both are now in metres, so the answer is in m².
2.5 × 0.8 = 2
(25 × 8 = 200, two decimal places, so 2.00)
Write the units. Metres times metres gives square metres.
Area = 2 m²

Check. In cm: 250 × 80 = 20 000 cm², and 20 000 ÷ 10 000 = 2 m². ✓

Watch out. Never multiply 2.5 m by 80 cm directly: change one so the units match.

Worked example 18.5.4

A floor is 3.2 m by 2.4 m. It is covered with square tiles of side 40 cm. How many tiles are needed?

Method. Find how many tiles fit along each side (in the same unit), then multiply. Or divide the floor area by the area of one tile.

Change the floor to centimetres. The tile is in cm, so use cm for everything.
3.2 m = 320 cm and 2.4 m = 240 cm
Tiles along the length. Divide the length by the tile side.
320 ÷ 40 = 8
Tiles along the width. Divide the width by the tile side.
240 ÷ 40 = 6
Total tiles. 8 columns of 6 tiles.
8 × 6 = 48 tiles

Check. Areas: floor 320 × 240 = 76 800 cm²; tile 40 × 40 = 1600 cm²; 76 800 ÷ 1600 = 48. ✓

Watch out. Do not mix metres and centimetres in the same division.

Worked example 18.5.5

A wall is 5 m wide and 2.4 m high. One litre of paint covers 8 m². The wall needs two coats. Paint is sold in 2.5 litre tins costing £18.50. Work out the total cost.

Method. Find the area, double it for two coats, divide by the cover per litre, work out how many tins, and multiply by the price.

Area of the wall. length × height.
5 × 2.4 = 12 m²
Area to paint with two coats. Each coat covers the whole wall.
12 × 2 = 24 m²
Litres needed. One litre covers 8 m².
24 ÷ 8 = 3 litres
Tins needed. Each tin holds 2.5 litres. 3 litres is more than one tin, so find 3 ÷ 2.5 and round up.
3 ÷ 2.5 = 1.2, so 2 tins
Total cost. Multiply by the price of one tin.
2 × 18.50 = £37

Check. 2 tins hold 5 litres, enough for the 3 litres needed; 1 tin (2.5 litres) is not enough. ✓

Watch out. Round up the number of tins. 1.2 tins means you must buy 2.

18.6

Circles: vocabulary and circumference

Edexcel 1MA1: G9, G14, G17
Grade 3-6

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

Video lesson: Circles: vocabulary and circumferenceComing soon

Before calculating with circles we need the right words. These are examined often (Edexcel G9), so learn them.

The circumference is the distance all the way round the circle: the perimeter of a circle.
The centre is the point in the middle. Every point on the circle is the same distance from it.
The radius is a straight line from the centre to the circumference.
The diameter is a straight line from one side to the other through the centre. It is twice the radius: d = 2r.
A chord is a straight line joining two points on the circle. (The diameter is the longest chord.)
A tangent is a straight line that touches the circle at exactly one point.
An arc is a part of the circumference.
A sector is a slice of the circle between two radii and an arc, like a slice of pizza.
A segment is the region between a chord and an arc.

Left: centre, radius, diameter, chord and tangent. Middle: a sector (two radii and an arc). Right: a segment (shaded, between a chord and an arc).

The number π

Measure round any circle and divide by the distance across it. You always get a little more than 3, exactly the same number every time. That number is called π (pi) and it is 3.14159265… (the digits never end or repeat). So the circumference is a little more than 3 diameters.

Unrolled, the circumference of a circle is about 3.14 diameters long.

Circumference C = πd or C = 2πr

The two forms are the same because d = 2r. Use πd when you know the diameter and 2πr when you know the radius.

Two ways to give the answer

With a calculator: use the π button (not 3.14), then round as the question says (decimal places or significant figures). In terms of π (an exact answer): multiply the numbers and leave the π symbol, for example C = 2 × π × 7 = 14π cm. No rounding is needed and it works without a calculator.

Perimeters involving circles

The perimeter of a semicircle is half the circumference plus the straight edge (the diameter). The perimeter of a quarter circle is a quarter of the circumference plus the two radii.

