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Forming Equations from Shapes –

GCSEMathsAI Team·7 min read·23 May 2026

Forming Equations from Shapes

One of the most common exam question styles combines algebra with geometry. You are given a shape with sides or angles written as algebraic expressions and asked to form an equation, then solve it. These questions test whether you can translate geometric rules into algebra. They appear on both Foundation and Higher papers and are worth several marks, so learning the method is essential.

What Is Forming Equations from Shapes?

Forming equations from shapes means using known geometric facts — such as angles in a triangle summing to 180°, or opposite sides of a rectangle being equal — to write an equation involving an unknown. You then solve the equation to find the unknown and, often, calculate a measurement.

Key Formulas

Angles in a triangle add up to 180°
Angles in a quadrilateral add up to 360°
Perimeter = sum of all side lengths
For a rectangle, opposite sides are equal

Step-by-Step Method

  1. Read the question carefully and identify which geometric fact applies (angles in a triangle, perimeter, area, etc.).
  2. Write an equation using the algebraic expressions given in the diagram and the geometric rule.
  3. Simplify the equation by collecting like terms.
  4. Solve for the unknown variable.
  5. Substitute back to find the required measurement (angle, side length, area, etc.).
  6. Check your answer makes sense — side lengths must be positive, angles must be sensible.

Worked Example 1 — Foundation Level

Question: A triangle has angles (2x + 10)°, (3x)°, and (x + 50)°. Find the value of x.

Working:

Angles in a triangle add up to 180°.

(2x + 10) + 3x + (x + 50) = 180

6x + 60 = 180

6x = 120

x = 20

Check: angles are 2(20) + 10 = 50°, 3(20) = 60°, 20 + 50 = 70°. Sum = 50 + 60 + 70 = 180° ✓

Answer: x = 20

Worked Example 2 — Higher Level

Question: A rectangle has length (3x + 1) cm and width (x + 5) cm. The perimeter is 44 cm. Find the area of the rectangle.

Working:

Perimeter of a rectangle = 2(length + width).

2[(3x + 1) + (x + 5)] = 44

2(4x + 6) = 44

8x + 12 = 44

8x = 32

x = 4

Length = 3(4) + 1 = 13 cm. Width = 4 + 5 = 9 cm.

Area = 13 × 9 = 117 cm².

Check perimeter: 2(13 + 9) = 2 × 22 = 44 cm ✓

Answer: 117 cm²

Worked Example 3 — Exam Style

Question: An isosceles triangle has two equal sides of length (2x + 3) cm and a base of (x + 1) cm. The perimeter is 27 cm. Find the length of each side.

Working:

Perimeter = 2(2x + 3) + (x + 1) = 27

4x + 6 + x + 1 = 27

5x + 7 = 27

5x = 20

x = 4

Equal sides = 2(4) + 3 = 11 cm each. Base = 4 + 1 = 5 cm.

Check: 11 + 11 + 5 = 27 cm ✓

Answer: Two sides of 11 cm and a base of 5 cm

Common Mistakes

  • Forgetting the geometric rule. Students write the expressions but do not set them equal to anything. You need a fact like "angles sum to 180°" or "perimeter = given value" to form the equation.
  • Mixing up perimeter and area. Read the question carefully — perimeter is the total of all sides; area involves multiplication.
  • Accepting negative or zero side lengths. If your value of x produces a negative side length, recheck your equation. Side lengths must be positive.

Exam Tips

  • Always state the geometric rule you are using — this earns a method mark even if your algebra goes wrong.
  • Show every step of your working. These questions carry 3–5 marks and part marks are available.
  • After finding x, substitute back to answer the specific question asked (the question might ask for a side length or an angle, not x itself).
  • Diagrams in exams are not drawn to scale, so do not estimate from the picture.

Practice Questions

Q1 (Foundation): A triangle has angles (x + 30)°, (2x + 10)°, and (x + 20)°. Find x and state each angle.

Answer: 4x + 60 = 180, 4x = 120, x = 30. Angles are 60°, 70°, 50°

Q2 (Foundation): A square has side length (2x − 1) cm and perimeter 28 cm. Find x.

Answer: 4(2x − 1) = 28, 8x − 4 = 28, 8x = 32, x = 4

Q3 (Higher): A rectangle has length (5x − 2) cm and width (2x + 1) cm. Its area is 85 cm². Find x.

Answer: (5x − 2)(2x + 1) = 85. 10x² + x − 2 = 85. 10x² + x − 87 = 0. (10x + 29)(x − 3) = 0. x = 3 (reject negative). Length = 13 cm, width = 7 cm

Practise forming equations from shapes with instant feedback free on GCSEMathsAI.

Summary

  • Use geometric rules (angle sums, perimeter, area) to set up equations from diagrams.
  • Collect like terms, solve for x, and substitute back to answer the question.
  • Always check that your answers are positive and make geometric sense.
  • State the geometric fact you are using to secure method marks in the exam.
  • These questions appear frequently and combine algebra with shape knowledge.

Test your understanding

5 quick MCQs to identify any misconceptions on this topic.

Take Diagnostic Quiz
§Academic References

Further reading from leading academic institutions — free and open-access.

N
Equations & IdentitiesNRICH

Cambridge challenges on forming and solving equations.

University of Cambridge · Free · Open Access
C
Solving EquationsCorbett Maths

Step-by-step methods for linear and more complex equations.

Corbett Maths · Free · Open Access
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