યામ-સમતલમાં $(-4,5),(0,7) (5,-5)$  અને $(-4-2)$ શિરોબિંદુઓવાળો ચતુષ્કોણ દોરો અને તેનું ક્ષેત્રફળ શોધો. 

Let $ABCD$ be the given quadrilateral with vertices $A(-4,5), B(0,7), C(5,-5),$ and $D(-4,- 2)$.

Then, by plotting $A, B, C$ and $D$ on the Cartesian plane and joining $A B, B C, C D,$ and $D A,$ the given quadrilateral can be drawn as

To find the area of quadrilateral $ABCD$, we draw one diagonal, say $AC$.

Accordingly, area $(ABCD)$ $=$ area $(\Delta ABC )+$ area $(\Delta ACD )$

We know that the area of a triangle whose vertices are $\left( x _{1}, y _{1}\right),\left( x _{2}, y _{2}\right),$ and $\left( x _{3}, y _{3}\right)$ is

$\frac{1}{2}\left|x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right|$

Therefore, area of $\Delta ABC$

$=\frac{1}{2} |-4(7+5)+0(-5-5)+5(5-7)$ unit$^{2} |$

$=\frac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)|$ unit$^{2}$

$=\frac{1}{2}|-4(12)+5(-2)|$ unit$^{2}$

$=\frac{1}{2}|-48-10|$ unit$^{2}$

$=\frac{1}{2}|-58|$ unit$^{2}$

$=\frac{1}{2} \times 58$ unit$^{2}$

$=29$ unit$^{2}$

Area of $\triangle ACD$

$=\frac{1}{2}|-4(-5+2)+5(-2-5)+(-4)(5-5)|$ unit$^{2}$

$=\frac{1}{2}|-4(-3)+5(-7)-4(10)|$ unit$^{2}$

$=\frac{1}{2}|12-35-40|$ unit$^{2}$

$=\frac{1}{2}|-63|$ unit$^{2}$

$=\frac{63}{2}$ unit$^{2}$

Thus, area $( ABCD )=\left(29+\frac{63}{2}\right)$ unit$^{2}=\frac{58+63}{2}$ unit$^{2}=\frac{121}{2}$ unit$^{2}$