The figure of quadrilateral whose vertices are A(– 4, 5), B(0, 7), C(5, –5) and D(–4, –2) is shown in the below figure.

Area of the quadrilateral ABCD
= area of ΔABC + area of ΔADC ... (i)
Now, Area of ΔABC =

1

2

|

−4	5	1
0	7	1
5	−5	1

|
=

1

2

|- 4 (7 + 5) - 0 (5 + 5) + 5 (5 - 7)| =

1

2

|- 4 (12) + 5 (- 2)| =

1

2

|- 58| = 29 sq. units
Also, Area of ΔADC =

1

2

|

−4	5	1
−4	−2	1
5	−5	1

| =

1

2

|63| =

63

2

sq.units
∴ Area of quadrilateral ABCD = 29 +

63

2

=

121

2

sq. units [From (i)]