How To Calculate The Area Of An Irregular Trapezoid

While it might seem like finding the area of various shapes and polygons is limited to a math class in school, the fact is that finding the area of polygons is something that applies to almost all parts of life. From agricultural calculations to understanding the area of a certain ecosystem in biology to computer science, calculating areas of complex shapes is an essential skill to master.

It's usually easier to measure the area of shapes with all equal sides and straightforward formulas. However, "irregular" shapes such as an irregular trapezium, also known as an irregular trapezoid, are common and need to be calculated as well. Thankfully, there are irregular trapezoid area calculators and a trapezoid area formula that makes the process simple.

A trapezoid is a four-sided polygon, also known as a quadrilateral, that has at least ​one set of parallel sides​. This differentiates a trapezoid from a parallelogram since parallelograms always have ​two​ sets of parallel sides. This is why you can consider all parallelograms to be trapezoids, but not all trapezoids are parallelograms.

The parallel sides of a trapezoid are called ​bases​ while the non-parallel sides of a trapezoid are called ​legs​. A regular trapezoid, also called an isosceles trapezoid, is a trapezoid where the non-parallel sides (the legs) are equal in length.

An irregular trapezoid, also called an irregular trapezium, is a trapezoid where the non-parallel sides are not equal in lengths. Meaning, they have legs of two different lengths.

In order to find the area of a trapezoid, you can use the following equation:

\(\text{Area } = \bigg(\frac{b_1 + b_2}{2}\bigg) × h\)

​​b​​_​1​ and ​**​b​₂**​ are the lengths of the two bases on the trapezoid; ​​h​​ is equal to the height of the trapezoid, which is the length from the bottom base to the top base line.

You're not always given the height of the trapezoid. If this is the case, you can often figure out the height using the Pythagorean Theorem.

This first example is going to represent a problem when you know all of the values of the trapezoid.

\(b_1 = 4 \text{ cm}\)
\(b_2 = 12 \text{ cm}\)
\(h = 8 \text{ cm}\)

Simply plug the numbers into the trapezoid area formula and solve.

\(\begin{aligned}\)
\(A &= \bigg(\frac{b_1 + b_2}{2}\bigg) × h\)
\(&= \bigg(\frac{4 \text{ cm} +12 \text{ cm}}{2}\bigg) × 8 \text{ cm}\)
\(&= \bigg(\frac{16 \text{ cm}}{2}\bigg) × 8 \text{ cm}\)
\(&= 8 \text{ cm} × 8 \text{ cm} = 64 \text{ cm}^2\)
\(\end{aligned}\)

In other problems or situations with irregular trapezoids, you're often only given the measurements of the bases and the legs of the trapezoid along with some of the trapezoid angles, which leaves you to calculate the height on your own before you can calculate the area.

You can then use the lengths and angles in order to calculate the height of the trapezoid using common triangular angle rules.

Think about it . . . when you draw in a line of height on a trapezoid at the endpoint of the smaller base length down to the longer base length, you create a triangle with that line as one side, the leg of the trapezoid as the second side and the distance from the point where the height line touches the larger base to the point where that base meets the leg as the third side (see a detailed picture here).

Let's say you have the following values (see image on schoolmath-help/how-to-find-the-area-of-a-trapezoid'>this page):

\(b_1 = 16 \text{ cm}\)
\(b_2 = 25 \text{ cm}\)
\(\text{leg }2 = 12 \text{ cm}\)
\(\text{Angle between } b_2 \text{ and leg } 2 = 30 \text{ degrees}\)

Knowing the angles and one of the side length values means that you can then use sin and cos rules to find the height. The hypotenuse would be equal to leg 2 (12 cm) and we have the angles to calculate the height.

Let's use sin to find the height using the given 30 degree angle, which would make height is equal to "opposite" in the sin equation:

\(\sin(\text{angle}) = \frac{\text{height}}{\text{hypotenuse}} \
\,\)
\(\sin(30) = \frac{ \text{height} }{12 \text{ cm}} \
\,\)
\(\sin(30) × 12 \text{ cm} = \text{height} = 6 \text{ cm}\)

Now that you have the height value, you can calculate the area using the area formula:

\(\begin{aligned}\)
\(A &= \bigg(\frac{b_1 + b_2}{2}\bigg) × h\)
\(&= \bigg(\frac{b_1 + b_2}{2} \bigg) × h\)
\(&= \bigg(\frac{16 \text{ cm} + 25 \text{ cm}}{2}\bigg) × 6 \text{ cm}\)
\(&= \bigg(\frac{41 \text{ cm}}{2}\bigg) × 6 \text{ cm}\)
\(&= 20.5 \text{ cm} × 6 \text{ cm} = 123 \text{ cm}^2\)
\(\end{aligned}\)