The Area of any Irregular Quadrilateral

A plane figure bounded by four straight line segments is called an irregular quadrilateral. The area of any irregular quadrilateral can be calculated by dividing it into triangles.

 

Example:

Find the area of a quadrilateral ABCD whose sides are 9 m, 40 m, 28 m and 15 m respectively and the angle between the first two sides is a right angle.

 

Solution:


irregular-quadrilateral-01

From the figure we notice that ABD is a right-angled triangle, in which AB = 40 m, AD = 9 m. Also

\begin{gathered} B{D^2} = A{B^2} + A{D^2} \\ B{D^2} = {40^2} + {9^2} \\ BD = \sqrt {{{40}^2} + {9^2}} = \sqrt {1681} = 41 \\ \end{gathered}

Now, the area of \Delta ABD = \frac{1}{2} \times 40 \times 9 = 180 m
In \Delta BCD, BD = a = 41 m, DC = b = 28 m, CB = c = 15 m
\therefore s = \frac{{a + b + c}}{2} = \frac{{41 + 28 + 15}}{2} = 42 m

Now,

\begin{gathered} {\text{Area}}\,{\text{of}}\,\Delta BCD = \sqrt {s(s - a)(s - b)(s - c)} \, \\ {\text{Area}}\,{\text{of}}\,\Delta BCD = \sqrt {42(42 - 41)(42 - 28)(42 - 15)} = \sqrt {42 \times 14 \times 27} = 126\,sq\,m \\ \end{gathered}

The area of the quadrilateral ABCD = area of \Delta ABD  + area of \Delta BCD
The area of the quadrilateral ABCD  = (180 + 126) = 306 square meters.

 

Example:

In a quadrilateral the diagonal is 42 cm and the two perpendiculars on it from the other vertices are 8 cm and 9 cm respectively. Find the area of the quadrilateral.


irregular-quadrilateral-02
 

Solution:

Given that from the figure AC = 42 m, BN = 9 m, DM = 8 m
The area of ABCD = area of \Delta ABC + area of \Delta ACD
The area of ABCD = \frac{1}{2} \times 9 \times 42 + \frac{1}{2} \times 8 \times 42 = 189 + 168 = 357 square meters.