Area of polygons by triangulation

A wooden floor tile, a farmer's oddly-shaped field, a kite-shaped garden bed — none of these are triangles, yet every one of their areas can be found using only the triangle area formula you already know. The trick is triangulation: slice the straight-edged figure into non-overlapping triangles that together cover it exactly, find each triangle's area separately, and add. No new formula is needed — just a new way of looking at an old one. This idea sits in the Area chapter of CBSE Class 8 Mathematics, right after the areas of triangles, parallelograms, and quadrilaterals have been developed. It is the general-purpose tool that handles any polygon a specific formula does not already cover: irregular quadrilaterals, pentagons, hexagons, and real field or floor-plan shapes met in exam problems and in surveying. Master triangulation and you can find the area of any straight-sided figure, however irregular, using nothing beyond $\frac12 \times \text{base} \times \text{height}$ .

The core insight

Splitting any straight-edged polygon into non-overlapping triangles that exactly cover it, and adding their individual areas, gives the area of the whole polygon.

The core idea

Any polygon can be triangulated. Choose one vertex and draw diagonals from it to every other non-adjacent vertex, or choose any convenient mix of diagonals — the only rules are that the resulting triangles must not overlap and must together cover the entire figure with no gaps. Since every triangle's area is already known to us,

A_{\triangle} = \frac12 \times \text{base} \times \text{height}

where: base is any one side of the triangle and height is the perpendicular distance from the opposite vertex to that side (or to that side extended).

Adding the pieces gives the whole. Once a polygon of $n$ triangles has been cut, its total area is simply

A_{\text{polygon}} = A_{\triangle_1} + A_{\triangle_2} + \cdots + A_{\triangle_n}

where: each $A_{\triangle_i}$ is computed independently using its own base and perpendicular height, and the sum is taken over every triangle in the chosen split.

The quadrilateral case is the simplest and most common. A single diagonal, say $AC$ in quadrilateral $ABCD$ , splits the four-sided figure into exactly two triangles, $\triangle ABC$ and $\triangle ACD$ , sharing the diagonal as a common base:

A_{ABCD} = A_{\triangle ABC} + A_{\triangle ACD}

where: $A_{\triangle ABC}$ uses the perpendicular distance from $B$ to $AC$ as its height, and $A_{\triangle ACD}$ uses the perpendicular distance from $D$ to $AC$ — the two vertices $B$ and $D$ normally lie on opposite sides of the diagonal, so both triangles' heights are measured independently even though they share the same base length.

Bigger polygons just mean more pieces. A pentagon triangulated from one vertex splits into three triangles; a hexagon into four; in general an $n$ -sided polygon triangulated from a single vertex splits into $n-2$ triangles. For fields and floor plans where diagonals are awkward to draw fully, surveyors often use the offset method: pick one diagonal as a baseline, then measure the perpendicular "offset" distance from each remaining vertex to that baseline, turning the whole figure into a stack of triangles (and sometimes trapeziums) whose areas are found and added. This is exactly the same "area is conserved when a figure is cut into pieces" principle that underlies the parallelogram, trapezium, and rhombus area formulas developed earlier in this chapter — there, a shape is cut and its pieces are rearranged into a simpler shape of equal area; here, the pieces are simply added up where they already lie, without any rearranging.

Figure

A quadrilateral ABCD split by diagonal AC into two triangles with heights from B and D shown as dashed perpendiculars

Derivation AArea of a quadrilateral by triangulation

The goal is to show, from first principles and using only the triangle area formula, that the area of any quadrilateral $ABCD$ split by diagonal $AC$ into $\triangle ABC$ and $\triangle ACD$ equals the sum of those two triangle areas — and to make the role of the two independent heights explicit, since mixing them up is the single most common error students make.

Step 1. Draw diagonal $AC$ . This diagonal, and only this diagonal, splits quadrilateral $ABCD$ into exactly two triangles: $\triangle ABC$ (vertices $A$ , $B$ , $C$ ) and $\triangle ACD$ (vertices $A$ , $C$ , $D$ ). These two triangles do not overlap, since $B$ and $D$ lie on opposite sides of line $AC$ , and together they cover the entire quadrilateral with no gaps — this is the "no overlap, no gap" requirement that makes any triangulation valid.

Step 2. Identify the common base and the two separate heights. Both triangles share diagonal $AC$ as a base, of length $d$ . But their heights are different: $h_1$ is the perpendicular distance from $B$ to line $AC$ , and $h_2$ is the perpendicular distance from $D$ to line $AC$ . These are measured independently — one from each vertex — because $B$ and $D$ are on opposite sides.

