Area of plane figures

A farmer's field is rarely a neat rectangle. It has a slanted boundary here, a bulge there — four straight sides that meet at four different angles, nothing like the tidy squares and rectangles of earlier classes. Yet its area still needs to be found, for splitting land, pricing a plot, or laying a floor. Class 8 hands you the tool for exactly this: instead of inventing a new formula for every odd shape, you learn to cut the shape into pieces you already know how to measure — triangles above all — and add the pieces back up. This article builds on the triangle, rectangle, square, parallelogram and rhombus formulas from Class 7 and extends them in two directions: the trapezium, a four-sided figure with just one pair of parallel sides, and the general quadrilateral, any four-sided figure at all, handled by drawing a diagonal and dropping perpendiculars onto it. The same cutting idea then scales up to irregular polygons with five, six, or more sides. Two worked examples anchor the two new formulas in numbers, and the common-traps section flags the single mistake — using a slant length where a perpendicular height belongs — that costs the most marks in this topic.

The core insight

Any quadrilateral or irregular polygon can be split by a diagonal into triangles whose areas add up, and this single cutting idea produces both the trapezium formula $\tfrac12(a+b)h$ and the general quadrilateral formula $\tfrac12 d(h_1+h_2)$ .

The core idea

Start from what Class 7 already gives you. With $b$ = base, $h$ = height, $l$ = length, $a$ = a square's side, and $d_1, d_2$ = the diagonals of a rhombus:

\text{Triangle} = \frac{1}{2}bh, \qquad \text{Rectangle} = l \times b, \qquad \text{Square} = a^2

\text{Parallelogram} = b \times h, \qquad \text{Rhombus} = \frac{1}{2}d_1d_2

where: $b$ and $h$ are always measured perpendicular to each other — the height is never a slanting side.

A trapezium has only one pair of parallel sides. Call the two parallel sides $a$ and $b$ , and let $h$ be the perpendicular distance between them:

\text{Area of trapezium} = \frac{1}{2}(a+b)h

where: $a, b$ are the lengths of the two parallel sides and $h$ is the perpendicular distance between them, never a slanted edge joining them.

A general quadrilateral has no parallel sides guaranteed at all, so it is handled differently: draw one diagonal $d$ , then drop perpendiculars — called offsets — of lengths $h_1$ and $h_2$ from the other two vertices onto that diagonal.

\text{Area of quadrilateral} = \frac{1}{2}d(h_1+h_2)

where: $d$ is the chosen diagonal and $h_1, h_2$ are the two perpendicular offsets dropped onto it from the remaining vertices.

Irregular polygons follow the same idea one step further. A pentagon, hexagon, or any many-sided irregular figure is cut, by drawing several diagonals from one vertex or across the shape, into a collection of triangles and trapeziums whose areas are already known formulas — the total area is simply their sum.

Figure

A quadrilateral ABCD with diagonal AC drawn, and perpendicular offset lines from B and D dropping onto the diagonal at right angles, splitting the quadrilateral into two triangles

Derivation AWhere the trapezium area formula comes from

The goal is to show why splitting a trapezium with one diagonal into two triangles that share the same perpendicular height, but sit on the two different parallel sides as their bases, forces the trapezium area to equal exactly $\tfrac12(a+b)h$ — no separate formula needs to be memorised from scratch, because it falls straight out of the triangle area formula applied twice.

Step 1. Take a trapezium $ABCD$ where $AB = a$ and $DC = b$ are the two parallel sides, and $h$ is the perpendicular distance between them. Draw the diagonal $AC$ . This single line cuts the trapezium into two triangles: $\triangle ABC$ and $\triangle ACD$ .

Step 2. Look at $\triangle ABC$ . Its base is $AB = a$ , and because $DC \parallel AB$ , the perpendicular distance from $C$ down to line $AB$ is the same $h$ that separates the two parallel sides. So $\text{Area}(\triangle ABC) = \tfrac{1}{2} \times a \times h$ .

Step 3. Look at $\triangle ACD$ . Its base is $DC = b$ , and by the same parallel-sides reasoning, the perpendicular distance from $A$ to line $DC$ is also $h$ . So $\text{Area}(\triangle ACD) = \tfrac{1}{2} \times b \times h$ .

