Year 7 · Chapter 11

Perimeter, area, volume and Pythagoras

By the end of this chapter you will be able to

Find the perimeter of rectilinear shapes, and of circles in terms of π
Find the area of triangles, parallelograms and trapezia using the formulae
Find the area of a circle and a composite shape, in terms of π where needed
Find the volume of a cuboid and a prism, and of a cylinder, in terms of π where needed
Count the faces, edges and vertices of common solids
Use Pythagoras' theorem to find a missing side, and check whether a triangle is right-angled

You should already know

Area of a rectangle and perimeter (Year 5 and Year 6, measures chapters)
Area of a triangle and parallelogram (Year 6, measures chapter)
Squares, square roots and the order of operations (Chapter 1)

Where this leads

Trigonometry in right-angled triangles in full, and the sine rule in Year 8
Circle theorems and proof with Pythagoras in Year 8

National Curriculum for England, Key Stage 3 mathematics, Year 7. Content: G1 formulae for the perimeter and area of triangles, parallelograms and trapezia, and the volume of cuboids and prisms (including cylinders), G2 the perimeter of 2-D shapes including circles, the area of a circle, and the area of composite shapes, G13 Pythagoras' theorem (introduced; its use in simple proofs is left to Year 8), G14 sine, cosine and tangent in right-angled triangles, with the standard angles 30, 45 and 60 degrees (full use is in Year 8), G15 the faces, surfaces, edges and vertices of cubes, cuboids, prisms, cylinders, pyramids, cones and spheres.

Exercise P Do you remember?

Non-calculator

FWhat is the area of a rectangle 5 cm by 3 cm?

Done

FWhat is the perimeter of a square with side 4 cm?

Done

FWork out 4² + 3².

Done

FWhat is the square root of 49?

Done

FWork out 2 × 3 × 5.

Done

11.1

Perimeter, including circles

Year 7
G2

Video lesson: Perimeter, including circlesComing soon

A rectangular garden is fenced along its edge. The fence length is the perimeter: the total distance all the way round. For a circular pond, the distance round is the circumference, which is a perimeter too.

The perimeter of a shape is the sum of the lengths of its sides. For a circle, the circumference is π × diameter, or 2π × radius. We often leave π in the answer, so 10π cm is exact.

Key ideas

Perimeter = the total length of all the sides.
Rectangle: perimeter = 2 × (length + width).
Circle: circumference = π × diameter = 2π × radius.
Leave π in the answer when the question asks for an exact answer.

Worked example 11.1.1

Find the perimeter of a rectangle 8 cm by 5 cm.

Method. Add all four sides.

Add the sides. Two lengths and two widths.
8 + 5 + 8 + 5
Add. The total is 26.
26
Write the answer. Give the unit with the answer.
26 cm

Check. 2 × (8 + 5) = 26. ✓

Watch out. Count all four sides.

Worked example 11.1.2

Find the circumference of a circle of radius 5 cm, in terms of π.

Method. Use circumference = 2π × radius.

Write the formula. Circumference = 2πr.
2 × π × 5
Multiply. 2 × 5 = 10.
10π
Write the answer. Give the unit with the answer.
10π cm

Check. The diameter is 10, and π × 10 = 10π. ✓

Watch out. Keep π in the answer.

Worked example 11.1.3

A rectangle 10 cm by 6 cm has a 4 cm by 2 cm corner removed. Find the new perimeter.

Method. A corner cut out does not change the outer edge.

Find the outer edge. The removed corner replaces two sides with two others of the same total length.
2 × (10 + 6)
Add. The outer perimeter is the same as the rectangle.
32
Write the answer. Give the unit with the answer.
32 cm

Check. The removed corner keeps the outside length. ✓

Watch out. Think about the outline, not the area.

Watch out

Watch out: a perimeter is a length, so its unit is cm, not cm². Also, the circumference is not 2π on its own: it is 2π times the radius.

Exam tip

Grown-up tip. Ask the pupil to walk round the edge of a table with a finger and count the sides, to check the perimeter.

11.2

Area of triangles, parallelograms and trapezia

Year 7
G1

Video lesson: Area of triangles, parallelograms and trapeziaComing soon

A sail is shaped like a triangle, a patio is a parallelogram, and a roof panel is a trapezium. Each area is found with a formula, using the perpendicular height.

The area of a triangle is ½ × base × height. A parallelogram has area base × height. A trapezium has parallel sides a and b: its area is ½(a + b) × height.

Key ideas

Triangle: area = ½ × base × perpendicular height.
Parallelogram: area = base × perpendicular height. Do not use the slanted side.
Trapezium: area = ½ × (sum of the parallel sides) × height.
The height is always perpendicular (at a right angle) to the base.

