The Area of a Trapezoid

A trapezoid is a quadrilateral with two sides parallel and the other two unparallel. The parallel sides of a trapezoid are called the bases. The altitude is the perpendicular distance between the bases.

Let ABCD be a trapezoid whose sides AB and CD are parallel, and CB and AD are the unparallel sides. Let a and b be the lengths of the parallel sides and h be the height of the trapezoid.


The diagonal of a trapezoid divides it into two triangles that have the same altitude and have as their bases the bases of the trapezoid. Therefore, in the trapezoid ABCD, the diagonal BD divides it into triangles ABD and BCD.

The area of the trapezoid  = {\text{Area of }}\Delta ABD + {\text{Area of }}\Delta BDC
The area of the trapezoid  = \frac{1}{2}AB \times h + \frac{1}{2}DC \times h = \frac{1}{2}(AB + DC) \times h = \frac{1}{2}(a + b) \times h

The area of the trapezoid  = \frac{{{\text{Sum of parallel sides}}}}{2} \times {\text{height}}



The section of an open channel is a trapezoid, the vertical height of which is 3 m and the breadth of the bottom is 2 m. The inclination of the sides to the horizontal is 60^\circ outwards. Find the area.



Given that AD = NM = 2 m and AM = 3 m


The area of the trapezoid  = \frac{{{\text{Sum of parallel sides}}}}{2} \times h = \frac{{AD + BC}}{2} \times AM
But the parallel sides are AD and CB


BC = BM + MN + NC…(i)

Now \frac{{BM}}{{AM}} = \cot 60^\circ
MB = AM\cot 60^\circ = 3 \times \frac{1}{{\sqrt 3 }} = 1.732

Similarly, NC = 1.732 and (i) gives

BC = 1.732 + 2 + 1.732 = 5.464

The area of the trapezoid  = \frac{{{\text{2}} \times {\text{5}}{\text{.464}}}}{2} \times 3 = \frac{{7.464}}{2} \times 3 = 3.732 \times 3 = 11.196 square m.