Question

In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD is 36 m, AL ⊥ BD and CM ⊥ BD such that AL = 19 m and CM = 11 m. Find the area of the field.

Answer 1

$\mathrm{Area}\mathrm{of}\mathrm{quadrilateral}\mathrm{ABCD}=\left(\mathrm{Area}\mathrm{of}∆\mathrm{ABD}\right)+\left(\mathrm{Area}\mathrm{of}∆\mathrm{BCD}\right)$
$=\left(\frac{1}{2}\times \mathrm{BD}\times \mathrm{AL}\right)+\left(\frac{1}{2}\times \mathrm{BD}\times \mathrm{CM}\right)$
$$\begin{gathered} =\left[\left(\frac{1}{2}\times 36\times 19\right)+\left(\frac{1}{2}\times 36\times 11\right)\right]{\mathrm{m}}^2 \\ =\left(342+198\right){\mathrm{m}}^2 \\ =540{\mathrm{m}}^2 \end{gathered}$$

$\mathrm{Hence},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{field}\mathrm{is}540{\mathrm{m}}^2.$