Worked example 11.2.1

Find the area of a triangle with base 6 cm and height 4 cm.

Method. Use area = ½ × base × height.

Write the formula. Triangle area is half the base times the height.
½ × 6 × 4
Multiply. Half of 6 times 4 is 12.
12
Write the answer. Give the unit with the answer.
12 cm²

Check. The triangle is half of a 6 by 4 rectangle: 24 ÷ 2 = 12. ✓

Watch out. Halve the answer.

Worked example 11.2.2

Find the area of a parallelogram with base 8 cm and perpendicular height 5 cm.

Method. Area = base × perpendicular height.

Multiply. Use the height at a right angle to the base.
8 × 5
Write the answer. The area is 40 square centimetres.
40
Write the answer. Give the unit with the answer.
40 cm²

Check. 40 ÷ 8 = 5. ✓

Watch out. Do not use the slanted side.

Worked example 11.2.3

Find the area of a trapezium with parallel sides 3 cm and 7 cm, and height 4 cm.

Method. Area = ½(a + b) × height.

Add the parallel sides. Add 3 and 7.
3 + 7 = 10
Multiply and halve. Area = ½ × 10 × 4.
20
Write the answer. Give the unit with the answer.
20 cm²

Check. The average width is 5, and 5 × 4 = 20. ✓

Watch out. Halve the sum of the parallel sides, then multiply by the height.

Watch out

Watch out: a parallelogram with base 6 and slanted side 8 does not have area 48. The height is the perpendicular distance, which may be 5 here.

Exam tip

Grown-up tip. Ask the pupil to cut a parallelogram from paper and straighten it into a rectangle, to see why base × height is the rule.

11.3

Area of circles and composite shapes

Year 7
G2

Video lesson: Area of circles and composite shapesComing soon

A circular pond is put in the middle of a rectangular patio. To find the paving needed, the area of the rectangle must be found, and the circle taken away. Shapes made from two or more simpler shapes are called composite shapes.

The area of a circle is πr², where r is the radius. A semicircle is half of this, and a quarter circle is a quarter. We usually leave π in the answer.

Key ideas

Area of a circle = π × radius².
A semicircle has half the area; a quarter circle has a quarter.
For a composite shape, split it into simpler shapes and add (or subtract the hole).
Area is in square units; leave π in the answer for an exact value.

Worked example 11.3.1

Find the area of a circle with radius 3 cm, in terms of π.

Method. Use area = πr².

Square the radius. 3² = 9.
9
Multiply by π. Area = 9π.
9π
Write the answer. Give the unit with the answer.
9π cm²

Check. 9π ÷ π = 9 = 3². ✓

Watch out. Square the radius first.

Worked example 11.3.2

A rectangle 6 cm by 4 cm has a semicircle of diameter 4 cm on top. Find the total area in terms of π.

Method. Split into a rectangle and a semicircle.

Rectangle. Area = 6 × 4 = 24.
24
Semicircle. Radius 2, so area = ½ × π × 2² = 2π.
2π
Add. Total = 24 + 2π.
24 + 2π
Write the answer. Give the unit with the answer.
24 + 2π cm²

Check. 2π is about 6.28, so the total is about 30.3 cm². ✓

Watch out. Add the two parts.

Worked example 11.3.3

A ring has an outer radius of 5 cm and an inner radius of 3 cm. Find its area in terms of π.

Method. Subtract the hole from the outside.

Outer area. π × 5² = 25π.
25π
Inner area. π × 3² = 9π.
9π
Subtract. Ring = 25π − 9π.
16π
Write the answer. Give the unit with the answer.
16π cm²

Check. 16π + 9π = 25π. ✓

Watch out. Subtract the inner circle.

Watch out

Watch out: the area of a circle is πr², not πd or 2πr. Square the radius, not the diameter.

Exam tip

Grown-up tip. Ask the pupil to draw the composite shape and shade the parts they add and the parts they take away, before any calculation.

11.4

Volume of cuboids, prisms and cylinders

Year 7
G1, G15

Video lesson: Volume of cuboids, prisms and cylindersComing soon

A fish tank is 60 cm long, 30 cm wide and 40 cm high. Its volume tells us how much water it holds. Volume is measured in cubic units, such as cm³.

The volume of a cuboid is length × width × height. A prism has the same cross-section all the way along: its volume is the area of the cross-section × the length. A cylinder is a prism with a circle as its cross-section: its volume is πr² × height.

