Question

Find the area of the quadrilateral whose vertices, taken in order, are $(-4,-2),(-3,-5),(3,-2)$ and $(2,3)$ .

Answer 1

Find the area of given quadrilateral

Given vertices are $A(-4,-2),B(-3,-5),C(3,-2)\&D(2,3)$

If the vertices of quadrilateral is given as $A(x_1,y_1),B(x_2,y_2),A(x_3,y_3)\&\hspace{0.17em}D(x_4,y_4)$
then area of quadrilateral $=\frac{1}{2}\left\{(x_1{y}_2+x_2{y}_3+x_3{y}_4+x_4{y}_1)-(x_2{y}_1+x_3{y}_2+x_4{y}_3+x_1{y}_4)\right\}$
Area of given quadrilateral

$=\frac{1}{2}\left\{\left(\right(-4\left)\right(-5)+(-3\left)\right(-2)+(3\left)\right(3)+2(-2\left)\right)-\left(\right(-3\left)\right(-2)+(3\left)\right(-5)+(2\left)\right(-2)+(-4\left)\right(3\left)\right)\right\}$
$$\begin{gathered} =\frac{1}{2}\left\{(20+6+9-4)-(6-15-4-12)\right\} \\ =\frac{1}{2}\times (31+27) \\ =28\mathrm{sq}.\mathrm{unit} \end{gathered}$$

Hence the required area is $28\mathrm{sq}.\mathrm{unit}$