question_answer

If \[A(-4,8),\text{ }B(-3,-4),\text{ }C(0,-5)\] and \[D(5,6)\]are the vertices of a quadrilateral ABCD, find its area.

Answer:

We have, \[A(-4,8),\text{ }B(-3,-4),\text{ }C(0,-5)\] and \[D(5,6)\]are the vertices of a quadrilateral.

Join A and C. Then, area of quadrilateral ABCD= (area of \[\Delta \,ABC\]) + (area of \[\Delta \,ACD\])

Area of \[\Delta \,ABC\]  \[=\frac{1}{2}[-4(-4+5)-3(-5-8)+0(8+4)]\]

\[=\frac{1}{2}[-4+39]=\frac{35}{2}\] sq. units

And, area of \[\Delta \,ACD\] \[=\frac{1}{2}[-4(-5-6)+0(6-8)+5(8+5)]\]

\[=\frac{1}{2}[44+65]\]

\[=\frac{109}{2}\] sq. units

\[\therefore \] Area of quadrilateral ABCD = Area of \[\Delta \text{ }ABC\] + Area of \[\Delta \text{ }ACD\]

\[=\frac{35}{2}+\frac{109}{2}\]

\[=\frac{144}{2}\] sq. units \[=72\] sq. units.