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 ca be calculated by dividing it into triangles.

Example:

Find the area of a quadrilateral ABCD whose sides are 9, 40, 28 and 15m 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 = 40m, AD = 9m. 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, Area of \Delta ABD = \frac{1}{2}  \times 40 \times 9 = 180m
In \Delta BCD, BD = a = 41m, DC = b = 28m, CB = c = 15m
\therefore s = \frac{{a + b + c}}{2} = \frac{{41 +  28 + 15}}{2} = 42m
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}

Area of Quadrilateral ABCD = Area of \Delta ABD  + Area of \Delta BCD
Area of Quadrilateral ABCD  = (180 + 126) = 306 Square meter

Example:

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


irregular-quadrilateral-02

Solution:

Given that from the figure AC = 42m, BN = 9m, DM = 8m
Area of ABCD = Area of \Delta ABC + Area of \Delta ACD
Area of ABCD = \frac{1}{2}  \times 9 \times 42 + \frac{1}{2} \times 8 \times 42 = 189 + 168 = 357 Square meter

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