Question

Find the area of the quadrilaterals, the coordinates of whose vertices are (–3, 2), (5, 4), (7, –6) and (–5, –4).

Answer 1

Let the vertices of the quadrilateral be A (−3, 2), B (5, 4), C (7, −6), and D (−5, −4). Join AC to form two triangles ΔABC and ΔACD.

Area  of triangle `=1/2{x_1(y_2-y_3) +x_2(y_3-y_1)+x_3(y_1-y_2)}` 

Area of ΔABC `=1/2{-3(4+6)+5(-6-2)+7(2-4)}`

 `=1/2(-30-40-14)=-42`

∴ Area of ΔABC = 42 square units 

Area of ΔACD `=1/2{-3(-6+4)+7(-4-2)-5(2+6)}` 

`=1/2 {6-42-40}=-38` 

∴ Area of ΔACD = 38 square units 

Area of `square`ABCD = Area of ΔABC + Area of ΔACD 

= (42 + 38) square units = 80 square units