Step 4. Add the two triangles back together, since they make up the whole trapezium with no overlap: $\text{Area}(ABCD) = \tfrac{1}{2}ah + \tfrac{1}{2}bh = \tfrac{1}{2}(a+b)h$ . The same cutting logic, applied to a diagonal of any quadrilateral with offsets $h_1$ and $h_2$ instead of a shared $h$ , gives $\tfrac{1}{2}d(h_1+h_2)$ by exactly the same two-triangle argument.

Worked example

Find the area of a trapezium whose parallel sides are 15 cm and 9 cm, and whose height (perpendicular distance between them) is 8 cm. Then find the area of a quadrilateral field $ABCD$ with diagonal $AC = 40$ m and perpendicular offsets from $B$ and $D$ of 12 m and 18 m.

Step 1. Identify the trapezium formula and substitute. With $a = 15$ cm, $b = 9$ cm, $h = 8$ cm:

\text{Area} = \frac{1}{2}(15+9) \times 8

Step 2. Calculate. $15 + 9 = 24$ , and $\tfrac12 \times 24 = 12$ , so $\text{Area} = 12 \times 8 = 96\ \text{cm}^2$ . The trapezium's area is $96\ \text{cm}^2$ .

Step 3. Now the quadrilateral field, using $d = 40$ m, $h_1 = 12$ m, $h_2 = 18$ m:

\text{Area} = \frac{1}{2} \times 40 \times (12+18) = \frac{1}{2} \times 40 \times 30

Step 4. Calculate. $\tfrac12 \times 40 = 20$ , so $\text{Area} = 20 \times 30 = 600\ \text{m}^2$ . Splitting along the diagonal gives $\triangle ABC = \tfrac12 \times 40 \times 12 = 240\ \text{m}^2$ and $\triangle ACD = \tfrac12 \times 40 \times 18 = 360\ \text{m}^2$ , and $240 + 360 = 600\ \text{m}^2$ checks against the total. The trapezium's area is $96\ \text{cm}^2$ , and the quadrilateral field's area is $600\ \text{m}^2$ , split into plots of $240\ \text{m}^2$ and $360\ \text{m}^2$ .

Common traps

Using a slant side instead of the perpendicular height in the trapezium or quadrilateral formula — $h$ , $h_1$ , $h_2$ must always be measured at right angles to the base or diagonal, never along a slanting boundary.
Forgetting the $\tfrac12$ in the trapezium and quadrilateral formulas, which silently doubles the computed area.
Treating a trapezium like a parallelogram and using $b \times h$ — a trapezium has only one pair of parallel sides, so the two parallel sides must first be averaged before multiplying by the height.
Halving each offset separately instead of adding $h_1$ and $h_2$ first, then multiplying by $d$ , then halving the whole product once.

Ask the tutor: stuck on which diagonal or offset to use in a quadrilateral problem, or how to split an irregular polygon into pieces? Ask the AI tutor to walk through it with you, one triangle at a time.

Why does the trapezium formula average the two parallel sides instead of just multiplying them like a rectangle

A trapezium is not uniformly wide the way a rectangle is — one parallel side is longer than the other, so no single width represents the whole shape. Splitting it with a diagonal shows it is really two triangles, one sitting on each parallel side but sharing the same height $h$ . Adding $\tfrac12 ah + \tfrac12 bh$ and factoring out $\tfrac12 h$ leaves $\tfrac12(a+b)h$ , which is exactly averaging the two parallel sides and then multiplying by the height — the average naturally falls out of adding two triangles, it is not an assumption bolted on.

How do I choose which diagonal to draw in a general quadrilateral problem

Usually the problem gives you the length of one specific diagonal along with the two offsets measured onto it, so the choice is already made for you — use that diagonal. If a problem instead gives you both diagonals and other measurements, pick the diagonal for which you can actually measure or are given the perpendicular offsets from the two remaining vertices, since the formula only works once you know $d$ , $h_1$ , and $h_2$ for the same diagonal.

Do I need to memorise a separate formula for every kind of irregular polygon, like a pentagon or a hexagon

No, and that is the whole point of this topic. There is no separate area formula for a general pentagon or hexagon. Instead, draw diagonals from one vertex, or across the figure, to cut it into triangles and trapeziums whose areas you already know how to compute, then simply add up the pieces. The same triangle and trapezium formulas you have just learned handle any irregular polygon this way, no matter how many sides it has.

Area of plane figures

The core idea

Worked example

Common traps

Before this

Where this leads