Working backwards

If you know the circumference, divide by π to get the diameter, then halve it for the radius.

Key ideas

Radius = half the diameter. Diameter = 2 × radius.
Circumference C = πd = 2πr. π = 3.14159…
Use the calculator π button, and round only at the end.
“In terms of π” means leave π as a symbol, e.g. 14π.
Semicircle perimeter = ½πd + d. Quarter circle perimeter = ¼(2πr) + 2r.
To find the radius from C: divide by π (that gives d), then halve.

Watch out

Do not mix up the radius and the diameter. If the circle shows 5 cm to the edge from the centre, that is the radius, and C = 2 × π × 5 = 10π, not 5π. And do not forget the straight edge on a semicircle: it is part of the perimeter.

Worked example 18.6.1

Work out the circumference of the circle. Give your answer to 1 decimal place.

Key words. The circumference is the distance round a circle. The diameter is the distance across through the centre; the radius is half of that. π (pi) is the number 3.14159… you get by dividing any circumference by its diameter.

Method. The circumference of a circle is π × diameter, or 2 × π × radius. Find the diameter or radius, substitute, and calculate.

Write down what is given. The diameter goes right across the circle through the centre. (The radius is half of it.)
diameter = 9 cm
Choose the formula. The circumference is the distance round the circle. It is always about 3.14 times the diameter, and that number is π. Using the radius it is 2πr.
C = πd
Substitute the number. Replace the letter with the value you know.
C = π × 9
Use the π button on your calculator. Type it in exactly as written. Do not use 3.14 unless you are told to; the button is more accurate.
C = 28.2743… cm
Round to 1 decimal place. Look at the next digit: 5 or more rounds up, 4 or less leaves it.
C = 28.3 cm

Check. Estimate: π is about 3, so C is about 3 × 9 = 27 cm, close to 28.3. ✓

Watch out. If you are given the diameter use πd. If you are given the radius, double it (2πr). Do not mix them up.

Worked example 18.6.2

Work out the circumference of the circle. Give your answer in terms of π.

Method. The circumference of a circle is π × diameter, or 2 × π × radius. Find the diameter or radius, substitute, and calculate.

Write down what is given. The radius is the distance from the centre to the edge. The diameter goes right across the circle through the centre, so it is twice the radius.
radius = 7 cm
Choose the formula. The circumference is the distance round the circle. It is always about 3.14 times the diameter, and that number is π. Since d = 2r, this can also be written 2πr.
C = 2πr
Substitute the number. Replace the letter with the value you know.
C = 2 × π × 7
Multiply the numbers and keep the π. The question wants an exact answer in terms of π, so do not use a calculator for π. Put the number in front of π.
2 × 7 = 14
C = 14π

Check. With π ≈ 3.14: 14 × 3.14 = 43.96, a sensible circumference for a circle of diameter 14 cm. ✓

Watch out. If you are given the diameter use πd. If you are given the radius, double it (2πr). Do not mix them up.

Worked example 18.6.3

Work out the perimeter of the semicircle. Give your answer to 1 decimal place.

Method. The perimeter of a semicircle is the curved edge (half a circumference) plus the straight edge (the diameter). Do not forget the straight edge.

Curved edge. Half of the circumference πd.
½ × π × 10 = 15.7080… cm
Straight edge. The diameter closes the shape.
diameter = 10 cm
Combine the pieces. Add (or subtract) the full calculator values. Do not round the pieces first, or the answer can be a little out.
Perimeter = 15.7080… + 10 = 25.7080… cm
Round to 1 decimal place. Look at the next digit: 5 or more rounds up.
Perimeter = 25.7 cm

Check. Estimate with π ≈ 3: the answer should be about 25 cm, which is close to the value found. ✓

Watch out. The straight edge is part of the perimeter. Half a circle’s curve alone is not enough.

Worked example 18.6.4

The circumference of a circle is 50 cm. Work out the radius. Give your answer to 2 decimal places.

Method. Use C = πd and rearrange: divide the circumference by π to find the diameter (then halve it for the radius).

