The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. The area of the field is?

Answer: Option A

Let ABCD is quadrilateral and its BD diagonal
BD = 24 metres
And, AM = 8 metres
CN = 13 metres

$$\begin{array}{rl}& \text{Area of }\square ABCD=\text{ar}\left(\Delta ABD\right)+\text{ar}\left(\Delta BCD\right)\\ & =\left(\frac{1}{2}\times BD\times AM\right)+\left(\frac{1}{2}\times BD\times CN\right)\\ & =\frac{1}{2}\times BD\left[AM+CN\right]\\ & =\frac{1}{2}\times 24\left[8+13\right]\\ & =12\times 21\\ & \text{Area of }\square ABCD=252\text{ metr}{\text{e}}^2\end{array}$$