Question

If A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area. [CBSE 2014]

Answer 1

It is given that A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD.



Area of the quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD

$$\begin{gathered} \mathrm{ar}\left(∆\mathrm{ABC}\right)=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ =\frac{1}{2}\left|-3\left[-7-\left(-8\right)\right]+\left(-2\right)\left(-8-5\right)+1\left[5-\left(-7\right)\right]\right| \\ =\frac{1}{2}\left|-3+26+12\right| \\ =\frac{35}{2}\mathrm{square}\mathrm{units} \end{gathered}$$

$$\begin{gathered} \mathrm{ar}\left(∆\mathrm{ACD}\right)=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ =\frac{1}{2}\left|-3\left(-8-3\right)+1\left(3-5\right)+6\left[5-\left(-8\right)\right]\right| \\ =\frac{1}{2}\left|33-2+78\right| \\ =\frac{109}{2}\mathrm{square}\mathrm{units} \end{gathered}$$

∴ Area of the quadrilateral ABCD = $\frac{35}{2}+\frac{109}{2}=\frac{144}{2}=72$ square units

Hence, the area of the given quadrilateral is 72 square units.