Key ideas

Cuboid: volume = length × width × height.
Prism: volume = area of the cross-section × length.
Cylinder: volume = πr² × height.
1 cm³ = 1 ml, and 1000 cm³ = 1 litre.

Worked example 11.4.1

Find the volume of a cuboid 4 cm by 3 cm by 2 cm.

Method. Volume = length × width × height.

Choose the formula. Decide which formula the question needs before calculating.
24
Multiply. 4 × 3 × 2.
24
Write the answer. Give the unit with the answer.
24 cm³

Check. 24 ÷ 2 = 12 = 4 × 3. ✓

Watch out. Multiply all three.

Worked example 11.4.2

A prism has a triangular cross-section of area 6 cm² and length 10 cm. Find its volume.

Method. Volume = cross-section area × length.

Choose the formula. Decide which formula the question needs before calculating.
60
Multiply. 6 × 10.
60
Write the answer. Give the unit with the answer.
60 cm³

Check. 60 ÷ 10 = 6. ✓

Watch out. Use the area of the cross-section.

Worked example 11.4.3

Find the volume of a cylinder with radius 2 cm and height 5 cm, in terms of π.

Method. Volume = πr² × height.

Find the area of the base. π × 2² = 4π.
4π
Multiply by the height. 4π × 5.
20π
Write the answer. Give the unit with the answer.
20π cm³

Check. 20π ÷ 5 = 4π. ✓

Watch out. Find the base area first.

Watch out

Watch out: the volume of a cylinder is πr² × h, not 2πr × h. Square the radius, and use the height.

Exam tip

Grown-up tip. Ask the pupil to fill a box with cubes, layer by layer, to see that volume = base area × height.

11.5

Faces, edges and vertices of solids

Year 7
G15

Video lesson: Faces, edges and vertices of solidsComing soon

A gift box is a cuboid, a tent is a pyramid, and an ice cream cone is a cone. To describe a solid, count its faces (flat or curved surfaces), its edges (where two faces meet), and its vertices (corners).

A prism has two identical ends and flat sides. A pyramid has a base and triangular faces that meet at one point, the apex. A sphere has no edges, vertices or flat faces.

Key ideas

Faces: flat or curved surfaces. Edges: where faces meet. Vertices: corners.
Cube: 6 faces, 12 edges, 8 vertices. Cuboid: the same counts as a cube.
Triangular prism: 5 faces, 9 edges, 6 vertices. Square pyramid: 5 faces, 8 edges, 5 vertices.
Cylinder: 2 circular faces and a curved surface, with 2 circular edges and no vertices.

Worked example 11.5.1

How many edges does a cube have?

Method. Count the edges where two faces meet.

Choose the formula. Decide which formula the question needs before calculating.
4 + 4 + 4 = 12
Count. A cube has 4 edges on the top, 4 on the bottom and 4 vertical edges.
4 + 4 + 4 = 12
Write the answer. Give the unit with the answer.
12

Check. Each of the 6 faces has 4 edges, and each edge is shared by 2 faces: 6 × 4 ÷ 2 = 12. ✓

Watch out. Each edge is counted once.

Worked example 11.5.2

How many vertices does a square-based pyramid have?

Method. Count the corners.

Choose the formula. Decide which formula the question needs before calculating.
4 + 1 = 5
Count. The square base has 4 corners, and the apex is one more.
4 + 1 = 5
Write the answer. Give the unit with the answer.
5

Check. Four base corners and one apex. ✓

Watch out. Include the apex.

Worked example 11.5.3

How many faces does a triangular prism have?

Method. Count the flat faces.

Choose the formula. Decide which formula the question needs before calculating.
2 + 3
Count. Two triangular ends and three rectangular sides.
2 + 3
Write the answer. Give the unit with the answer.
5

Check. 2 + 3 = 5 faces. ✓

Watch out. Count the ends and the sides separately.

Watch out

Watch out: a cylinder is not a polyhedron, because it has a curved surface. A cone also has a curved face.

Exam tip

Grown-up tip. Ask the pupil to make a cube from straws and a bag of marshmallows, counting the straws (edges) and marshmallows (vertices).

11.6

Pythagoras' theorem and right-angled triangles

Year 7
G13, G14

Video lesson: Pythagoras' theorem and right-angled trianglesComing soon

A ladder 13 m long leans against a wall. Its foot is 5 m from the base of the wall. How high up the wall does it reach? In a right-angled triangle, the sides are linked by a simple rule, Pythagoras' theorem.

In a right-angled triangle, the longest side, opposite the right angle, is the hypotenuse. Pythagoras' theorem says: a² + b² = c², where c is the hypotenuse and a and b are the other two sides.

