Question

(i) Find the area of a quadrilateral ABCD whose vertices are
A(−4, −2), B(−3, −5), C(3, −2) and D(2, 3)

(ii) Find the area of the quadrilateral ABCD whose vertices are
A(0, 0), B(6, 0), C(4, 3) and D(0, 3)

(iii) Find the area of the quadrilateral ABCD whose vertices are
A(1, 0), B(5, 3), C(2, 7) and D(−2, 4).

Answer 1

(i) The vertices are A(−4, −2), B(−3, −5), C(3, −2) and D(2, 3).
Join AC. Then,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC \\ =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ =\frac{1}{2}\left\{\left(-4\right)\left(-5-\left(-2\right)\right)+\left(-3\right)\left(\left(-2\right)-\left(-2\right)\right)+3\left(-2-\left(-5\right)\right)\right\} \\ =\frac{1}{2}\left\{-4\left(-3\right)-3\left(0\right)+3\left(3\right)\right\} \\ =\frac{1}{2}\left\{12+9\right\} \\ =\frac{1}{2}\left(21\right) \\ =10.5\mathrm{sq}.\mathrm{units} \end{gathered}$$
Now,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}ACD \\ =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ =\frac{1}{2}\left\{\left(-4\right)\left(-2-3\right)+3\left(3-\left(-2\right)\right)+2\left(-2-\left(-2\right)\right)\right\} \\ =\frac{1}{2}\left\{-4\left(-5\right)+3\left(5\right)+2\left(0\right)\right\} \\ =\frac{1}{2}\left\{20+15\right\} \\ =\frac{1}{2}\left(35\right) \\ =17.5\mathrm{sq}.\mathrm{units} \end{gathered}$$
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 10.5 sq. units + 17.5 sq. units = 28 sq. units

(ii) The vertices are A(0, 0), B(6, 0), C(4, 3) and D(0, 3).
Join AC. Then,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC \\ =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ =\frac{1}{2}\left\{0\left(0-3\right)+6\left(3-0\right)+4\left(0-0\right)\right\} \\ =\frac{1}{2}\left\{0+18+0\right\} \\ =\frac{1}{2}\left\{18\right\} \\ =9\mathrm{sq}.\mathrm{units} \end{gathered}$$
Now,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}ACD \\ =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ =\frac{1}{2}\left\{0\left(3-3\right)+4\left(3-0\right)+0\left(0-3\right)\right\} \\ =\frac{1}{2}\left\{0+12+0\right\} \\ =\frac{1}{2}\left\{12\right\} \\ =6\mathrm{sq}.\mathrm{units} \end{gathered}$$
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 9 sq. units + 6 sq. units = 15 sq. units

(iii) The vertices are A(1, 0), B(5, 3), C(2, 7) and D(−2, 4).
Join AC. Then,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC \\ =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ =\frac{1}{2}\left\{1\left(3-7\right)+5\left(7-0\right)+2\left(0-3\right)\right\} \\ =\frac{1}{2}\left\{1\left(-4\right)+5\left(7\right)+2\left(-3\right)\right\} \\ =\frac{1}{2}\left\{-4+35-6\right\} \\ =\frac{1}{2}\left(25\right) \\ =12.5\mathrm{sq}.\mathrm{units} \end{gathered}$$
Now,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}ACD \\ =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ =\frac{1}{2}\left\{1\left(7-4\right)+2\left(4-0\right)-2\left(0-7\right)\right\} \\ =\frac{1}{2}\left\{1\left(3\right)+2\left(4\right)-2\left(-7\right)\right\} \\ =\frac{1}{2}\left\{3+8+14\right\} \\ =\frac{1}{2}\left(25\right) \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}ACD=12.5\mathrm{sq}.\mathrm{units} \end{gathered}$$
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 12.5 sq. units + 12.5 sq. units = 25 sq. units