The Area of a Polygon from Coordinates

A triangle with no base or height

A triangle has its corners at (1, 1), (5, 2) and (3, 4). None of its sides runs along a grid line, so there is no base with a height that can be read off the grid, and ½ × base × height cannot be used directly. The coordinates of the corners are enough to find the area another way.

The triangle with corners (1, 1), (5, 2) and (3, 4). The arrows go round it in that order, counterclockwise, and back to (1, 1).

Box and subtract

Draw the smallest rectangle around the triangle with its sides along the grid lines. Its left side is at x = 1 and its right side at x = 5, so it is 4 wide. Its bottom is at y = 1 and its top at y = 4, so it is 3 tall. Its area is 4 × 3 = 12.

The rectangle is the triangle plus three right triangles in its corners, and each of those has its two shorter sides along the grid, so its area is ½ × base × height. In the bottom right corner, below the side from (1, 1) to (5, 2): ½ × 4 × 1 = 2. In the top right corner, above the side from (5, 2) to (3, 4): ½ × 2 × 2 = 2. In the top left corner, left of the side from (3, 4) to (1, 1): ½ × 2 × 3 = 3. The fourth corner of the rectangle, (1, 1), is a corner of the triangle itself, so nothing is left over there.

Subtract them from the rectangle: 12 − 2 − 2 − 3 = 5. The area of the triangle is 5.

The triangle inside a 4 by 3 rectangle, with a right triangle left over in three corners. Their areas are 2, 2 and 3, and 12 − 7 = 5.

The shoelace method

The shoelace method gets the same area from the coordinates alone. List the corners in order around the shape, and write the first corner again at the end to close the loop: (1, 1), (5, 2), (3, 4), (1, 1).

Multiply each x-coordinate by the y-coordinate of the next corner in the list, and add: 1 × 2 + 5 × 4 + 3 × 1 = 2 + 20 + 3 = 25. These are the down-products.

Multiply each y-coordinate by the x-coordinate of the next corner, and add: 1 × 5 + 2 × 3 + 4 × 1 = 5 + 6 + 4 = 15. These are the up-products.

The area is half of the difference: ½ × (25 − 15) = ½ × 10 = 5, the same as box and subtract. The name comes from the criss-cross pattern the multiplications make when the list is written in a column, like the laces of a shoe.

Why half the difference is the area

Start with a triangle that has one corner at the origin: O(0, 0), A(4, 1) and B(1, 3). Its box is 4 by 3, with area 12. The three corner triangles are ½ × 4 × 1 = 2, ½ × 3 × 2 = 3 and ½ × 1 × 3 = 1.5, so the area is 12 − 6.5 = 5.5.

The shoelace method gives ½ × (4 × 3 − 1 × 1) = ½ × 11 = 5.5, because every product that uses the origin is 0. That is no coincidence. With A(a, b) and B(c, d), where c is less than a and b is less than d as here, the box is a by d and the corner triangles are ½ab, ½(a − c)(d − b) and ½cd. Expanding (a − c)(d − b) = ad − ab − cd + bc and subtracting everything from ad leaves ½ad − ½bc = ½(ad − bc).

So a side from one corner to the next, together with the origin, makes a triangle of area ½(x of the first × y of the next − y of the first × x of the next), and the shoelace method adds one of these for each side of the shape. When a side runs the other way around the origin, the same working gives its area with a minus sign.

The triangle as origin triangles

For the triangle with corners (1, 1), (5, 2) and (3, 4), the three sides give 1 × 2 − 5 × 1 = −3, then 5 × 4 − 3 × 2 = 14, then 3 × 1 − 1 × 4 = −1. Their total is 10, and half of it is 5.

The side from (5, 2) to (3, 4) makes the large origin triangle, of area ½ × 14 = 7. It covers the whole triangle and also two thin triangles, one between the origin and the side from (1, 1) to (5, 2), of area ½ × 3 = 1.5, and one between the origin and the side from (3, 4) to (1, 1), of area ½ × 1 = 0.5. Those two sides run the other way around the origin, so their terms are negative and the thin triangles are taken away: 7 − 1.5 − 0.5 = 5.

Lines from the origin, at the bottom left, to each corner of the triangle. The large triangle from the origin to (5, 2) and (3, 4) has area 7; it holds the shape, of area 5, and the two thin triangles next to the origin, of areas 1.5 and 0.5.

Order matters

The corners must be listed in order around the edge of the shape, going one way. Listed counterclockwise, (1, 1), (5, 2), (3, 4), (1, 1), the difference is positive. Listed clockwise, (1, 1), (3, 4), (5, 2), (1, 1), the down-products come to 4 + 6 + 5 = 15 and the up-products to 3 + 20 + 2 = 25, so the difference is −10. The size is the same, so take the positive value: the area is 5.

The method works for any number of corners. For the rectangle (0, 0), (4, 0), (4, 3), (0, 3), the down-products are 0 × 0 + 4 × 3 + 4 × 3 + 0 × 0 = 24 and the up-products are 0 × 4 + 0 × 4 + 3 × 0 + 3 × 0 = 0, so the area is ½ × 24 = 12, which is 4 × 3.

The usual mistakes

Forgetting to halve. ½ × 10 = 5 is the area; 10 is twice the area, because each product pair measures twice an origin triangle.

Adding the two rows instead of subtracting. ½ × (25 + 15) = 20 is more than the whole 4 by 3 box, which is impossible. The area comes from the difference.

Not closing the loop. Without the first corner written again at the end, the last side of the shape is left out of the sums.

Listing the corners out of order. A list that jumps across the shape traces a path that crosses itself, and the sums measure a different shape.