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Question

Question 4
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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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 trianglesΔABC and ΔACD.
Area of a triangle
=12{x1(y2y3)+x2(y3y1)+x3(y1y2)}
Area of ΔABC
=12[(4){(5)(2)}+(3){(2)(2)}
+3{(2)(5)}]
=12(12+0+9)
=212square units
Area of ΔACD
=12[(4){(2)(3)}+3{(3)(2)}
+2{(2)(2)}]
=12(20+15+0)
=352 square units
Area of ΔABCD
=Area of ΔABC+ Area of ΔACD
=(212+352) square units = 28 square units

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