Write the formula. Circumference = π × diameter.
C = πd
Substitute the circumference. Put the number in for C.
50 = π × d
Divide both sides by π. To undo “multiply by π” we divide by π (use the calculator button).
d = 50 ÷ π = 15.91549…
Halve it to get the radius. The radius is half the diameter.
r = 15.91549… ÷ 2 = 7.95775…
Round to 2 decimal places. Round only at the end.
r = 7.96 cm

Check. Check: 2 × π × 7.96 = 50.01, close to 50. ✓

Watch out. Divide by π, do not multiply by it. Check that your answer is about a third of the circumference.

Worked example 18.6.5

The circumference of a circle is 26π cm. Work out the radius.

Method. Use C = πd and divide by π to find the diameter, then halve it.

Write the formula. Circumference = π × diameter.
C = πd
Substitute the circumference. The circumference is given in terms of π, so both sides contain π.
26π = π × d
Divide both sides by π. This undoes the multiplication by π and the π’s cancel.
d = 26
Halve the diameter to get the radius. The radius is half the diameter.
r = 26 ÷ 2 = 13 cm

Check. 2 × π × 13 = 26π. ✓

Watch out. Do not divide by 3.14 here: the π cancels exactly.

18.7

Area of a circle, semicircles and composite shapes

Edexcel 1MA1: G9, G17, G14
Grade 4-6

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

Video lesson: Area of a circle, semicircles and composite shapesComing soon

To see why the area of a circle is πr², cut the circle into thin sectors, like slices of pizza, and lay them alternately point up and point down. The pieces fit together to make a shape that is almost a rectangle. The more slices you cut, the closer it is to a rectangle.

The circle cut into 8 sectors, re-arranged. With more sectors the shape becomes a rectangle.

The long side of this rectangle is half of the circumference, ½ × 2πr = πr. The short side is the radius r. So the area is πr × r:

Area of a circle A = πr²

Always use the radius. If you are given the diameter, halve it first. Then square the radius first (BIDMAS: powers before multiplying) and multiply by π. As before, you may be asked for an answer with a calculator (rounded) or in terms of π, such as 36π cm².

Semicircles and quarter circles

A semicircle is half a circle, so its area is ½πr². A quarter circle has area ¼πr².

Composite shapes

As in Section 18.4, split the shape into pieces whose area you know (rectangles, triangles, semicircles...) and add, or subtract a piece that has been cut out (a hole, a corner, a middle circle). Keep full calculator values until the very end and round once. If the question says “in terms of π”, keep π as a symbol and add numbers to numbers and π-terms to π-terms.

Working backwards from the area

If you know the area: divide by π to get r², then take the square root to get r.

Left: a rectangle with a semicircle on top (split and add). Right: a ring (big circle minus small circle).

Key ideas

Area of a circle A = πr². Use the radius; halve a diameter first.
Square the radius first, then multiply by π.
Semicircle: ½πr². Quarter circle: ¼πr².
Composite shapes: split and add, or subtract a piece that is removed.
Do not round until the final answer. “In terms of π” means keep the π symbol.
From the area: divide by π, then square root, to find the radius.

Watch out

The most common mistakes are using the diameter instead of the radius, and multiplying π by the radius before squaring. For a circle of diameter 10 cm the area is π × 5² = 25π cm² = 78.5 cm², not π × 10².

Worked example 18.7.1

Work out the area of the circle, in terms of π.

Key words. The area of a circle is the space inside it, in square units. The radius is the distance from the centre to the edge.

Method. The area of a circle is πr². Make sure you use the radius, square it first, then multiply by π.

Write down the radius. The radius is the distance from the centre to the edge.
radius = 6 cm
Choose the formula. The area of a circle is π times the radius squared. “Squared” means radius × radius.
A = πr²
Substitute and square the radius first. BIDMAS: powers come before multiplying. Square the radius, then multiply by π.
A = π × 6²
6² = 6 × 6 = 36
A = π × 36
Write the answer in terms of π. Leave the π as a symbol and put the number in front.
A = 36π cm²

Check. With π ≈ 3.14: 36 × 3.14 = 113.04 cm², a sensible size. ✓

Watch out. Square the radius first: πr² is π × r × r. It is not (π × r) squared, and it is not 2 × r.

