Question

Find the area of quadrilateral ABCD whose vertices are A(−3, −1), B(−2, −4), C(4, −1) and D(3, 4).
[CBSE 2013C]

Answer 1

By joining A and C, we get two triangles ABC and ACD.
Let $A\left(x_1,y_1\right)=A\left(-3,-1\right),B\left(x_2,y_2\right)=B\left(-2,-4\right),C\left(x_3,y_3\right)=C\left(4,-1\right)\mathrm{and}D\left(x_4,y_4\right)=D\left(3,4\right)$. Then
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ABC=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ =\frac{1}{2}\left[-3\left(-4+1\right)-2\left(-1+1\right)+4\left(-1+4\right)\right] \\ =\frac{1}{2}\left[9-0+12\right]=\frac{21}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ACD=\frac{1}{2}\left[x_1\left(y_3-y_4\right)+x_3\left(y_4-y_1\right)+x_4\left(y_1-y_3\right)\right] \\ =\frac{1}{2}\left[-3\left(-1-4\right)+4\left(4+1\right)+3\left(-1+1\right)\right] \\ =\frac{1}{2}\left[15+20+0\right]=\frac{35}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$
So, the area of the quadrilateral ABCD is $\frac{21}{2}+\frac{35}{2}=28\mathrm{sq}.\mathrm{units}$ sq. units.