Question

Find the area of quadrilateral ABCD whose vertices are A(3, −1), B(9, −5), C(14, 0) and D(9, 19). [CBSE 2012]

Answer 1

By joining A and C, we get two triangles ABC and ACD.
Let $A\left(x_1,y_1\right)=A\left(3,-1\right),B\left(x_2,y_2\right)=B\left(9,-5\right),C\left(x_3,y_3\right)=C\left(14,0\right)\mathrm{and}D\left(x_4,y_4\right)=D\left(9,19\right)$. Then
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ABC=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ =\frac{1}{2}\left[3\left(-5-0\right)+9\left(0+1\right)+14\left(-1+5\right)\right] \\ =\frac{1}{2}\left[-15+9+56\right]=25\mathrm{sq}.\mathrm{units} \end{gathered}$$
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ACD=\frac{1}{2}\left[x_1\left(y_3-y_4\right)+x_3\left(y_4-y_1\right)+x_4\left(y_1-y_3\right)\right] \\ =\frac{1}{2}\left[3\left(0-19\right)+14\left(19+1\right)+9\left(-1-0\right)\right] \\ =\frac{1}{2}\left[-57+280-9\right]=107\mathrm{sq}.\mathrm{units} \end{gathered}$$
So, the area of the quadrilateral is 25 + 107 = 132 sq. units.