Question

A(−4, −2), B(−3, −5) C(3, −2) and D(2, k) are the vertices of a quadrilateral ABCD. Find the value of k, if the area of the quadrilateral is 28 sq units.

Answer 1

Let A(−4, −2), B(−3, −5), C(3, −2) and D(2, k) be the vertices of quadrilateral ABCD.
Join AC. Then, area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD.

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{triangle}ABC \\ =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ =\frac{1}{2}\left\{\left(-4\right)\left(-5-\left(-2\right)\right)+\left(-3\right)\left(\left(-2\right)-\left(-2\right)\right)+3\left(-2-\left(-5\right)\right)\right\} \\ =\frac{1}{2}\left\{-4\left(-3\right)-3\left(0\right)+3\left(3\right)\right\} \\ =\frac{1}{2}\left\{12+9\right\} \\ =\frac{1}{2}\left(21\right) \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{triangle}ABC=10.5\mathrm{sq}.\mathrm{units}. \end{gathered}$$
$$\begin{gathered} \mathrm{Therefore},\mathrm{area}\mathrm{of}\mathrm{triangle}ACD=\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{quadrilateral}ABCD-\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{triangle}ABC \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}ACD=28\mathrm{sq}\mathrm{units}-10.5\mathrm{sq}\mathrm{units}=17.5\mathrm{sq}.\mathrm{units}. \end{gathered}$$
To find the value of k, let us consider the area of triangle ACD.
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{triangle}ACD=\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ 17.5=\frac{1}{2}\left\{\left(-4\right)\left(-2-k\right)+3\left(k-\left(-2\right)\right)+2\left(-2-\left(-2\right)\right)\right\} \\ 17.5=\frac{1}{2}\left\{-4\left(-2-k\right)+3\left(k+2\right)+2\left(0\right)\right\} \\ 17.5=\frac{1}{2}\left\{8+4k+3k+6\right\} \\ 17.5=\frac{1}{2}\left(7k+14\right) \\ 35=7k+14 \\ 7k=35-14 \\ k=\frac{21}{7}=3 \end{gathered}$$
Therefore, the value of k = 3.