Question

Find the area of quadrilateral ABCD whose vertices are A(−5, 7), B(−4, −5), C(−1, −6) and D(4, 5).

Answer 1

By joining A and C, we get two triangles ABC and ACD.
Let $A\left(x_1,y_1\right)=A\left(-5,7\right),B\left(x_2,y_2\right)=B\left(-4,-5\right),C\left(x_3,y_3\right)=C\left(-1,-6\right)\mathrm{and}D\left(x_4,y_4\right)=D\left(4,5\right)$. Then
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ABC=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ =\frac{1}{2}\left[-5\left(-5+6\right)-4\left(-6-7\right)-1\left(7+5\right)\right] \\ =\frac{1}{2}\left[-5+52-12\right]=\frac{35}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆ACD=\frac{1}{2}\left[x_1\left(y_3-y_4\right)+x_3\left(y_4-y_1\right)+x_4\left(y_1-y_3\right)\right] \\ =\frac{1}{2}\left[-5\left(-6-5\right)-1\left(5-7\right)+4\left(7+6\right)\right] \\ =\frac{1}{2}\left[55+2+52\right]=\frac{109}{2}\mathrm{sq}.\mathrm{units} \end{gathered}$$
So, the area of the quadrilateral ABCD is $\frac{35}{2}+\frac{109}{2}=72\mathrm{sq}.\mathrm{units}$.