Question

Find the area of the quadrilateral ABCD whose vertices are ${\text{A}}( - 5,3),{\text{B}}( - 4, - 6),{\text{C}}(2, - 3)$ and ${\text{D}}(1,2)$

Answer 1

Hint: Use the diagonal method multiplication of vertices to get the area. For getting the area of any figure with n diagonals, we use diagonal method multiplication by
\[\left( \dfrac{{{\text{x}}_{\text{1}}}}{{{\text{y}}_{\text{1}}}}\dfrac{{{\text{x}}_{\text{2}}}}{{{\text{y}}_{\text{2}}}}\dfrac{{{\text{x}}_{\text{3}}}}{{{\text{y}}_{\text{3}}}}\dfrac{{{\text{x}}_{\text{4}}}}{{{\text{y}}_{\text{1}}}}\dfrac{{{\text{x}}_{\text{1}}}}{{{\text{y}}_{\text{1}}}} \right)\]
where it is equal to half of the consecutive sum of products of $x_1$×$y_2$ +$x_2$×$y_3$ + .... and subtraction of the sum of $y_1$×$x_2$ + $y_2$×$x_3$ ….

Complete step-by-step answer:

Value of ($x_1$,$y_1$)=(-5,3),($x_2$,$y_2$)=(-4,6),($x_3$,$y_3$)=(2,-3) and ($x_4$,$y_4$)=(1,2)
As we know that the quadrilateral area can be calculated by their vertices using the formula
\[\dfrac{1}{2}\,\left[ {{x_1}{Y_2} + \,{x_2}{Y_3}\, + \,{x_3}{Y_4}\, + \,{x_4}{Y_1}\, - \,({x_2}{Y_1}\, + \,{x_3}{Y_2}\, + {x_4}{Y_3}\, + \,{x_1}{Y_4})} \right]\]
Where x and y are the coordinates of the vertices
So put the value of those in the formula we get
Area = \[
   = \dfrac{1}{2}[( - 5)( - 6) + ( - 4)( - 3) + (2)(2) + (1)(3) - (( - 4)(3) + (2)( - 6) + (1)( - 3) + ( - 5)(2))] \\
   = \dfrac{1}{2}[30 + 12 + 4 + 3 - ( - 12 - 12 - 3 - 10)] \\
   = \dfrac{1}{2}[49 + 37] = \dfrac{{86}}{2} = 43{\text{unit sq}} \\
\]

Note: We can also solve this problem in another way by dividing the quadrilateral (ABCD) into two triangles (ABC & ACD) and adding the area of both triangles. We will get the same answer.