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

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what is the best solution for coordinate geometry of exercise 7.3. please help me to find the best and easy way to solve the coordinate geometry with tick.Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3)

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1. Let the vertices of the quadrilateral be A (- 4, â€“ 2), B ( â€“ 3, â€“ 5), C (3, â€“ 2), and D (2, 3).

Join AC and divide the quadrilateral into two triangles.

We have two triangles Î”ABC and Î”ACD.

Area of a triangle =Â 1/2 Ã— [x1(y2Â â€“ y3) + x2(y3Â â€“ y1) + x3(y1Â â€“ y2)]

Area of Î”ABCÂ = 1/2 [(-4) {(-5) â€“ (-2)} + (-3) {(-2) â€“ (-2)} + 3 {(-2) â€“ (-5)}]

= 1/2 (12 + 0 + 9)

= 21/2 square units

Area of Î”ACDÂ = 1/2 [(-4) {(-2) â€“ (3)} + 3{(3) â€“ (-2)} + 2 {(-2) â€“ (-2)}]

= 1/2 (20 + 15 + 0)

= 35/2 square units

Area of quadrilateral ABCD = Area of Î”ABC + Area of Î”ACD

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

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