The sine, cosine and tangent ratios link an angle with two sides. For the standard angles, sin 30° = ½, cos 60° = ½ and tan 45° = 1.

Key ideas

Pythagoras: a² + b² = c², with c the hypotenuse (opposite the right angle).
To find the hypotenuse: add the squares of the two short sides, then square root.
To find a short side: subtract the square of the known short side from c², then square root.
Check if a triangle is right-angled: if the squares of the two shorter sides add to the square of the longest, it is.

Worked example 11.6.1

A right-angled triangle has sides 6 cm and 8 cm. Find the hypotenuse.

Method. Use c² = a² + b².

Square and add. 6² + 8² = 36 + 64.
100
Square root. c = √100.
c = 10
Write the answer. Give the unit with the answer.
10 cm

Check. 10² = 100 = 36 + 64. ✓

Watch out. The hypotenuse is the longest side.

Worked example 11.6.2

The hypotenuse is 13 cm and one short side is 5 cm. Find the other short side.

Method. Rearrange: b² = c² − a².

Subtract. 13² − 5² = 169 − 25.
144
Square root. b = √144.
b = 12
Write the answer. Give the unit with the answer.
12 cm

Check. 5² + 12² = 25 + 144 = 169 = 13². ✓

Watch out. Subtract the squares, not the sides.

Worked example 11.6.3

In a right-angled triangle, the hypotenuse is 10 cm and the angle is 30°. Find the side opposite the 30° angle.

Method. Use sin 30° = opposite ÷ hypotenuse.

Write the ratio. sin 30° = opposite ÷ 10.
opposite = 10 × sin 30°
Use sin 30° = ½. Half of 10 is 5.
10 × ½ = 5
Write the answer. Give the unit with the answer.
5 cm

Check. sin 30° = 5 ÷ 10 = ½. ✓

Watch out. Use the exact value for 30°.

Watch out

Watch out: Pythagoras only works for right-angled triangles. The hypotenuse is always the longest side, and it is always opposite the right angle.

Exam tip

Grown-up tip. Ask the pupil to test the rule with a 3, 4, 5 triangle drawn on squared paper, and check that the squares add up.

Chapter summary

Skill	Method
Perimeter	Add all the sides; circle circumference = π × diameter
Area of triangle and parallelogram	½ × base × height; base × perpendicular height
Area of trapezium	½(a + b) × height
Area of circle	πr²; semicircle half; quarter a quarter; composite shapes by adding or subtracting
Volume	Cuboid: l × w × h; prism: cross-section × length; cylinder: πr² × h
Solids	Faces, edges and vertices; cube 6, 12, 8
Pythagoras	a² + b² = c², c is the hypotenuse; check a triangle with the squares
Trigonometry (standard angles)	sin 30° = ½; cos 60° = ½; tan 45° = 1

Can you…?

I can find the perimeter of shapes, including circles in terms of π
I can find the area of triangles, parallelograms and trapeziums
I can find the area of circles and composite shapes
I can find the volume of cuboids, prisms and cylinders
I can count the faces, edges and vertices of solids
I can use Pythagoras' theorem, and check whether a triangle is right-angled

Key terms

perimeter: the total distance round the outside of a shape
circumference: the perimeter of a circle
composite shape: a shape made from two or more simpler shapes
prism: a solid with the same cross-section all the way along its length
hypotenuse: the longest side of a right-angled triangle, opposite the right angle
Pythagoras' theorem: in a right-angled triangle, a² + b² = c²
face: a flat or curved surface of a solid
vertex: a corner of a solid (plural vertices)

Total 21 marks. Allow about 21 minutes.

← Chapter 10: Transformations, congruence and similarity Chapter 12: Probability →

Perimeter, area, volume and Pythagoras

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

Perimeter, including circles

Key ideas

Watch out

Exam tip

Exercise 11A Perimeter

Area of triangles, parallelograms and trapezia

Key ideas

Watch out

Exam tip

Exercise 11B Area of triangles, parallelograms and trapezia

Area of circles and composite shapes

Key ideas

Watch out

Exam tip

Exercise 11C Area of circles and composite shapes

Volume of cuboids, prisms and cylinders

Key ideas

Watch out

Exam tip

Exercise 11D Volume of cuboids, prisms and cylinders

Faces, edges and vertices of solids

Key ideas

Watch out

Exam tip

Exercise 11E Faces, edges and vertices

Pythagoras' theorem and right-angled triangles

Key ideas

Watch out

Exam tip

Exercise 11F Pythagoras and right-angled triangles

Chapter summary

Can you…?

Key terms

Exercise 11Z Check-up

Exercise 11X Practice test questions