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?

Solution (By Examveda Team)

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

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