Question

The coordinates of the vertices of a rectangle are $\left(-3,5\right),\left(0,-4\right),\left(3,7\right)$ and $\left(6,-2\right).$ How do you find the area of this figure?

Answer 1

Step 1: Find the area of triangle $ABD$

The four vertices are $A\left(-3,5\right),B\left(0,-4\right),C\left(3,7\right)$ and $D\left(6,-2\right)$

We know that the area of triangle joining three points $\left(x_1,y_1\right),\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$ is given by

$\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

$\therefore$Area of $\begin{array}{rcl}∆ABD& =& \frac{1}{2}\left|0\left(7-5\right)+3\left(5-\left(-4\right)\right)+\left(-3\right)\left(-4-7\right)\right|\end{array}$

$\begin{array}{rcl}& =& \frac{1}{2}\left|27+33\right|\\ & =& 30\mathrm{sq}.\mathrm{units}\end{array}$

Step 2: Find the area of triangle $BDC$

Area of $∆BDC=\frac{1}{2}\left|0\left(-2-7\right)+6\left(7-\left(-4\right)\right)+3\left(-4-\left(-2\right)\right)\right|$

$$\begin{gathered} =\frac{1}{2}\left|66-6\right| \\ =30\mathrm{sq}.\mathrm{units} \end{gathered}$$

Step 3: Find the area of rectangle $ABCD$

Area of rectangle $ABCD$$=$Area of $ABD$$+$Area of $BDC$

$$\begin{gathered} =30+30 \\ =60sq.units \end{gathered}$$

Hence, the required area of rectangle is $60\mathrm{sq}.\mathrm{units}$