Worked example 18.7.2

Work out the area of the circle. Give your answer to 1 decimal place.

Key words. The area of a circle is the space inside it, in square units. The radius is the distance from the centre to the edge.

Method. The area of a circle is πr². Make sure you use the radius, square it first, then multiply by π.

Find the radius. The formula for area needs the radius. The radius is half the diameter.
radius = 10 ÷ 2 = 5 cm
Choose the formula. The area of a circle is π times the radius squared. “Squared” means radius × radius.
A = πr²
Substitute and square the radius first. BIDMAS: powers come before multiplying. Square the radius, then multiply by π.
A = π × 5²
5² = 5 × 5 = 25
A = π × 25
Use the π button on your calculator. Multiply the squared radius by π.
A = 78.5398… cm²
Round to 1 decimal place. Look at the next digit: 5 or more rounds up.
A = 78.5 cm²

Check. Estimate: π is about 3, so A is about 3 × 25 = 75 cm², close to 78.5. ✓

Watch out. Square the radius first: πr² is π × r × r. It is not (π × r) squared, and it is not 2 × r.

Worked example 18.7.3

Work out the area of the quarter circle, in terms of π.

Method. A quarter circle is one quarter of a full circle, so its area is ¼ × πr².

Write down the radius. Both straight edges of a quarter circle are radii, so the radius is the length shown.
radius = 8 cm
Area of the quarter circle. A quarter of πr². Square the radius first.
¼ × π × 8² = ¼ × π × 64 = 16π cm²
Write the answer with its units. Leave π as a symbol; the units are square units for an area and plain units for a length.
Area = 16π cm²

Check. Estimate with π ≈ 3: the answer should be about 48 cm², which is close to the value found. ✓

Watch out. Quarter means divide by 4 (not by 2).

Worked example 18.7.4

Work out the area of this window shape. Give your answer to 2 decimal places.

Method. Split the shape into a rectangle and a semicircle, find each area and add.

Find the radius of the semicircle. The semicircle sits across the top, so its diameter is the width of the rectangle. The radius is half of that.
diameter = 6 m, radius = 6 ÷ 2 = 3 m
Area of the rectangle. length × width.
6 × 8 = 48 m²
Area of the semicircle. Half of πr², using the radius you found.
½ × π × 3² = ½ × π × 9 = 14.13717… m²
Combine the pieces. Add (or subtract) the full calculator values. Do not round the pieces first, or the answer can be a little out.
Area = 48 + 14.13717… = 62.13717… m²
Round to 2 decimal places. Look at the next digit: 5 or more rounds up.
Area = 62.14 m²

Check. Estimate with π ≈ 3: the answer should be about 61.5 m², which is close to the value found. ✓

Watch out. The semicircle’s diameter is the width of the rectangle: halve it to get the radius.

Worked example 18.7.5

Work out the area of the shaded ring. Give your answer in terms of π.

Method. The shaded ring is the big circle with the small circle taken away: subtract the two areas.

Area of the big circle. Use πR² with the outer radius.
π × 10² = π × 100 = 100π cm²
Area of the small circle. Use the inner radius.
π × 6² = π × 36 = 36π cm²
Combine the pieces. Subtract the parts that are taken away. Whole numbers go with whole numbers and the π terms go with π terms. A number and a π term cannot be merged into one, so leave them as a sum.
Area = 100π − 36π = 64π cm²

Check. Estimate with π ≈ 3: the answer should be about 192 cm², which is close to the value found. ✓

Watch out. Square each radius separately and then subtract. (R − r)² is not the same as R² − r².

Worked example 18.7.6

The area of a circle is 100 cm². Work out the diameter, to 1 decimal place.

Method. Use A = πr² and rearrange: divide the area by π, then take the square root.

