Question

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

Answer 1

Let the vertices of the quadrilateral be A(-4,8), B(-3,-4), C(0,-5) and D(5,6)
Join AC to form two triangles, namely $\Delta$ ABC and $\Delta$ ACD.
Area of quadrilateral ABCD = Area of $\Delta$ ABC + Area of $\Delta$ ACD
We know
Area of triangle having vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)=\frac{1}{2}\left(x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$
Now,
Area of $\Delta ABC=\frac{1}{2}\left(\right(-4\left)\right(-4+5)+(-3\left)\right(-5-8)+0(8+4)$
$=\frac{1}{2}(-4+39)$
$=\frac{35}{2}$
Area of $\Delta ACD=\frac{1}{2}(-4)(-6-5)+5(-5-8)+0(8-6)$
$=\frac{1}{2}(-44-65)$
$=\frac{109}{2}$
$\therefore$ Area of quadrilateral ABCD = Area of $\Delta$ ABC + Area of $\Delta$ ACD $=\frac{35}{2}+\frac{109}{2}=72sq.units$
Thus, the area of quadrilateral ABCD is 72 sq.units.