Q29 of 47 Page 1

If A(-4,8), B(-3,-4), C(0,-5) and D(5,6) are the vertices of a quadrilateral ABCD, find its area.


To find the area of the quadrilateral ABCD, we divide the quadrilateral into two triangles by joining A and C.


So, Area(ABCD) = Area(ΔABC) + Area(ΔACD)


Now,


Area(ΔABC) = [x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]


= [(-4)(-4-(-5)) + (-3)(-5-8) + 0(8-(-4))]


= [(-4) × 1 + (-3) × (-13) + 0] = 35/2 = 17.5sq. units


Area(ΔACD) = [x1(y2-y3) + x2(y3-y1) + x3(y1-y2)]


= [(-4)(-5-6) + 0(6-8) + 5(8-(-5))]


= [44 + 0 + 65]


= [109] = 54.5 sq. units


Area(ABCD) = Area(ΔABC) + Area(ΔACD)


⇒ Area(ABCD) = 17.5 + 54.5 = 72sq.units


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