When a shape has sides given as expressions in x, write a formula for the area using those expressions, set it equal to the given value, rearrange into ax² + bx + c = 0 and solve. Reject any solution that gives a negative length.
Why do geometry problems lead to quadratic equations?
Area formulas multiply two lengths together. If each length contains x, the product produces an x² term — hence a quadratic. Perimeter formulas add lengths together, giving a linear equation; it is area (and sometimes the cosine rule or Pythagoras) that produces quadratics at GCSE.
Worked example 1 — rectangle area
A rectangle has length (x + 2) cm and width x cm. Its area is 35 cm².
Form an equation:
x(x + 2) = 35
x² + 2x = 35
x² + 2x − 35 = 0
Factorise: (x + 7)(x − 5) = 0
Solutions: x = −7 or x = 5.
Since x is a length, x > 0, so x = 5.
Dimensions: length = 7 cm, width = 5 cm. Check: 7 × 5 = 35 ✓.
Worked example 2 — right-angled triangle using Pythagoras
A right-angled triangle has legs of length x cm and (x + 7) cm, and a hypotenuse of 13 cm.
By Pythagoras:
x² + (x + 7)² = 13²
x² + x² + 14x + 49 = 169
2x² + 14x − 120 = 0
x² + 7x − 60 = 0 ÷ 2
Factorise: (x + 12)(x − 5) = 0
Solutions: x = −12 or x = 5. Since x > 0, x = 5.
Legs: 5 cm and 12 cm. Check: 5² + 12² = 25 + 144 = 169 = 13² ✓.
Worked example 3 — area of trapezium
A trapezium has parallel sides of length x cm and (2x − 1) cm, and a perpendicular height of (x + 1) cm. Its area is 40 cm².
Area of trapezium = ½(a + b) × h:
½(x + 2x − 1)(x + 1) = 40
½(3x − 1)(x + 1) = 40
(3x − 1)(x + 1) = 80
3x² + 3x − x − 1 = 80
3x² + 2x − 81 = 0
This does not factorise neatly, so use the quadratic formula:
x = (−2 ± √(4 + 972)) / 6 = (−2 ± √976) / 6 ≈ (−2 ± 31.24) / 6
Taking the positive root: x ≈ 29.24 / 6 ≈ 4.87 cm (to 3 s.f.)
Check: discard the negative root x ≈ −5.54 since lengths must be positive.
Worked example 4 — angles in a triangle
The angles of a triangle are x°, (2x − 15)° and (x + 35)°.
Sum of angles in a triangle = 180°:
x + (2x − 15) + (x + 35) = 180
4x + 20 = 180
4x = 160
x = 40
Wait — this gives a linear equation, not a quadratic. Quadratic angle equations arise when an angle is expressed as a product or a square. For example:
The angles of a triangle are x², (2x + 15)° and (x + 15)°. Form and solve.
x² + 2x + 15 + x + 15 = 180
x² + 3x + 30 = 180
x² + 3x − 150 = 0
Discriminant: 9 + 600 = 609. No integer solution. Adjust: angles x², 3x and (180 − 4x)°.
A cleaner example: two angles are x° and (x + 30)°; the third is x²°.
x² + x + x + 30 = 180
x² + 2x − 150 = 0
x = (−2 ± √(4 + 600)) / 2 = (−2 ± √604) / 2 ≈ (−2 + 24.58) / 2 ≈ 11.3°
Steps for any geometry quadratic problem
- Label all sides or angles using x.
- Write the geometric formula (area, perimeter, Pythagoras).
- Substitute the expressions and expand.
- Rearrange into the form ax² + bx + c = 0.
- Solve by factorising, completing the square or the quadratic formula.
- Reject negative or geometrically impossible solutions and state the physical answer with units.
| Geometry context | Formula used | Equation type |
|---|---|---|
| Rectangle area | length × width = area | Quadratic |
| Right-angled triangle | a² + b² = c² | Quadratic |
| Trapezium area | ½(a+b)h = area | Quadratic |
| Triangle perimeter | sum of sides = perimeter | Linear |
| Circle area | πr² = area | Quadratic (in r) |
Frequently asked questions
What if both solutions to the quadratic are positive?
This can happen when the context allows multiple valid configurations. Draw a diagram and check both solutions: does each one give a valid (positive) set of dimensions? If both are geometrically valid, report both; if only one makes sense in context, state why you rejected the other.
Should I always reject negative solutions?
Reject any solution that makes a side length or a length measurement negative. However, if x represents a coordinate or a signed quantity, a negative value may be valid — read the question carefully. For dimensions, areas and lengths, always require x > 0.
How do I know when to use the quadratic formula vs factorising?
Try factorising first: write pairs of factors of the constant term and check which pair gives the correct middle coefficient. If no integer factor pair works within about 30 seconds, switch to the quadratic formula. The formula always works; factorising is faster when it applies.
Can the area of a non-standard shape give a cubic equation?
Yes, in principle — for example, the volume of a box made from a square sheet (with squares cut from corners) leads to a cubic. At GCSE Higher, these problems are usually set up so the equation is quadratic, not cubic. If you get a cubic, check your expansion for an error.
Try geometry algebra problems with Professor Pi's step-by-step hints at aitutors.me.