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

For the quadrilateral

Step-by-step explanation:

The vertices of quadrilateral be given as A (-4,8), B (-3,-4), C (0,5) and D (5,6)

Area of quadrilateral ABCD is  a = Area of triangle ABC+area of triangle ACD

Area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3)

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

Area of triangle ABC=1/2[(-4)(-4-(-5))+(-3)[(-5)-(-8)]+(0)[(8)-(-4)]

=1/2(-4+39)=35/2

similarly,

Area of triangle ACD=1/2(44+65)=109/2

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

35/2+109/2=72 sq. units