Q22 of 55 Page 1

Find the area of the quadrilateral ABCD, whose vertices are A(-3,-1), B(-2,-4), C(4,-1), and D(3,4).


The quadrilateral is divided into two triangles ABC and ACD.


Area of the quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD.


Area of triangle ABC =


1/2 [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]


= 1/2 [-3(-4 + 1) + -2(-1 + 1) + 4(-1 + 4)]


= 1/2 [9 + 0 + 12]


= 21/2 sq. units


Area of triangle ACD =


1/2 [x1(y3 – y4) + x3(y4 – y1) + x4(y1 – y3)]


= 1/2 [-3(-1 - 4) + 4(4 + 1) + 3(-1 + 1)]


= 1/2 [15 + 20 + 0]


= 35/2 sq. units


∴ Area of the quadrilateral = (21/2) + (35/2)


= (56/2)


= 28 sq. units


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