# 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 $15$m respectively and the angle between the first two sides is a right angle.

Solution:

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

Now, 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,

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

Solution:

Given that from the figure $AC = 42$m, $BN = 9$m, $DM = 8$m
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