Question

Draw a quadrilateral in the Cartesian plane, whose vertices are $(–4,5),(0,7),(5,–5)\text{and}(–4,–2)$. Also, find its area.

Answer 1

Drawing quadrilateral.
Let points be $A(-4,5),B(0,7),C(5,-5),D(-4,-2)$


Area of quadrilateral $ABCD=$Area of $△ABD+$ Area of $△BCD$

Step $:1$ Finding Area of $△ABD$ and Area of $△BCD$ separately.

Calculating area of $△ABD$
Here
$x_1=-4,y_1=5$
$x_2=0,y_2=7$
$x_3=-4,y_3=-2$
Area of $△ABD$
$=\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$
$=\frac{1}{2}|-4(7-(-2))+0(-2-5)+(-4\left)\right(5-7\left)\right|$
$=\frac{1}{2}|-4(7+2)+0-(4\left)\right(-2\left)\right|$
$=\frac{1}{2}|-4(9)+8|$
$=\frac{1}{2}|-28|$
$=\frac{1}{2}\times 28$
$=14$

Calculating Area of $△BCD$
Here $x_1=0,y_1=7$
$x_2=5,y_2=-5$
$x_3=-4,y_3=-2$

Area of $△BCD$
$=\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$
$=\frac{1}{2}\left|0\right(-5-(-2))+5(-2-7)-4(7-(-5)\left)\right|$
$=\frac{1}{2}|0+5(-9)-4(7+5\left)\right|$
$=\frac{1}{2}|0-45-4(12\left)\right|$
$=\frac{1}{2}|-45-48|$
$=\frac{1}{2}|-93|$
$=\frac{93}{2}$

Step $2:$ Area of quadrilateral
Now, Area of quadrilateral $ABCD=\text{Area of}△ABD+\text{Area of}△BCD$
$=14+\frac{93}{2}$
$=\frac{28+93}{2}$
$=\frac{121}{2}$
$=60.5$square units.

Final answer :
Therefore, area of quadrilateral is $=60.5\text{square units}$ .