Step 3. Apply the triangle area formula to each piece. $A_{\triangle ABC} = \frac12 \times d \times h_1$ and $A_{\triangle ACD} = \frac12 \times d \times h_2$ .

Step 4. Add the two areas, since together they exactly reconstruct the whole quadrilateral with nothing missing and nothing double-counted. $A_{ABCD} = A_{\triangle ABC} + A_{\triangle ACD} = \frac12 d h_1 + \frac12 d h_2 = \frac12 d (h_1 + h_2)$ . This last form — half the diagonal times the sum of the two offset heights — is the compact version of the formula used directly in field-survey and exam problems.

Worked example

Quadrilateral $ABCD$ has diagonal $AC = 18\ \text{cm}$ . The perpendicular distance from $B$ to $AC$ is $6\ \text{cm}$ , and the perpendicular distance from $D$ to $AC$ is $4\ \text{cm}$ , with $B$ and $D$ on opposite sides of the diagonal. Find the area of $ABCD$ , then find the total area of a pentagon-shaped field $PQRST$ triangulated from vertex $P$ into three pieces of area $40\ \text{m}^2$ , $55\ \text{m}^2$ , and $30\ \text{m}^2$ .

Step 1. Split quadrilateral $ABCD$ by diagonal $AC$ into $\triangle ABC$ and $\triangle ACD$ , both with base $AC = 18\ \text{cm}$ .

Step 2. Find each triangle's area separately, using each vertex's own perpendicular height to the diagonal. $A_{\triangle ABC} = \frac12 \times 18 \times 6 = 54\ \text{cm}^2$ and $A_{\triangle ACD} = \frac12 \times 18 \times 4 = 36\ \text{cm}^2$ .

Step 3. Add the two pieces. $A_{ABCD} = 54 + 36 = 90\ \text{cm}^2$ .

Step 4. Now triangulate the pentagon field from vertex $P$ : adding the three given triangle areas gives $A_{PQRST} = 40 + 55 + 30 = 125\ \text{m}^2$ . An equal-area square plot would need side $s$ with $s^2 = 125$ ; since $11^2 = 121$ and $12^2 = 144$ , $s = \sqrt{125} \approx 11.18\ \text{m}$ .

The quadrilateral's area is $\mathbf{90\ cm^2}$ , and the pentagon field's area is $\mathbf{125\ m^2}$ (equal-area square side $\approx 11.18\ \text{m}$ ).

Common traps

Choosing diagonals that make triangles overlap, which double-counts part of the area, or leaving gaps between triangles, which under-counts it.
Using the wrong perpendicular height for a triangle in the split — the height must be measured perpendicular to the chosen base (usually the diagonal), never along a slanted side.
Applying only one height to both triangles sharing a diagonal — $B$ and $D$ each have their own separate perpendicular distance to $AC$ , and mixing them up gives a wrong sum.
In many-sided polygons, forgetting to add every triangle piece — it is easy to lose one when the split has four or five triangles.

Ask the tutor: Have an irregular field or floor plan you are not sure how to split into triangles — ask the AI tutor to help you sketch a triangulation and work through the area step by step.

Does it matter which diagonal I choose to split a quadrilateral?

No, as long as the diagonal actually lies inside the quadrilateral and splits it into two non-overlapping triangles that together cover the whole figure, either diagonal works and both give the same total area. In a general quadrilateral either diagonal is usually valid; the practical choice depends on which perpendicular heights are easiest for you to measure or are already given in the problem.

What if the polygon has five or six sides — do I still just use one diagonal?

Not just one. For a pentagon you typically draw two diagonals from a single chosen vertex, producing three triangles; for a hexagon, three diagonals from one vertex give four triangles. The general pattern is that triangulating an $n$ -sided polygon from one vertex always produces $n-2$ triangles. You then find each triangle's area separately and add them all, exactly as with the two-triangle quadrilateral case.

Can I split a polygon into shapes other than triangles, like rectangles?

Yes — the "cut into pieces and add" principle is not limited to triangles. For polygons with some right angles or convenient straight edges, splitting into a mix of triangles and rectangles (or trapeziums) is often faster, as long as the pieces still do not overlap and still cover the entire figure exactly. Triangulation using only triangles is the most general method because a triangle is the simplest shape whose area formula always applies, but any valid non-overlapping, gap-free decomposition gives the correct total area.

Area of polygons by triangulation

The core idea

Worked example

Common traps

Before this

Where this leads