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Pearson Edexcel 4MA1 IGCSE  ·  Topic G2.11

G2.11 Area and Perimeter

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01 — Explanation

What's the difference between area and perimeter?

Perimeter is the total distance around the outside of a shape — a one-dimensional length (mm, cm, m). Area is the amount of space enclosed within that boundary — a two-dimensional measure (cm², m²). Confusing the two (applying an area formula where perimeter is needed, or vice versa) is the single most common error on this topic.

02 — Explanation

Which formulas must I memorise, and which are given to me?

The Foundation Tier formulae sheet provides the trapezium area formula, but it deliberately omits the rectangle, triangle and parallelogram formulas — these must be memorised:

ShapeAreaStatus
RectangleA=lwA = lwMemorise
TriangleA=12bhA = \tfrac{1}{2}bhMemorise
ParallelogramA=bhA = bhMemorise
TrapeziumA=12(a+b)hA = \tfrac{1}{2}(a+b)hGiven on formula sheet

Common Mistake

using a triangle's slanted side instead of its perpendicular height. The height is always the shortest distance from the base to the opposite vertex, which may not be a labelled side at all.

03 — Explanation

How do I find the area of a compound shape?

Split the shape into rectangles and triangles you recognise, find each area separately, then add them together. For shapes with a missing length, express the area as an algebraic expression in terms of that length, set it equal to the total area given, and solve.

04 — Explanation

How do I link perimeter and area in a multi-step problem?

A classic Foundation question gives one shape's perimeter and asks about a different shape's area.

Example: a rectangle measuring 10 cm×6 cm10\text{ cm} \times 6\text{ cm} has the same perimeter as a square. Find the square's area, then find by how much it exceeds the rectangle's area.

  1. Rectangle's perimeter: 2(10)+2(6)=32 cm2(10)+2(6) = 32\text{ cm}.
  2. Square's side: 32÷4=8 cm32 \div 4 = 8\text{ cm}, so its area is 82=64 cm28^2 = 64\text{ cm}^2.
  3. Rectangle's area: 10×6=60 cm210 \times 6 = 60\text{ cm}^2.
  4. Difference: 64−60=4 cm264 - 60 = 4\text{ cm}^2.

Exam Tip

always write the formula first, then substitute — e.g. "A=lw=10×6A = lw = 10 \times 6" — before simplifying. This secures method marks even if the final arithmetic goes wrong, and makes multi-stage problems like this one much easier to check back through.

05 — Worked example

Worked exam-style question

A triangle has sides 3x−53x-5, 2x−12x-1 and x+1x+1 (in cm), and a perimeter of 3131 cm. Find xx, then find the length of the shortest side.

  1. Sum the sides: (3x−5)+(2x−1)+(x+1)=6x−5(3x-5)+(2x-1)+(x+1) = 6x - 5.
  2. Set equal to the perimeter: 6x−5=316x - 5 = 31.
  3. Solve: 6x=366x = 36, so x=6x = 6.
  4. Side lengths: 3(6)−5=133(6)-5=13, 2(6)−1=112(6)-1=11, 6+1=76+1=7.

Answer: x=6x=6; the shortest side is 7 cm.

Exam Tip

check your solution by substituting x=6x=6 back into all three side expressions and confirming they sum to the given perimeter (13+11+7=3113+11+7=31) — this catches sign errors in setting up the equation before you commit to a final answer.