In the given figure, ABCD is a quadrilateral-shaped field in which diagonal BD is 36 m,

Area of quadrilateral $\mathrm{ABCD}=($ Area of $\Delta \mathrm{ABD})+($ Area of $\Delta \mathrm{BCD})$

$=\left(\frac{1}{2} \times \mathrm{BD} \times \mathrm{AL}\right)+\left(\frac{1}{2} \times \mathrm{BD} \times \mathrm{CM}\right)$

$=\left[\left(\frac{1}{2} \times 36 \times 19\right)+\left(\frac{1}{2} \times 36 \times 11\right)\right] \mathrm{m}^{2}$

$=(342+198) \mathrm{m}^{2}$

$=540 \mathrm{~m}^{2}$

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