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Cambridge IGCSE Mathematics — 0580 Extended

Topic 5.2: Mensuration — Area and Perimeter

Rectangle: perimeter and area

Perimeter is the distance once round the boundary, measured in cm. Area is the space inside, measured in \(\text{cm}^2\).

A rectangle is \(8\,\text{cm}\) long and \(5\,\text{cm}\) wide. Find its area and perimeter.

Rectangle drawn to scale with length 8 cm and width 5 cm, labelled area 40 square centimetres and perimeter 26 cm
\(A = \ell w = 8 \times 5 = 40\,\text{cm}^2\) and \(P = 2(\ell + w) = 2(13) = 26\,\text{cm}\).

Not on the formula sheet. Except for the area of a triangle, none of the 2D area formulas in this topic are given. Learn them.

Triangle

\(A = \tfrac{1}{2}bh\) is on the formula sheet. What the sheet cannot tell you is which length is \(h\): it is the perpendicular height from the chosen base to the opposite vertex.

A triangle has base \(10\,\text{cm}\) and perpendicular height \(6\,\text{cm}\). Find its area.

Scalene triangle with base 10 cm and a dashed perpendicular height of 6 cm meeting the base at a right angle
\(A = \tfrac{1}{2} \times 10 \times 6 = 30\,\text{cm}^2\). The dashed line, not a sloping side, is the height.

Method

  1. Pick a side to be the base — any of the three will do.
  2. Find the perpendicular distance from that base to the opposite vertex.
  3. Multiply and halve: \(A = \tfrac{1}{2}bh\).

Parallelogram

\[ A = bh \]

A parallelogram is a rectangle with a triangle slid from one end to the other, so the area is base times perpendicular height — exactly the rectangle rule.

A parallelogram has base \(10\,\text{cm}\) and perpendicular height \(6\,\text{cm}\). Find its area.

Parallelogram with base 10 cm, a dashed perpendicular height of 6 cm, the slanted side labelled slant, and arrow marks showing the parallel sides
\(A = 10 \times 6 = 60\,\text{cm}^2\). The slanted side is longer than \(6\,\text{cm}\) and must not be used as \(h\).

For the perimeter, you do need the slanted side: add all four sides, or \(P = 2(\text{base} + \text{slant})\).

Trapezium

\[ A = \tfrac{1}{2}(a + b)h \]

Here \(a\) and \(b\) are the two parallel sides and \(h\) is the perpendicular gap between them. In words: average the parallel sides, then multiply by the height.

A trapezium has parallel sides \(12\,\text{cm}\) and \(8\,\text{cm}\), and height \(5\,\text{cm}\). Find its area.

Isosceles trapezium with parallel sides of 12 cm and 8 cm marked with arrows and a dashed perpendicular height of 5 cm
\(A = \tfrac{1}{2}(12 + 8) \times 5 = \tfrac{1}{2} \times 20 \times 5 = 50\,\text{cm}^2\).

Paper 2 (non-calculator)

Add the parallel sides first. \(\tfrac{1}{2}(12+8)\times 5\) becomes \(10 \times 5\) in one step; multiplying \(12 \times 5\) and \(8 \times 5\) separately takes longer and invites slips.

The four formulas at a glance

Four small labelled shapes — rectangle, triangle, parallelogram and trapezium — each with its area formula and a note on whether the formula is given
Only \(A = \tfrac{1}{2}bh\) is printed on the formula sheet. The other three must be memorised.
ShapeAreaOn the sheet?
Rectangle\(A = \ell w\)No
Triangle\(A = \tfrac{1}{2}bh\)Yes
Parallelogram\(A = bh\)No
Trapezium\(A = \tfrac{1}{2}(a+b)h\)No

Try this

A trapezium has area \(84\,\text{cm}^2\). Its parallel sides are \(9\,\text{cm}\) and \(15\,\text{cm}\). Find the perpendicular height.

Show answer
Answer

  1. Substitute into \(A = \tfrac{1}{2}(a+b)h\)

    \[ 84 = \tfrac{1}{2}(9 + 15)h = 12h \]

  2. Divide by \(12\)

    \[ h = 7\,\text{cm} \]

Exam Traps

  • Do not use a slanted side as \(h\) in \(bh\) or \(\tfrac{1}{2}(a+b)h\). The height is always the perpendicular distance, and a diagram often gives you the slant as a distractor.
  • In a trapezium, \(a\) and \(b\) must be the parallel pair. Averaging the two sloping sides scores nothing.
  • Area answers take \(\text{cm}^2\) and perimeter answers take \(\text{cm}\). Marks are given for the unit in "show that" and problem-solving questions.

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