Find the area of the quadrilateral whose vertices taken in order are

Solution:

Join A and C. The given points are

$\mathrm{A}(-4,-2), \mathrm{B}(-3,-5), \mathrm{C}(3,-2)$ and $\mathrm{D}(2,3)$



Area of $\triangle \mathrm{ABC}$

$=\frac{\mathbf{1}}{\mathbf{2}}[(-4)(-5+2)-3(-2+2)+3(-2+5)]$

$=\frac{\mathbf{1}}{\mathbf{2}}[12+0+9]=\frac{\mathbf{2 1}}{\mathbf{2}}=10.5$ sq. units

Area of $\Delta \mathrm{ACD}$

$=\frac{1}{2}[(-4)(-2-3)+3(3+2)+2(-2+2)]$

$=\frac{\mathbf{1}}{\mathbf{2}}\left[20+15 \mid=\frac{\mathbf{3 5}}{\mathbf{2}}=17.5\right.$ sq. units.

Area of quadrilateral ABCD

$=\operatorname{ar}(\Delta \mathrm{ABC})+\operatorname{ar}(\Delta \mathrm{ACD})$

$=(10.5+17.5) \mathrm{sq} .$ units $=28 \mathrm{sq} .$ units