The diagonal of a quadrilateral shaped field is 24m and the perpendiculars dropped on it from the remaining opposite vertices are 8m and 12m. Find the area of the field in $m^2$.

The correct option is D. $240m^2$

We can calculate the area of this quadrilateral as the sum of two triangles taking diagonal length 24m as the base for both triangles and perpendicular length 8m as the height of one triangle and perpendicular length 12m as the height of the other triangle.

Area of triangle = $\frac{1}{2}\times base\times height$

Area of the first triangle = $\frac{1}{2}\times 24\times 8$

Area of the second triangle = $\frac{1}{2}\times 24\times 12$

Area of quadrilateral = Area of the first triangle + Area of the second triangle

Area of quadrilateral = $\frac{1}{2}\left(24\right)(8+12)=\frac{1}{2}\left(24\right)\left(20\right)=240m^2$