Question

Find the area of quadrilateral PQRS whose vertices are P(−5, −3), Q(−4, −6), R(2, −3) and S(1, 2).
[CBSE 2015]

Answer 1

By joining P and R, we get two triangles PQR and PRS.
Let $P\left(x_1,y_1\right)=P\left(-5,-3\right),Q\left(x_2,y_2\right)=Q\left(-4,-6\right),R\left(x_3,y_3\right)=R\left(2,-3\right)\mathrm{and}S\left(x_4,y_4\right)=S\left(1,2\right)$. Then
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆PQR=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ =\frac{1}{2}\left[-5\left(-6+3\right)-4\left(-3+3\right)+2\left(-3+6\right)\right] \\ =\frac{1}{2}\left[15-0+6\right]=\frac{21}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆PRS=\frac{1}{2}\left[x_1\left(y_3-y_4\right)+x_3\left(y_4-y_1\right)+x_4\left(y_1-y_3\right)\right] \\ =\frac{1}{2}\left[-5\left(-3-2\right)+2\left(2+3\right)+1\left(-3+3\right)\right] \\ =\frac{1}{2}\left[25+10+0\right]=\frac{35}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$
So, the area of the quadrilateral PQRS is $\frac{21}{2}+\frac{35}{2}=28\mathrm{sq}.\mathrm{units}$ sq. units.