(Q27) Find the area of the quadrilateral whose vertices, taken in order, are (-4,-2),(-3,-5),(3,-2) and (2,3).

A(-4,-2)B(-3,-5)C(3,-2)D(2,3)

Given : A(-4,-2), B(-3,-5), C(3,-2) and D(2,3)

are the vertices of the quadrilateral ABCD

Area of ABCD =∆ABC + ∆ACD

Area of a triangle

=
1
2
 |x1(y2-y3)+x2(y3-y1)+x3(y1-y2)= 0|
Area of triangle ABC =
1
2
 |-4×(-+)-3(-+)+3(-+)|
         =
1
2
 || =
2
 sq.units
Area of triangle ACD =
1
2
 |-4×(--)+3(+)+2(-+)|
         =
1
2
 || =
2
 sq.units
ABCD =
2
 +
2
 =
2
 ∴Area of Quadrilateral = sq.units .