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Area of any triangle

Free Printable Area of any triangle PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 7)

Download free area of any triangle worksheets (PDF), GCSE maths past paper questions and worked solutions for Area of any triangle (7L). This is a Grade 7 geometry topic from the GCSE questions by topic collection for Higher tier GCSE Maths (Edexcel, AQA, OCR and IGCSE).

The free PDF worksheets table below brings together printable question booklets (Qs) and worked answers or mark schemes (Ans) from Maths Genie, Maths4Everyone, Corbett Maths, Dr Frost, 1st Class Maths, Maths Takeaway, Maths DIY, Dr Austin, E Classroom and PMT, along with past paper question booklets.

Below the worksheet links you can read our area of any triangle revision notes and worked examples with diagrams, try real GCSE past paper area of any triangle questions with step-by-step solutions, or download a PDF of all area of any triangle worksheet links.

Free Area of any triangle PDF Worksheets & Answers

Grade 7 • Topic 7L
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Grade 7 topics
7A - Surds7B - Bounds
7C - Proportion (harder)7D - Algebraic fractions
7E - The quadratic formula7F - Factorising harder quadratics
7G - Trigonometric & exp. Graphs7H - Rearranging (harder)
7J - Functions (harder)7K - Iteration
7L - Area of any triangle7M - The sine rule
7N - The cosine rule7P - Congruent triangles
7Q - 3D pythagoras / trig.7R - Histograms
7S - Conditional probability7T - Exact trig. values (harder)
7U - Area of a segment
GCSE Maths Grade 7 Revision Notes & Method

Guide to Area of any triangle: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

The area of any triangle can be found using Area = 12ab sin C, where a and b are two sides and C is the included angle sandwiched directly between them. If you are given two angles instead of the included angle, use angles in a triangle (180°) to find the included angle first.

When working backwards from a given area to find an unknown angle C, remember that sin(180° − θ) = sin θ: if the question states the angle is obtuse, subtract the calculator's acute angle from 180°.

Example 1: Find the area of a triangle with sides 8.5 cm and 11.2 cm and included angle 64° (3 s.f.).

64° 8.5 cm 11.2 cm

Solution:

Substitute a = 8.5, b = 11.2, C = 64°:
Area = 12 × 8.5 × 11.2 × sin 64°
Area = 47.6 × sin 64° = 42.7826... cm2
Answer: 42.8 cm2 (3 s.f.)

Example 2: A triangle has sides 9 cm and 12 cm and area 45 cm2. Given angle C is obtuse, find C (1 d.p.).

Solution:

12 × 9 × 12 × sin C = 45 ⇒ 54 sin C = 45
sin C = 4554 = 56
Acute angle = sin−1(56) = 56.442...°
Obtuse angle C = 180° − 56.442...° = 123.557...°
Answer: 123.6° (1 d.p.)
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Area of any triangle Questions & Answers

Past Paper Example 1 GCSE Jun 2019 3H Q14 4 marks
Full Paper (PDF) • Mark Scheme (PDF)
The diagram shows a rectangle, ABDE, and two congruent triangles, AFE and BCD.
30°30°24 cm24 cmABCDEF
area of rectangle ABDE = area of triangle AFE + area of triangle BCD
AB : AE = 1 : 3
Work out the length of AE.
Worked Solution & Mark Scheme (4 marks)
Area of 2 triangles = 2 × (1/2 × 24 × AE × sin 30°) = 12 × AE M2
Area of rectangle = AB × AE = 12 × AE ⇒ AB = 12 cm M1
Since AB : AE = 1 : 3, AE = 3 × 12 = 36 cm A1
Answer: 36 cm
Past Paper Example 2 GCSE Jun 2025 3H Q14 2 marks
Full Paper (PDF) • Mark Scheme (PDF)
Here is triangle ABC.
35°ABC8.3 cm9.8 cm
Calculate the area of triangle ABC.
Give your answer correct to 3 significant figures.
Worked Solution & Mark Scheme (2 marks)
Area = 1/2 × b × c × sin A = 1/2 × 8.3 × 9.8 × sin(35°) = 40.67 × 0.573576... = 23.327... M1
⇒ 23.3 cm² (3 s.f.). A1
Answer: 23.3 cm²
Past Paper Example 3 GCSE Jun 2017 3H Q15 5 marks
Full Paper (PDF) • Mark Scheme (PDF)
ABC45°(2x – 1) metres(x + 3) metres
The area of triangle ABC is 6√(2) m2.
Calculate the value of x.
Give your answer correct to 3 significant figures.
The area of triangle ABC is 6√(2) m2.
Calculate the value of x.
Give your answer correct to 3 significant figures.
Worked Solution & Mark Scheme (5 marks)
Area = 1/2 × (x + 3)(2x – 1) × sin 45° = 6√2. Since sin 45° = √2 / 2, (1/2)(√2 / 2)(2x² + 5x – 3) = 6√2 M2
⇒ 2x² + 5x – 3 = 24 M2
⇒ 2x² + 5x – 27 = 0. Using quadratic formula: x = (–5 + √(25 – 4(2)(–27))) / 4 = (–5 + √241) / 4 = 2.631... = 2.63. A1
Answer: 2.63

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