The diagonal of a quadrilateral shaped field is 24 m

Given:

Diagonal of a quadrilateral shaped field $=24 \mathrm{~m}$

Perpendiculars dropped on it from the remaining opposite vertices are $8 \mathrm{~m}$ and $13 \mathrm{~m}$.

Now, we know :

Area $=\frac{1}{2} \times \mathrm{d} \times\left(\mathrm{h}_{1}+\mathrm{h}_{2}\right)$

$\therefore$ Area of the field $=\frac{1}{2} \times 24 \times(8+13)$

$=12 \times 21$

$=252 \mathrm{~m}^{2}$