Area Of A Triangle | National 5 Maths - ot

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National 5 MathsArea of a Triangle For textbooks and tutoring, we recommend:

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• Topic content
• Key ideas
• Worked examples
• Past paper questions
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• Notes and videos

Topic content

• Calculate the area of a triangle using the trigonometric formula:

$$ \large Area = \small \frac{1}{2} \small\ \large ab\ sin\ C \normalsize $$

Textbook page numbers

• Zeta National 5+ Maths pp.224-225
• TeeJay Maths Book N5 pp.73-75
• Leckie National 5 Maths pp.293-298

Buy our favourite textbook

Zeta National 5+ MathsClear and comprehensive.Progressive exercises.Includes answers. Buy from Zeta Press

Key ideas

• This formula works in any triangle: right-angled or non-right angled.
• This formula only needs us to know two the lengths of two sides and the angle enclosed between them.
• Unlike the other formula for the area of a triangle, \(A\!\,=\!\,\frac{1}{2}bh \), we don't need to know the vertical height.

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Example 1 (calculator)

In the triangle below, BC = 9 mm, AC = 15 mm and angle C = 60°.

Calculate the area of triangle ABC.

Show answer

In this example, the vertex letters A, B and C exactly match the formula. So:

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2} \ \normalsize ab\ sin\ C \\[6pt] &=& \small\frac{1}{2}\normalsize \times 9 \times 15\ sin\ 60^\circ \\[6pt] &\approx& 58.5\ \small\textsf{mm²} \\[12pt] \end{eqnarray} $$

Video solution by Clelland Maths

Example 2 (non-calculator)

In the triangle below, AB = 5 cm, BC = 3 cm and sin B = \( \frac{1}{\sqrt{2}} \).

Calculate the area of triangle ABC, expressing your answer as a fully simplified surd with a rational denominator.

Show answer

This question would be suitable for Paper 1. We have been told sin B, not B itself.

In this example, the vertex letters A, B and C don't match the formula. So we can rewrite the formula using the different letters.

The SQA has never combined area of a triangle with the surds topic, but there is no reason why they shouldn't!

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\small \ \normalsize ac\ sin\ B\:\ \small\textsf{(same formula really!)}\normalsize\\[6pt] &=& \small\frac{1}{2}\normalsize \times 3 \times 5\ \times \small \frac{1}{\sqrt{2}}\normalsize \\[6pt] &=& \frac{15}{2 \sqrt{2}} \\[6pt] &=& \frac{15}{2 \sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}} \ \small\textsf{(rationalising)}\normalsize\\[6pt] &=& \frac{15\sqrt{2}}{4}\ \small\textsf{cm²} \end{eqnarray} $$

Video solution by Clelland Maths

Example 3 (calculator)

Calculate the area of an equilateral triangle with side lengths of 5 cm.

Show answer

This question relies on our knowledge that equilateral triangles contain three 60° angles.

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\small \ \normalsize ab\ sin\ C \\[6pt] &=& \small\frac{1}{2}\normalsize \times 5 \times 5 \times sin\ 60^\circ \\[6pt] &\approx& 10.8\ \small\textsf{cm²} \end{eqnarray} $$

Video solution by Clelland Maths

Example 4 (calculator)

Calculate the area of a regular hexagon with side length 10 cm, giving your answer correct to 3 significant figures.

Show answer

This question relies on our knowledge that a regular hexagon consists of 6 equilateral triangles.

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& 6 \times \small\frac{1}{2}\normalsize\small \ \normalsize ab\ sin\ C\\[6pt] &=& 6 \times \small\frac{1}{2}\normalsize \times 10 \times 10 \times sin\ 60^\circ \\[6pt] &=& 259.807... \\[6pt] &\approx& 260\ \small\textsf{cm²}\ \small\textsf{(correct to 3 s.f.)}\normalsize \end{eqnarray} $$

Video solution by Clelland Maths

Example 5 (calculator)

A triangle of area 50 cm2 has two sides with lengths 12 cm and 10 cm. The angle between these two sides is acute. Calculate this angle.

Show answer

This question requires us to work backwards. For convenience, let's call the unknown angle C°.

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\small \ \normalsize ab\ sin\ C\\[6pt] 50 &=& \small\frac{1}{2}\normalsize \times 12\times 10\ sin\ C\\[6pt] 100 &=& 120\small \ \normalsize sin\ C \\[6pt] sin\ C&=& \small\frac{100}{120}\normalsize \\[6pt] C&=& sin^{-1}\ \small\ \frac{100}{120}\normalsize \\[6pt] &\approx& 56.4^\circ\ \end{eqnarray} $$

Video solution by Clelland Maths

Example 6 (calculator)

A triangle of area 50 cm2 has two sides with lengths 12 cm and 10 cm. The angle between these two sides is obtuse. Calculate this angle.

Show answer

This is almost the same as the previous example, except now the angle is obtuse, not acute.

The working is exactly the same, all the way down to \(C \approx 56.4^\circ\), which cannot be correct as \(C\) is obtuse.

However, the final step is simple if your knowledge of trig equations is solid. \(C\) is the 2nd quadrant (obtuse) angle related to \(56.4^\circ\), so it's \(180-56.4^\circ=123.6^\circ\).

Video solution by Clelland Maths

Example 7 (calculator)

SQA National 5 Maths 2015 P2 Q11

The top of a table is in the shape of a regular hexagon. The three diagonals of the hexagon which are shown as dotted lines in the diagram below each have length 40 centimetres.

Calculate the area of the top of the table.

