Area of Trapezoid

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

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


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

Area of the trapezoid  = {\text{Area of }}\Delta  ABD + {\text{Area of }}\Delta BDC
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

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 = 2m and AM = 3m


Area of the trapezoid  = \frac{{{\text{Sum of  parallel sides}}}}{2} \times h = \frac{{AD + BC}}{2} \times AM
But 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

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