Find the area of a quadrilateral whoose vertices taken an order, are A (-3,0) B(2,0),C( 2,4), D(-3,5)
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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 =21∣x1(y2−y3)+x2(y3−y2)+x3(y1−y2)∣
Area of △ABC
=21[−4(−5+2)+−3(−2+2)+3(−2+5)]
=21[12+9]
=221sq.units
Area of △ACD=21[−4(3+2)+−2(−2+2)+3(−2−3)]
=21[−20−15]
=235sq.units
∴Area of quadilateral=221+235=256=28 sq.unit
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