Question

Find the area of the quadrilateral whose vertices, taken in order, are (–4, –2), (–3, –5), (3, –2) and (2, 3).

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

Answer 1

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

Area of a 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 [(-4) {(-5) - (-2)} + (-3) {(-2) - (-2)} + 3 {(-2) - (-5)}]`

= `1/2 (12+0+9)`

= `21/2` square units

Area of ΔACD = `1/2 [(-4) {(-2) - (3)} + 3{(3) - (-2)} + 2 {(-2) - (-2)}]`

= `1/2 (20+15+0)`

= `35/2` square units

Area of ☐ABCD = Area of ΔABC + Area of ΔACD

=`(21/2 + 35/2)` square units = 28 square units