Area of a polygon

A survey field rarely comes as a neat rectangle or triangle — it is usually an irregular four, five or six-sided plot, and no single "area of a polygon" formula covers every shape at once. The trick used throughout this topic is to turn the hard, irregular figure into several easy ones. Draw a diagonal across a quadrilateral and it splits into two triangles. Draw two or three diagonals across a pentagon or hexagon and it splits into a handful of triangles and trapeziums, each with a formula you already know. Measure the right base and the right perpendicular height for every piece, add up the pieces, and the irregular field's area falls out. This article builds the one master formula for a general quadrilateral, $A = \tfrac12 d(h_1+h_2)$ , and then extends the same slicing idea to larger irregular polygons.

The core insight

Split any irregular polygon into triangles and trapeziums using its diagonals, find each simple piece's area, and add them with no gaps or overlaps.

The core idea

A quadrilateral splits into two triangles. Any quadrilateral $ABCD$ can be cut by one diagonal, say $AC$ , into triangle $ABC$ and triangle $ACD$ . If the diagonal has length $d$ and the perpendicular distances (offsets) from the other two vertices $B$ and $D$ onto that diagonal are $h_1$ and $h_2$ , each triangle's area is half base times height, and adding the two gives the whole quadrilateral:

A = \frac{1}{2}d\,h_1 + \frac{1}{2}d\,h_2 = \frac{1}{2}d(h_1+h_2)

where: $d$ is the length of the chosen diagonal, $h_1$ and $h_2$ are the perpendicular offsets from the remaining two vertices onto that diagonal, and $A$ is the quadrilateral's area.

A bigger polygon splits into several pieces. A pentagon, hexagon or any irregular field with five or more sides usually needs more than one diagonal. Drawing diagonals from convenient vertices divides the whole shape into a set of triangles and trapeziums that exactly tile it — no piece left out, no piece counted twice. Each piece's area is found with its own matching formula, and the pieces are added:

A_{\text{polygon}} = A_1 + A_2 + A_3 + \cdots

where: $A_1, A_2, A_3, \dots$ are the areas of the individual triangles or trapeziums the diagonals create. Two conditions make this work every time. First, the pieces must not overlap and must together cover the polygon exactly, with no gap. Second, every base and height used in a piece's formula must be measured from that same piece — a height belonging to one triangle can never be borrowed for another.

Choosing how to split. A pentagon can often be handled by drawing all the diagonals from a single vertex, which produces three triangles fanning out from that corner — this is usually the fastest choice when the figure is roughly convex. A more irregular hexagonal field might instead be split by one long diagonal into a triangle and a trapezium, or into two trapeziums, depending on which parts have parallel-looking sides. There is no single correct way to split a polygon; any valid set of non-overlapping diagonals that tiles the whole shape will give the same final area, since the total area does not depend on how it is sliced up.

Figure

Quadrilateral split by one diagonal into two triangles with perpendicular offsets marked

Derivation AArea of a general quadrilateral from a diagonal and two offsets

The goal is to show that splitting a quadrilateral with one diagonal and measuring only two perpendicular offsets is enough to find its whole area, without needing any angles or the lengths of its four sides, so that the working formula $A = \tfrac12 d(h_1+h_2)$ can be trusted and applied directly to any quadrilateral, however irregular its shape.

Step 1. Draw quadrilateral $ABCD$ and mark diagonal $AC$ , giving it length $d$ . This one diagonal cuts the quadrilateral into exactly two triangles, $ABC$ and $ACD$ , that share the side $AC$ and together make up the whole figure with no overlap.

Step 2. In triangle $ABC$ , treat $AC$ as the base. Drop a perpendicular from vertex $B$ onto line $AC$ ; call its length $h_1$ . The triangle's area is half base times height, so $\text{area}(ABC) = \tfrac12 \times d \times h_1$ .

Step 3. In triangle $ACD$ , use the same diagonal $AC = d$ as the base. Drop a perpendicular from vertex $D$ onto line $AC$ ; call its length $h_2$ . By the same rule, $\text{area}(ACD) = \tfrac12 \times d \times h_2$ .

Step 4. Add the two triangles, since they exactly tile the quadrilateral with no gap and no overlap: $A = \tfrac12 d h_1 + \tfrac12 d h_2 = \tfrac12 d(h_1+h_2)$ . Only one length ( $d$ ) and two perpendicular heights ( $h_1, h_2$ ) are needed — the four side lengths of the quadrilateral never enter the calculation.

