4.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. Find the area of the field.

As per the figure we have area of field=area of ∆ABD+area of ∆BCD

\(=\frac{1}2\times base \times height+\frac{1}2\times base \times height\)

\(=\frac{1}2\times 24 \times 13+\frac{1}2\times 24\times 8\)

\(=12\times 13+12\times 8\)

\(=12\times(13+8)\)

\(=12\times 21\)

\(=252m^2\)

Hence the required area of the given field=\(252m^2\)