Question

Find the area of quadrilateral ABCD, whose vertices are (2,3),(4,2),(5,4),(3,3)

Answer 1

Hey mate your answer is here,

A(-3,1), B(-2,-4), C(4,-1), D(3,4) are the vertices of quadrilateral ABCD

Let AC be the diagonal of quad ABCD

therefore, ar of quad ABCD= ar triangle ABC+ ar triangle ADC

ar 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

ar of triangle ADC= 1/2[ (-3) (4+1) + 3(-1+1) + 4(-1-4)]

=1/2[-15+0-20]

= -35/2 sq units (since area cant be negative so area of triangle ADC=35/2 sq units)

Hence ar of quadrilateral ABCD= 21/2 + ( 35/2)

=56/2= 28 sq units

plz mark as the brilliant