Question

Find the area of the quadrilateral whose vertices are taken in order are $\left(-4.-2\right),\left(-3,-5\right),\left(3,-2\right)$ and $\left(2,3\right)$

Answer 1

Solve for area of the given quadrilateral

Let the vertices of the quadrilateral be:

$$\begin{gathered} A\left(x_1,y_1\right)=\left(-4,-2\right) \\ B\left(x_2,y_2\right)=\left(-3,-5\right) \\ C\left(x_3,y_3\right)=\left(3,-2\right) \\ D\left(x_4,y_4\right)=\left(2,3\right) \end{gathered}$$

From the figure, we can say that,

Area of quadrilateral $ABCD$$=$Area of $△$$ABC$$+$ Area of $△$$ACD$

Step 1: Find area of $△$$ABC$

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

$\therefore$Area of $△ABC=\frac{1}{2}\left|-4\left(-5+2\right)-3\left(-2+2\right)+3\left(-2+5\right)\right|$

$=\frac{1}{2}\left|12-0+9\right|$

$\Rightarrow$Area of $△ABC=\frac{21}{2}sq.units$

Step 2: Find area of $△$$ACD$

$\therefore$Area of $△ACD=\frac{1}{2}\left|-4\left(-2-3\right)+3\left(3+2\right)+2\left(-2+2\right)\right|$

$=\frac{1}{2}\left|20+15+0\right|$

$\Rightarrow$Area of $△ACD=\frac{35}{2}sq.units$

Step 3: Find area of quadrilateral $ABCD$

Area of quadrilateral $ABCD$$=$$Areaof$$△$$ABC$$+$$Areaof$$△$$ACD$

$=\frac{21}{2}+\frac{35}{2}$

$\Rightarrow$Area of quadrilateral $ABCD=28sq.units$.

Hence, area of quadrilateral $ABCD$ is $28sq.units$