What happens if the vertices are listed clockwise instead of counterclockwise?

The value of $(S_1 - S_2)$ will be negative instead of positive. Because the final step takes the absolute value, the calculated area will still be correct.

Does this work for concave polygons?

Yes. The shoelace formula works for all simple polygons, convex or concave, provided the perimeter does not cross itself.

What if the polygon has collinear vertices along an edge?

Collinear vertices do not change the bounded area. The formula naturally handles them; they contribute a net zero addition to the final area.

How to find the area of a polygon from its vertices

The area of a polygon can be calculated from its vertex coordinates using the shoelace formula (also known as Surveyor's formula).

This method applies to any non-self-intersecting polygon, whether convex or concave, as long as the vertices are ordered sequentially along the perimeter.

The setup

Let the vertices of the polygon be $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$ , listed in consecutive order around the perimeter (either clockwise or counterclockwise). To use the formula, you must append the first vertex to the end of the list, creating an $n+1$ sequence: $(x_{n+1}, y_{n+1}) = (x_1, y_1)$ .

The steps

Write down the $x$ and $y$ coordinates in two columns, ending with the first coordinate pair repeated.
Multiply each $x_i$ by the following $y_{i+1}$ (diagonally down-right) and sum these products: $S_1 = (x_1 y_2 + x_2 y_3 + \dots + x_n y_1)$ .
Multiply each $y_i$ by the following $x_{i+1}$ (diagonally up-right) and sum these products: $S_2 = (y_1 x_2 + y_2 x_3 + \dots + y_n x_1)$ .
Subtract $S_2$ from $S_1$ .
Take the absolute value of the difference and divide by 2 to find the area: $A = \frac{1}{2} |S_1 - S_2|$ .

Checking the result

Verify that the area is strictly positive unless the polygon is degenerate. For simple shapes like triangles or rectangles, verify the result using basic geometric area formulas. Ensure you have not skipped any vertices and that the first vertex is repeated at the end.

Common errors

The most common error is listing the vertices out of order (e.g., zig-zagging across the polygon). Vertices must trace the perimeter. Another frequent error is forgetting to repeat the first vertex at the end of the calculation, which fails to close the polygon.

Worked example

Find the area of the quadrilateral with vertices at $(1, 1)$ , $(4, 2)$ , $(3, 5)$ , and $(-1, 4)$ .

First, list the vertices in order and repeat the first vertex at the end: $(1, 1)$ , $(4, 2)$ , $(3, 5)$ , $(-1, 4)$ , $(1, 1)$

Calculate $S_1$ (down-right products): $S_1 = (1)(2) + (4)(5) + (3)(4) + (-1)(1)$ $S_1 = 2 + 20 + 12 - 1 = 33$

Calculate $S_2$ (up-right products): $S_2 = (1)(4) + (2)(3) + (5)(-1) + (4)(1)$ $S_2 = 4 + 6 - 5 + 4 = 9$

Subtract $S_2$ from $S_1$ : $S_1 - S_2 = 33 - 9 = 24$

Multiply by $\frac{1}{2}$ and take the absolute value: $A = \frac{1}{2} |24| = 12$

The area of the quadrilateral is $12$ .

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References: Khan Academy: High School Geometry, Analytic Geometry · Stewart Calculus, 8th Edition · OpenStax Algebra and Trigonometry