Question

what is the area of triangle ADC if its vertices,taken in the order (-4,-2);(2,3);(3,-2)​

Answer 1

Answer:

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+