Area of Every Polygon (and Heron's Formula)

The One Brick Everything Stands On

Area answers a single, very physical question: how many unit squares would it take to tile this shape with no gaps and no overlaps? For a rectangle the answer is immediate. Lay the unit squares out in rows, count the rows and the columns, and you have the area of a rectangle as base times height, A = b times h. That product is not a definition pulled from nowhere — it is literally the rows-times-columns count of the tiling. Everything in this guide is built by reducing harder shapes to this one honest brick.

Before we cut anything up, separate two words that beginners often blur. The perimeter is the distance once around the boundary — add up the side lengths, full stop. The area is the amount of surface enclosed. They measure different things and live in different units: perimeter in plain length units, area in square units. Two shapes can share a perimeter yet enclose wildly different areas, and we will lean on that fact more than once.

Parallelogram, Triangle, Trapezoid: Shear, Halve, and Average

Start with a parallelogram. Slice a right-triangular wedge off one slanted end and slide it across to the other end, and the leaning shape stands up into a rectangle of the same base and the same height. Nothing was added or removed, so the area is unchanged: the area of a parallelogram is also A = b times h. The crucial honesty here is which h you use — not the slanted side length, but the perpendicular height, the straight-up distance between the two parallel base lines. Plug in the slant by mistake and you overshoot every time.

Now a triangle. Take any triangle, make a copy, flip the copy upside down, and snap the two together along a matching side. They lock into a parallelogram with the same base and height, so the triangle is exactly half of it. That gives the area of a triangle: A = (1/2) times b times h. Any of the three sides can play the role of base, as long as you pair it with the height dropped perpendicularly onto that same side — and reassuringly, all three choices give the same answer.

A trapezoid has two parallel sides of different lengths, call them b_1 and b_2, separated by a perpendicular height h. Glue a second flipped copy alongside the first and the pair becomes a parallelogram whose base is b_1 + b_2. Halving gives the area of a trapezoid: A = (1/2) times (b_1 + b_2) times h. Read it as the average of the two parallel sides times the height — the trapezoid behaves like a rectangle whose width is the midline. The rectangle and triangle are just the special cases where the two parallel sides are equal, or where one of them shrinks to a point.

Heron's Formula: Area From the Three Sides Alone

Every formula so far asked for a height. But often you are handed a triangle by its three side lengths a, b, c and nothing else — no angle, no altitude drawn. Can you still find the area? You can, and the answer is Heron's formula, one of the most quietly astonishing results in elementary geometry. It computes the area from the sides alone, with not a single height in the picture.

Let s = (a + b + c) / 2     (the semiperimeter: half the perimeter)

    Area = sqrt( s (s - a) (s - b) (s - c) )

Example: sides 13, 14, 15
    s = (13 + 14 + 15) / 2 = 21
    Area = sqrt( 21 * (21-13) * (21-14) * (21-15) )
         = sqrt( 21 * 8 * 7 * 6 )
         = sqrt( 7056 )
         = 84

Heron's formula and a worked example: the 13-14-15 triangle has area exactly 84, found without ever drawing a height.

The half-perimeter s = (a + b + c)/2 is called the semiperimeter, and it is the formula's lever. Notice the elegant symmetry: each of the three factors (s - a), (s - b), (s - c) leaves one side out, so no side is privileged. Notice too the built-in honesty check. If one side is longer than the other two combined — say sides 1, 1, 5 — then that side's factor (s - a) turns negative, the product under the square root goes negative, and the formula refuses to return a real number. That is not a glitch; it is the formula correctly reporting that no such triangle exists, which is exactly the triangle inequality making itself heard.

How could area depend only on the sides? Because three side lengths already pin a triangle's shape completely — that is the content of SSS congruence from an earlier rung. Once the three sides are fixed, every angle and every height is fixed too, so the area cannot be free to wander. Heron's formula is honestly not obvious to derive — the cleanest route runs through the Law of Cosines and a patient page of algebra, which we will not grind through here — but its truth rests squarely on SSS, and you can verify any instance with a ruler.

Any Polygon: Cut It Into Triangles

Here is the move that finishes the job for the whole family. Take any polygon, however many sides it has, and slice it into triangles by drawing diagonals from a single corner. A polygon with n sides splits into exactly n - 2 triangles this way — a pentagon into three, a hexagon into four. Find each triangle's area by base-times-height or by Heron's formula, add them up, and you have the area of the whole polygon. No new theory is needed; the triangle is the universal building block.

Pick a convex polygon and label its vertices in order, P_1, P_2, up to P_n.
Draw every diagonal from P_1: the segments P_1 P_3, P_1 P_4, and so on. This carves the polygon into n - 2 triangles that share the corner P_1.
Find each triangle's area — base times height if you have a height, Heron's formula if you only have side lengths.
Add the triangle areas. The total is the area of the polygon, since the pieces tile it exactly with no gap and no overlap.

Regular Polygons: The Apothem Shortcut

When the polygon is regular — all sides equal, all angles equal — the triangle-counting collapses into one clean formula. A regular polygon has a centre equidistant from every vertex and from every side. The perpendicular distance from that centre out to the midpoint of a side is the apothem, written a. Spoke lines from the centre to the corners split the polygon into n identical triangles, each with base equal to one side and height equal to the apothem.

Each of those n triangles has area (1/2) times (side) times a. Add all n of them, and since n copies of the side length is just the perimeter P, the whole regular polygon has area A = (1/2) times a times P — one half the apothem times the perimeter. It is a small, memorable result, and it is the direct ancestor of the circle's area you will meet in the very next guide: let the number of sides grow without bound and the apothem becomes the radius, the perimeter becomes the circumference, and (1/2) a P quietly turns into (1/2) r times circumference. The polygon formula does not break at the circle; it ripens into it.

Honest Edges: Units, Coordinates, and a Warning

A few cautions keep these formulas trustworthy. Always keep area in square units — square centimetres, square metres — and never let it borrow the plain length units that belong to perimeter. And resist the seductive false belief that a bigger perimeter forces a bigger area: a long thin sliver of a triangle can have an enormous perimeter and almost no area, while a fat near-square shape with a smaller perimeter encloses far more. Perimeter and area are genuinely independent measurements, and conflating them is one of the most common beginner errors.

Finally, a glimpse ahead. Once you reach coordinate geometry, you will not even need to draw diagonals: the shoelace formula reads a polygon's area straight off the coordinates of its vertices, by a tidy sum of cross-products. It is the same triangulation idea wearing algebra, and it relies on the same honest brick we started from. Master the cut-and-rearrange thinking here, and that later machinery will feel like an old friend rather than a new rule to memorise.