Area of a Quadrilateral from Its Vertices

The array (shoelace) method finds a polygon’s area from its vertices. List them in order around the shape, repeat the first at the end, and take half the size of (sum of down-products) minus (sum of up-products).

A quadrilateral has vertices $A(1, 1)$, $B(5, 2)$, $C(6, 5)$ and $D(2, 4)$, taken in order. Find its area. [4]

The working

Step 1, set up the array, listing the vertices in order and repeating $A$ at the end: $\frac{1}{2}\left|\begin{array}{ccccc} 1 & 5 & 6 & 2 & 1 \\ 1 & 2 & 5 & 4 & 1 \end{array}\right| \quad \text{(M1)}$

Step 2, sum the down-diagonal products ($x_i\,y_{i+1}$): $(1)(2) + (5)(5) + (6)(4) + (2)(1) = 2 + 25 + 24 + 2 = 53 \quad \text{(M1)}$

Step 3, sum the up-diagonal products ($y_i\,x_{i+1}$): $(1)(5) + (2)(6) + (5)(2) + (4)(1) = 5 + 12 + 10 + 4 = 31$

Step 4, take half the size of the difference: $\text{Area} = \tfrac{1}{2}\,|53 - 31| = \tfrac{1}{2}(22) = 11 \quad \text{(A1, A1)}$

The area is $11$ square units.

Where the marks are won and lost

• List the vertices consecutively around the perimeter ($A \to B \to C \to D$). Jumbled order gives a wrong, often self-intersecting, figure.
• Close the loop: repeat the first vertex at the end so every edge is counted.
• Take the absolute value and the factor of $\frac12$; a negative before the modulus just means you went clockwise.

Common mistakes

• Not repeating the first vertex, so the last edge is missed.
• Listing vertices out of order.
• Forgetting the $\frac12$ or the modulus.

Topic home: Straight-Line Graphs pillar. More: Worked examples.