Finding the area of a polygon from its vertices

This method finds the area of a triangle or any polygon when you know the coordinates of its corners, without needing a base and a height. It uses the shoelace array, a compact way of writing a determinant-style calculation, so that the area drops straight out of the vertex coordinates.

It answers questions such as 'the vertices of a triangle are $A$, $B$ and $C$; find its area', or 'show that the area of quadrilateral $PQRS$ is $k$ units$^2$'. Because the working is purely numerical once the coordinates go in, it is fast and reliable in the exam, and it extends to four, five or more vertices simply by adding more columns to the array.

Reach for this method whenever a question gives you the coordinates of the corners of a triangle or polygon and asks for its area, or asks you to prove an area equals a certain value or to find an unknown coordinate from a given area. The tell-tale sign is that you are given points, not a base and a height.

It is usually quicker than splitting the shape into right-angled triangles or using $\frac{1}{2} \times \text{base} \times \text{height}$, especially when no side is horizontal or vertical. If the area is given and one coordinate is unknown, set the array equal to twice the area and solve for the missing letter.

Just make sure you can list the vertices in order around the shape before you start.

1. 1

   List the vertices in cyclic order

   Go around the polygon in one direction (anticlockwise is standard) and write the first vertex again at the end so the loop closes.

2. 2

   Set up the shoelace array

   Write the $x$-coordinates along the top row and the matching $y$-coordinates along the bottom row.

3. 3

   Sum the down-products

   Multiply each top term by the bottom term one step to its right and add them: $x_1 y_2 + x_2 y_3 + \dots$.

4. 4

   Sum the up-products

   Multiply each bottom term by the top term one step to its right and add them: $y_1 x_2 + y_2 x_3 + \dots$.

5. 5

   Subtract, take the modulus, halve

   Compute $\frac{1}{2}\,|\,\text{down-sum} - \text{up-sum}\,|$. The absolute value keeps the area positive.

• Listing the vertices out of cyclic order (jumping across the shape), which produces the wrong area.
• Forgetting to repeat the first vertex at the end, so the final cross-product is dropped.
• Leaving out the absolute value and reporting a negative area, a negative result only means the vertices were listed clockwise.
• Forgetting the factor $\frac{1}{2}$, which doubles the answer.
• Sign slips when a coordinate is negative, especially inside the up-products.

The step students rush is the very first one, writing the vertices in cyclic order and repeating the first at the end. Skip it and every cross-product after is scrambled.

In one-to-one lessons our teachers make you draw a quick sketch and trace the loop with your pen before a single number enters the array, so the order is never in doubt. We then separate the down-sum and the up-sum on two clear lines and reunite them only at the modulus, which is where careless sign errors usually creep in.

Because SPM Add Math uses analytic marking, a correctly built array earns method marks even if one product is mis-multiplied. Every teacher on spmaddmath.com.my is experienced.

Lessons are online and taught in English, at your own pace.