Write the formula. Area = π × radius squared.
A = πr²
Substitute the area. Put the number in for A.
100 = π × r²
Divide both sides by π. To undo “multiply by π” divide by π.
r² = 100 ÷ π = 31.8310…
Take the square root. The opposite of squaring is the square root (use the calculator).
r = √31.8310… = 5.6419…
Double the radius. The diameter is twice the radius.
diameter = 2 × 5.6419… = 11.2838…
Round to 1 decimal place. Round only at the end.
diameter = 11.3 cm

Check. Check: π × 5.6² = 98.5, close to 100. ✓

Watch out. Divide by π first, then take the square root. Taking the square root first will give the wrong answer.

18.8

Arc lengths and areas of sectors

Edexcel 1MA1: G9, G18, G17
Grade 4-7

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

Video lesson: Arc lengths and areas of sectorsComing soon

A sector is a slice of a circle between two radii, like a slice of pizza. Its curved edge is an arc. The angle θ between the two radii (at the centre) tells you what fraction of the whole circle the slice is.

A full turn is 360°. So a sector with angle 90° is 90 out of 360, which is ¼ of the circle. A sector with angle 60° is 60/360 = ⅙ of the circle. In general the fraction is θ ÷ 360 (simplify it if you can). Because the slice is that fraction of the circle, both its arc length and its area are the same fraction of the whole circle’s values:

Arc length = (θ/360) × 2πr Sector area = (θ/360) × πr²

Notice what is the same and what is different. Both have the fraction θ/360 in front. The arc length starts from the circumference 2πr (a length), the area starts from πr² (an area). The perimeter of a sector is the arc plus the two straight radii: arc + 2r.

Left: a 60° sector is 60/360 = ⅙ of a circle. Right: a sector can be more than half a circle (a reflex angle at the centre).

Working backwards

If you know the arc length or area, write the formula as an equation and solve for the angle θ or the radius r. For example: if the arc length is 4π cm and r = 12 cm, then 4π = (θ/360) × 2π × 12. Divide by π and simplify to get θ.

Key ideas

Fraction of the circle = θ ÷ 360 (simplify it).
Arc length = (θ/360) × 2πr.
Sector area = (θ/360) × πr².
Perimeter of a sector = arc length + 2 × radius.
Use the same fraction for the arc and the area. Do not mix up 2πr (arc) and πr² (area).
To find θ or r: write the formula as an equation and undo it step by step.

Watch out

The perimeter of a sector is not just the arc. Do not forget the two straight edges (the radii). Also check that you divide the angle by 360, not by 180.

Worked example 18.8.1

Work out the arc length of the sector. Give your answer in terms of π.

Key words. An arc is part of the circumference. A sector is a slice of a circle between two radii and an arc. The angle θ is measured at the centre.

Method. An arc is part of the circumference. Its length is (θ ÷ 360) × 2πr, the fraction of the circle it covers times the whole circumference.

Write the angle as a fraction of a full turn. A whole circle is 360°. An arc with angle θ is the same fraction of the whole circumference as θ is of 360.
60360 = 16
Find the circumference of the whole circle. Use C = 2πr.
C = 2 × π × 6 = 12π
Multiply the fraction by the circumference. The arc is that fraction of the whole circumference.
arc length = 16 × 12π
Simplify and keep π. Multiply the number parts together; leave π.
arc length = 2π cm

Check. As a decimal: 2 × 3.14 = 6.28 cm. Compare with the circumference 37.68 cm: the arc is 16 of it. ✓

Watch out. Divide the angle by 360, not by 180 or 100. Use 2πr (the circumference), not πr² (the area).

Worked example 18.8.2

Work out the area of the sector. Give your answer in terms of π.

Method. The area of a sector is (θ ÷ 360) × πr²: the fraction of the circle it covers times the whole area.

Write the angle as a fraction of a full turn. A sector with angle θ is the same fraction of the whole circle as θ is of 360.
120360 = 13
Find the area of the whole circle. Use A = πr². Square the radius first.
A = π × 9² = 81π
Multiply the fraction by the area. The sector is that fraction of the circle.
sector area = 13 × 81π
Simplify and keep π. Multiply the number parts together.
sector area = 27π cm²

Check. The whole circle has area 81π. The sector is 13 of it, and 13 × 81 = 27. ✓

Watch out. Use πr² (area) for a sector’s area and 2πr (circumference) for an arc length. Do not mix them.

