Answer:
Explanation:
Let A(–4,–2), B(–3,–5), C(3,–2) and D(2,3) be the vertices of the quadrilateral ABCD.
Area of a quadrilateral ABCD= Area of △ABC+ Area of △ACD
By using a formula for the area of a triangle =
2
1
∣x
1
(y
2
−y
3
)+x
2
(y
3
−y
2
)+x
3
(y
1
−y
2
)∣
Area of △ABC
=
2
1
[−4(−5+2)+−3(−2+2)+3(−2+5)]
=
2
1
[12+9]
=
2
21
sq.units
Area of △ACD=
2
1
[−4(3+2)+−2(−2+2)+3(−2−3)]
=
2
1
[−20−15]
=
2
35
sq.units
∴Area of quadilateral=
2
21
+
2
35
= 2/ 56 =28 sq.unit