Areas of flat shapes · Plane mensuration

Area is the amount of flat surface a shape covers, measured in square units (cm², m²). Perimeter is the distance all the way round its edge, measured in ordinary units (cm, m). Keep the two apart: a question about fencing or a wire border wants a perimeter; one about tiles, paint or land wants an area.

Every formula in this lesson comes from one idea: a rectangle’s area is length × width, and other shapes can be cut up and rearranged into rectangles.

Try it

Where the area formulae come fromChange the shape, then show why

24base × height = 6 × 44height (perpendicular)

base = 6height = 4slant = 2

Move the slant: the shape leans, but the base and the perpendicular height stay the same, and so does the area. Press Show why.

Pick each shape and press Show why. Then move the slant slider: the shape leans, but as long as the base and the perpendicular height stay the same, so does the area.

The formulae

Rectangle

A = l \times w

Parallelogram

A = b \times h

Triangle

A = \frac12 b h

Trapezium

A = \frac12(a + b)h

Rhombus

A = \frac12 d_1 d_2

In every drawing the dashed line is the perpendicular height, meeting the base at a right angle. In a trapezium, $a$ and $b$ are the two parallel sides; in a rhombus, $d_1$ and $d_2$ are the diagonals.

Triangles when you know two sides and the angle between them

Two sides and the angle between

A = \frac12 ab\sin C

If you know two sides $a$ and $b$ and the angle $C$ between them, the height is $b\sin C$ (from trigonometry), so

\text{area} = \frac12 ab\sin C

Rhombuses: the diagonals

The diagonals of a rhombus cut each other in half at right angles. That gives four right-angled triangles, so you can find a side with Pythagoras.

Trapeziums

Worked example · WAEC 2015

WAEC 2015 · Paper 2 · Q5

A trapezium $PQRS$ is such that $PQ \parallel RS$ and the perpendicular from $P$ to $RS$ is $40\text{ cm}$ . If $|PQ| = 20\text{ cm}$ , $|SP| = 50\text{ cm}$ and $|SR| = 60\text{ cm}$ , calculate, correct to 2 significant figures, the:

area of the trapezium;

$\angle QRS$ .

Draw it
Put $SR$ (60 cm) at the bottom and $PQ$ (20 cm) at the top. Drop perpendiculars from $P$ and $Q$ to $SR$ , meeting it at $N$ and $M$ . Both are 40 cm: the distance between the parallel sides.
Think first. Which sides are parallel? Where does the 40 cm go?
(a) The area
$\frac12(20 + 60) \times 40 = 1600\text{ cm}^2$
Think first. Which formula, and which numbers?
Find MR
$|SN| = \sqrt{50^2 - 40^2} = 30$ cm. $NM = PQ = 20$ cm, so $|MR| = 60 - 30 - 20 = 10$ cm.
Think first. In triangle $PNS$ , the hypotenuse is 50 and one side is 40. Find $SN$ .
(b) The angle
The height 40 is opposite $\angle QRS$ and $MR = 10$ is next to it: $\tan\angle QRS = \frac{40}{10} = 4$ , so $\angle QRS \approx 76^\circ$ .
Think first. In triangle $QMR$ , which ratio links 40 and 10 to $\angle QRS$ ?

Your turn

WAEC 2018 · Paper 2 · Q3 (a)

(a)
The diagonals of a rhombus are $10.2\text{ cm}$ and $9.3\text{ cm}$ long. Calculate, correct to one decimal place, the perimeter of the rhombus.

Worked solution (try it first)

(a)

The diagonals of a rhombus cut each other in half at right angles.
So each side is the hypotenuse of a right-angled triangle with legs $\frac{10.2}{2} = 5.1$ cm and $\frac{9.3}{2} = 4.65$ cm.
Side $= \sqrt{5.1^2 + 4.65^2}$
$= \sqrt{26.01 + 21.6225}$
$= \sqrt{47.6325}$
$\approx 6.902$ cm.
The four sides are equal, so the perimeter is $4 \times 6.902 \approx 27.6$ cm.

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