Worked example 18.8.3

Work out the arc length of the sector to 1 decimal place.

Key words. An arc is part of the circumference. A sector is a slice of a circle between two radii and an arc. The angle θ is measured at the centre.

Method. An arc is part of the circumference. Its length is (θ ÷ 360) × 2πr, the fraction of the circle it covers times the whole circumference.

Write the angle as a fraction of a full turn. A whole circle is 360°. An arc with angle θ is the same fraction of the whole circumference as θ is of 360.
75360 = 524
Find the circumference of the whole circle. Use C = 2πr.
C = 2 × π × 8 = 50.2655…
Multiply the fraction by the circumference. The arc is that fraction of the whole circumference.
arc length = 524 × 50.2655…
Round to 1 decimal place. Look at the next digit.
arc length = 10.5 cm

Check. The arc should be 524 of the full circumference 50.27 cm: 524 × 50.27 = 10.47. ✓

Watch out. Divide the angle by 360, not by 180 or 100. Use 2πr (the circumference), not πr² (the area).

Worked example 18.8.4

Work out the perimeter of the sector. Give your answer in terms of π.

Method. The perimeter of a sector is the curved arc plus the two straight radii. Do not forget the two radii.

Arc length. The arc is 16 of the circumference 2πr.
16 × 2 × π × 6 = 2π cm
Two radii. The two straight edges are both radii.
2 × 6 = 12 cm
Combine the pieces. Add. Whole numbers go with whole numbers and the π terms go with π terms. A number and a π term cannot be merged into one, so leave them as a sum.
Perimeter = 2π + 12 = (12 + 2π) cm

Check. Estimate with π ≈ 3: the answer should be about 18 cm, which is close to the value found. ✓

Watch out. A sector has three edges: the arc and two radii. Adding only the arc is a very common slip.

Worked example 18.8.5

The arc length of a sector of radius 12 cm is 4π cm. Work out the angle of the sector.

Method. The arc is θ/360 of the circumference. Write that as an equation and solve for θ.

Write the equation. arc length = (θ ÷ 360) × 2πr.
4π = θ360 × 2 × π × 12
Simplify the right-hand side. Multiply the numbers 2 × r.
4π = θ360 × 24π
Divide both sides by π. The π’s cancel.
4 = θ360 × 24
Divide by 24. Undo the multiplication.
16 = θ360
Multiply by 360. Undo the division by 360.
θ = 16 × 360 = 60°

Check. 60 ÷ 360 = 16, and 16 of the circumference 24π is 4π. ✓

Watch out. Divide by π at the start so the π’s cancel. Then θ is found by multiplying by 360.

Chapter summary

Skill	Method
Perimeter and area of a rectangle / square	Perimeter = 2(l + w) or 4s (add the sides, unit cm). Area = l × w or s × s (unit cm²)
Area of a triangle	½ × base × perpendicular height. Use the dotted height, never a slanting side
Area of a parallelogram	base × perpendicular height
Area of a trapezium	½ × (a + b) × h, where a and b are the parallel sides
Compound shapes	Split and add, or complete a rectangle and subtract. Find missing lengths first. Perimeter: add every outside side
Missing side / height	Put the numbers in the formula and undo it: divide by the other length; double the area first for a triangle or trapezium
Units of area	1 m² = 10 000 cm²; 1 cm² = 100 mm²; 1 km² = 1 000 000 m². Square the length conversion
Area problems in context	Same units, find the area, divide by what one item covers, round UP, multiply by the price. Fencing uses perimeter
Circle words	Centre, radius, diameter (d = 2r), chord, tangent, arc, sector, segment, circumference
Circumference of a circle	C = πd = 2πr. Use the π button, or leave π in the answer (“in terms of π”)
Area of a circle	A = πr². Use the radius; square it first, then multiply by π
Semicircle and quarter circle	Area ½πr² or ¼πr². Perimeter = curved part + straight edges (diameter, or two radii)
Composite shapes with circles	Split and add or subtract; keep full calculator values, round once at the end; or keep π for exact answers
Working backwards from C or A	From C: divide by π to get d, halve for r. From A: divide by π, then square root
Arc length and sector area	Fraction = θ/360. Arc = fraction × 2πr. Area = fraction × πr². Sector perimeter = arc + 2r
(Higher) Segments	Area = sector − triangle (½r² sin θ). Exact: sin 30° = ½, sin 60° = √3/2, sin 90° = 1
(Higher) Algebraic and exact forms	Write the area/perimeter with x, simplify, solve. Keep π and √ as symbols. Show-that: finish on the stated expression

