Question

Find the area of quadrilateral ABCD whose vertices are A(−4, −2), B(−3, −5), C(3, −2) and D(2, 3).

Answer 1

Join A and C . Then,

Area of quadilateral ABCD$=\left(\text{area of}∆ABC\right)+\left(\text{area of}∆ACD\right)$
$$\begin{gathered} \text{Area of}∆ABC \\ =\frac{1}{2}\left|\left\{-4\left(-5+2\right)-3\left(-2+2\right)+3\left(-2+5\right)\right\}\right| \\ =\frac{1}{2}\left|-4\left(-3\right)-3\left(0\right)+3\left(3\right)\right| \\ =\frac{1}{2}\left|12-0+9\right| \\ =\frac{21}{2}\text{sq units} \end{gathered}$$

$$\begin{gathered} \text{Area of}∆ACD \\ =\frac{1}{2}\left|\left\{-4\left(-2-3\right)+3\left(3+2\right)+2\left(-2+2\right)\right\}\right| \\ =\frac{1}{2}\left|-4\left(-5\right)+3\left(5\right)+2\left(0\right)\right| \\ =\frac{1}{2}\left|20+15+0\right| \\ =\frac{35}{2}\text{sq units} \end{gathered}$$
Therefore, area of quadilateral ABCD $=\left(\frac{21}{2}+\frac{35}{2}\right)$
$=28\text{sq units}$