# 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 $= \frac{{{\text{Sum of parallel sides}}}}{2} \times {\text{height}}$

Example:

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.

Solution:

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.