Show answer

A hexagon comprises six equilateral triangles, which contain 60° angles.

Each diagonal of the hexagon is 40 cm so each side of the triangles is 20 cm.

Now we can find the area of each triangle:

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\small \ \normalsize ab\ sin\ C\\[6pt] &=& \small\frac{1}{2}\normalsize \times 20\times 20\ sin\ 60^\circ\\[6pt] &\approx& 173.205... \end{eqnarray} $$

So the area of the hexagon is \(6\times 173.205... \approx\ 1039.2\) cm2

Video solution by Clelland Maths

Example 8 (non-calculator)

SQA National 5 Maths 2017 P1 Q7

In triangle DEF: • DE = 8 centimetres • EF = 12 centimetres • sin E \(=\frac{2}{3}\)

Calculate the area of triangle DEF.

Show answer

The sides DE and EF enclose the angle E°, so we can find the area as follows:

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\times\small \ \normalsize \small\textsf{DE}\normalsize\times\small\textsf{EF}\normalsize\times sin\,E\\[6pt] &=& \small\frac{1}{2}\normalsize \times 8\times 12\times\small\frac{2}{3}\\[6pt] &=& \left(\small\frac{1}{2}\normalsize\times\small\frac{2}{3}\right) \times \left(8\times 12\right)\\[6pt] &=& \small\frac{1}{3}\normalsize \times 96\\[6pt] &=& 32\ \small\textsf{cm²} \end{eqnarray} $$

Video solution by Clelland Maths

Example 9 (calculator)

SQA National 5 Maths 2019 P2 Q3

The diagram shows triangle PQR.

• PR = 45 centimetres • PQ = 70 centimetres • Angle QPR = 129° Calculate the area of triangle PQR.

Show answer

This is a straightfoward 2-mark use of the area formula.

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\small \ \normalsize ab\ sin\ C\\[6pt] &=& \small\frac{1}{2}\normalsize \times 45\times 70\ sin\ 129^\circ\\[6pt] &=& 1224.0048...\\[6pt] &\approx& 1224.0\ \small\textsf{cm²} \end{eqnarray} $$

Video solution by Clelland Maths

Example 10 (calculator)

SQA National 5 Maths 2023 P2 Q15

In the diagram: • AC is perpendicular to BC • AB = 18 centimetres • BD = 6 centimetres • BC = 8 centimetres.

The area of triangle ADE is 160 square centimetres. Calculate the length of AE.

Show answer

This 4-mark question first needs SOH CAH TOA in triangle ABC to find the sine of the angle at A.

$$ \begin{eqnarray} sin\,A &=& \small\frac{8}{18}\normalsize\\[6pt] &=& \small\frac{4}{9}\normalsize \end{eqnarray} $$

Note that we don't need to find angle A itself, because the area formula uses sin A°, not A°.

Now we use the area formula for triangle ADE, as follows:

$$ \begin{eqnarray} \small\textsf{Area}\normalsize &=& \small\frac{1}{2}\normalsize\times\small\textsf{AD}\normalsize\times\small\textsf{AE}\normalsize\times sin\,A\\[8pt] 160&=& \small\frac{1}{2}\normalsize \times 24\times \small\textsf{AE}\normalsize\times\small\frac{4}{9}\\[8pt] 160&=& \small\frac{2}{9}\normalsize \times 24\times \small\textsf{AE}\normalsize\\[8pt] 160\times 9&=& 2 \times 24\times \small\textsf{AE}\normalsize\\[8pt] 1440 &=& 48\times\small\textsf{AE}\normalsize\\[8pt] \small\textsf{AE}\normalsize &=& \small\frac{1440}{48}\normalsize\\[8pt] &=& 30\ \small\textsf{cm} \end{eqnarray} $$

Video solution by Clelland Maths

Buy N5 Maths practice papers

Hodder: N5 Maths Questions & Papers CGP: N5 Maths SQA Exam Practice

Past paper questions

• All past paper questions by topic

• 2014 Paper 2 Q13 (with sector area) • 2015 Paper 2 Q11 (hexagon) • 2017 Paper 1 Q7 (non-calculator) • 2018 Paper 2 Q17 (with sector area) • 2019 Paper 2 Q3 • 2021 Paper 2 Q8 (with sector area) • 2022 Paper 2 Q6 • 2023 Paper 2 Q15 (finding a length) • 2025 Paper 1 Q5 (non-calculator)

Buy our favourite N5 textbook

Zeta National 5+ MathsClear and comprehensive.Progressive exercises.Includes answers. Buy from Zeta Press

Triangle area worksheets

National5Maths.co.uk worksheet 1. Triangle area (with solutions)

Maths Genie worksheet • Area of a triangle (Solutions)

Corbettmaths worksheet • With sine/cosine rules (Answers)

CJ Maths worksheet • Area of a triangle (no answers)

Larkhall Academy exercises • Pages 10-11 Ex 7 (no answers)

Buy N5 Maths revision guides

How to Pass: National 5 Maths BrightRED: N5 Maths Study Guide

Notes and videos

Videos - Maths180.com

Video - Tynecastle HS

Video - Mr Graham Maths

Videos - YouKenMaths 1. Area of a triangle 2. Hexagon, decagon etc

PowerPoint - MathsRevision.com

Notes and videos - Mistercorzi

• Revision notes - BBC Bitesize • Test yourself - BBC Bitesize

Notes - Maths4Scotland.co.uk

Theory guide - National 5 Maths

Worked examples - Maths Mutt

Notes - D R Turnbull

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