# Area Tutorial 5 – Area of a Trapezoid

In this tutorial, we are going to derive the area of a trapezoid. A trapezoid (sometimes called a trapezium) is a quadrilateral with exactly one pair of parallel sides. Trapezoid PQRS is shown below, with PQ parallel to RS.  We have learned that the area $A$ of the trapezoid with bases $b_1$ and $b_2$ and altitude $h$ is given by the formula $A_{PQRS} = \displaystyle\frac{(b_1 + b_2)h}{2}$.

Figure 1 - Trapezoid PQRS with PQ parallel to RS.

We are going to derive the area of a trapezoid in two ways: First by dividing into different sections and second by rotation.

Derivation 1: Area by Dividing into Regions

If we drop another line from Q, then we will have two altitudes namely PT and QU, which both have length $h$ units.

Figure 2 - Trapezoid PQRS divided into two triangles and a rectangle.

From Figure 2, it is clear that Area of PQRS = Area of PST + Area of PQUT + Area of QRU. We have learned that the area of a triangle is the product of its base and altitude divided by 2, and the area of a rectangle is the product of its length and width. Hence, we can easily compute the area of PQRS. It is clear that $A_{PQRS} = (ah/2) + b_1h + (ch/2)$.  Simplifying, we have  $A = \displaystyle\frac{ah + 2b_1 + ch}{2}$. Factoring we have, $A_{PQRS} = (a + 2b_1 + c) \frac{h}{2} = [(a + b_1 + c) + b_1] \frac{h}{2}.$ But, $a + b_1 + c$ is equal to $b_2$, the longer base of our trapezoid. Hence, $A_{PQRS}= (b_1 + b_2) \frac{h}{2}$.

Derivation 2: By Rotation

In the second derivation, we are going to duplicate the trapezoid and rotate it as shown below. It is evident that quadrilateral PS’P’S is a parallelogram (Why?). But we have learned that the area of the parallelogram is the product of its height and its base. Hence, $A_{PS'P'S} = (b_1 + b_2)h$.

Figure 3 - PQRS translated and rotated to form a parallelogram.

But the area of the trapezoid PQRS is half of the area of the parallelogram PS’P’S. Thus, $A_{PQRS} = \displaystyle\frac{(b_1 + b_2)h}{2}$.