Question

Find the area of quadrilateral whose vertices taken in order are $A(-3,2),B(5,4),C(7,-6)$ and $D(-5,-4)$ (in sq. ut)

Answer 1

Let the points be $A(-3,2),B(5,4),C(7,-6)andD(-5,-4)$

The quadrilateral ABCD can be divided into triangles ABC and ACD and hence the area of the quadrilateral is the sum of the areas of the two triangles.

Area of a triangle with vertices $(x_1,y_1)$ ; $(x_2,y_2)$ and $(x_3,y_3)$ is

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

Hence, Area of triangle ABC

$=\left|\frac{(-3)(4+6)+\left(5\right)(-6-2)+\left(7\right)(2-4}{2}\right|$

$=\left|\frac{-30-40-14}{2}\right|$

$=\frac{84}{2}=42squnits$

And, Area of triangle ACD

$=\left|\frac{(-3)(-6+4)+\left(7\right)(-4-2)+(-5)(2+6)}{2}\right|$

$=\left|\frac{6-42-40}{2}\right|$

$=\frac{76}{2}=38squnits$

Hence, Area of quadrilateral $=42+38=80squnits$