Can you…?

I can find the perimeter and area of rectangles, squares, triangles, parallelograms and trapezia, and say where the formulas come from.
I can split a compound shape into simple pieces (or subtract a piece) and find missing lengths.
I can convert between cm², m², mm² and km² and solve area problems about paint, tiles, turf and fencing.
I can name the parts of a circle: centre, radius, diameter, chord, tangent, circumference, arc, sector and segment.
I can find circumferences and areas of circles, semicircles and quarter circles, with a calculator and in terms of π.
I can find the area of composite shapes made with circles and work backwards to find a radius or diameter.
I can find arc lengths and areas of sectors, and the angle or radius of a sector.
(Higher) I can find the area and perimeter of segments, give exact answers with π and surds, and work with algebraic lengths.

Key terms

Perimeter: The distance all the way round the outside of a shape. Measured in length units such as cm or m.
Area: The amount of flat space inside a shape. Measured in square units such as cm² or m².
Perpendicular height: The distance from the base to the opposite corner (or the parallel side), measured at 90° to the base.
Parallelogram: A four-sided shape with two pairs of parallel sides.
Trapezium: A four-sided shape with exactly one pair of parallel sides.
Compound (composite) shape: A shape made by joining simple shapes, or by cutting a piece out of a simple shape.
Circumference: The distance all the way round a circle.
Radius: A straight line from the centre of a circle to its circumference.
Diameter: A straight line across a circle through the centre. It is twice the radius.
Chord: A straight line joining two points on a circle.
Tangent: A straight line that touches a circle at exactly one point.
Arc: A part of the circumference of a circle.
Sector: A slice of a circle between two radii and an arc.
Segment: The region of a circle between a chord and an arc.
π (pi): The number 3.14159… equal to the circumference of any circle divided by its diameter.
In terms of π: An exact answer that keeps the symbol π, for example 12π cm².
Surd: An exact square root that is not a whole number, such as √3.

Total 40 marks. Allow about 40 minutes.

← Chapter 17: Angles, polygons and symmetry Chapter 19: Volume, surface area and 3D shapes →

GCSE Maths Perimeter, Area and Circles (Foundation): 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?

Perimeter and area of rectangles and squares

Rectangles

Squares

Working backwards

Key ideas

Watch out

Exercise 18A Perimeter and area of rectangles and squares

Area of triangles and parallelograms

Triangles

Parallelograms

Key ideas

Watch out

Exercise 18B Triangles and parallelograms

Area of trapezia

Where the formula comes from

Key ideas

Watch out

Exercise 18C Trapezia

Compound shapes

Finding missing lengths

Perimeter

Triangles on top

Key ideas

Watch out

Exercise 18D Compound shapes

Units of area and area in context

Area problems in context

Key ideas

Watch out

Exercise 18E Units of area and area in context

Circles: vocabulary and circumference

The number π

Two ways to give the answer

Perimeters involving circles

Working backwards

Key ideas

Watch out

Exercise 18F Circles: vocabulary and circumference

Area of a circle, semicircles and composite shapes

Semicircles and quarter circles

Composite shapes

Working backwards from the area

Key ideas

Watch out

Exercise 18G Area of a circle, semicircles and composite shapes

Arc lengths and areas of sectors

Working backwards

Key ideas

Watch out

Exercise 18H Arc lengths and areas of sectors

Chapter summary

Can you…?

Key terms

Exercise 18Z Check-up (non-calculator)

Exercise 18X Exam-style questions