Question

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

Answer 1

Joining diagonal AC we get two triangles ΔADC and ΔABC .

Area of ΔADC :

$x_1=-4,x_2=5,x_3=0,y_1=8,y_2=6,y_3=-5$

$Area =\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_1)]\\\\=\frac{1}{2}[-4(6-(-5))+5(-5-8)+0(8-6)]\\\\=\frac{1}{2}[-44-65]\\\\=\frac{1}{2}[-109]\\\\=-54.5units$

Area of ΔABC :

$x_1=-4,x_2=-3,x_3=0,y_1=8,y_2=-4,y_3=-5$

$Area =\frac{1}{2}[-4(-4-(-5))+(-3)(-5-8)+0(8-(-4))]\\\\=\frac{1}{2}[-4+39]\\\\=\frac{1}{2}[35]\\\\=17.5units$

Total area = |-54.5| + |17.5| = 72 units