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