Find the area of quadrilateral formed by joining points (4, 2), (8, 2), (8, 14) and (4, 10).

Solution (By Examveda Team)

$$\eqalign{ & {\text{Area of }}\Delta ABC \cr & = \frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] \cr & = \frac{1}{2}\left[ {4\left( {2 - 14} \right) + 8\left( {14 - 2} \right) + 8\left( {2 - 2} \right)} \right] \cr & = \frac{1}{2}\left[ {4 \times \left( { - 12} \right) + 8\left( {12} \right) + 8\left( 0 \right)} \right] \cr & = \frac{1}{2}\left[ { - 48 + 96} \right] \cr & = \frac{1}{2} \times 48 \cr & = 24{\text{ sq}}{\text{. units}} \cr & {\text{Area of }}\Delta ACD \cr & = \frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] \cr & = \frac{1}{2}\left[ {4\left( {14 - 10} \right) + 8\left( {10 - 2} \right) + 4\left( {2 - 14} \right)} \right] \cr & = \frac{1}{2}\left[ {4 \times 4 + 8 \times 8 + 4 \times \left( { - 12} \right)} \right] \cr & = \frac{1}{2}\left[ {16 + 64 - 48} \right] \cr & = \frac{1}{2} \times 32 \cr & = 16{\text{ sq}}{\text{. units}} \cr & {\text{Hence Area of Quadrilateral }}ABCD \cr & = {\text{Area of }}\left( {\Delta ABC + \Delta ACD} \right) \cr & = 24 + 16 \cr & = 40{\text{ sq}}{\text{. units}} \cr} $$