Question

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

Answer 1

It is given that P(−5, −3), Q(−4, −6), R(2, −3) and S(1, 2) are the vertices of a quadrilateral PQRS.



Area of the quadrilateral PQRS = Area of ∆PQR + Area of ∆PRS

$$\begin{gathered} \mathrm{ar}\left(∆\mathrm{PQR}\right)=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ =\frac{1}{2}\left|-5\left[-6-\left(-3\right)\right]+\left(-4\right)\left[-3-\left(-3\right)\right]+2\left[-3-\left(-6\right)\right]\right| \\ =\frac{1}{2}\left|15+0+6\right| \\ =\frac{21}{2}\mathrm{square}\mathrm{units} \end{gathered}$$

$$\begin{gathered} \mathrm{ar}\left(∆\mathrm{PRS}\right)=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ =\frac{1}{2}\left|-5\left(-3-2\right)+2\left[2-\left(-3\right)\right]+1\left[-3-\left(-3\right)\right]\right| \\ =\frac{1}{2}\left|25+10+0\right| \\ =\frac{35}{2}\mathrm{square}\mathrm{units} \end{gathered}$$

∴ Area of the quadrilateral PQRS = $\frac{21}{2}+\frac{35}{2}=\frac{56}{2}=28$ square units

Hence, the area of the given quarilateral is 28 square units.