Worked example

Problem: In quadrilateral $PQRS$ , diagonal $PR = 24\ \text{cm}$ . The perpendicular offsets from $Q$ and $S$ onto this diagonal are $8\ \text{cm}$ and $6\ \text{cm}$ . A neighbouring irregular field $ABCDE$ is split from vertex $A$ into three triangles: $ABC$ (base $10\ \text{m}$ , height $6\ \text{m}$ ), $ACD$ (base $12\ \text{m}$ , height $9\ \text{m}$ ), and $ADE$ (base $8\ \text{m}$ , height $5\ \text{m}$ ). Find the area of $PQRS$ and the area of field $ABCDE$ .

Step 1. Find the quadrilateral's area using $A = \tfrac12 d(h_1+h_2)$ with $d = 24$ , $h_1 = 8$ , $h_2 = 6$ :

A = \frac{1}{2}\times 24\times(8+6) = 12\times 14 = 168\ \text{cm}^2

Step 2. Find each triangle's area in the field using $A = \tfrac12\times\text{base}\times\text{height}$ : triangle $ABC$ gives $\tfrac12\times10\times6 = 30\ \text{m}^2$ , triangle $ACD$ gives $\tfrac12\times12\times9 = 54\ \text{m}^2$ , and triangle $ADE$ gives $\tfrac12\times8\times5 = 20\ \text{m}^2$ .

Step 3. Add the three triangle areas, since together they tile the whole field with no gaps or overlaps:

A_{\text{field}} = 30 + 54 + 20 = 104\ \text{m}^2

Final answer: the quadrilateral $PQRS$ has area $168\ \text{cm}^2$ , and field $ABCDE$ has total area $104\ \text{m}^2$ .

Competitive stretch

If instead only the diagonal $PR = 24\ \text{cm}$ and total offset sum $h_1+h_2 = 14\ \text{cm}$ were given (without splitting them individually), the same formula $A = \tfrac12 d(h_1+h_2) = 12\times14 = 168\ \text{cm}^2$ still works — the two offsets never need to be equal or separately identified once their sum is known, since the formula only ever uses $(h_1+h_2)$ together.

Common traps

Measuring a slant side instead of the true perpendicular offset for $h_1$ or $h_2$ — only a right-angle distance onto the diagonal counts.
Choosing diagonals that leave a gap or make two triangles overlap when splitting a bigger polygon, which double-counts or misses part of the area.
Computing each triangle or trapezium's area correctly but forgetting to add every piece into the final total.
Swapping the base and height between two different sub-triangles instead of keeping each triangle's own matching pair.

Ask the tutor: If you are stuck deciding which diagonals to draw for an irregular field in your own textbook problem, ask your AI tutor to walk through marking the diagonals and offsets on that exact diagram with you.

Why does splitting with just one diagonal work for a quadrilateral but not for a pentagon or hexagon?

A single diagonal in a quadrilateral always creates exactly two triangles that fully tile the shape, so one diagonal and two offsets are enough. A pentagon or hexagon has more vertices and more area to cover, so one diagonal usually leaves leftover regions that are still irregular. You need enough diagonals, drawn from one or more chosen vertices, to break the whole shape down into pieces that are each a plain triangle or trapezium with a known formula.

Does it matter which diagonal I choose when splitting a quadrilateral?

No — either diagonal works, as long as you correctly measure the perpendicular offsets from the two vertices it does not pass through onto whichever diagonal you pick. The formula $A = \tfrac12 d(h_1+h_2)$ gives the same final area either way, because both diagonals divide the same quadrilateral into two triangles that together cover it exactly. Choose whichever diagonal makes the offsets easiest to measure or read off a diagram.

What if a diagonal I need to split a polygon lies outside the shape, like in a very irregular figure?

For simple convex-looking fields this rarely happens, but in a strongly irregular or reflex polygon, a diagonal joining two vertices can pass outside the boundary and is not usable for splitting. In that case, choose a different pair of vertices whose diagonal stays inside the shape, or split the polygon using more than one diagonal from a single interior-friendly vertex so every resulting triangle or trapezium sits fully inside the original figure.

Area of a polygon

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads