Question

Find area of a quadrilateral Abcd whose vertices have coordinates a(_5,7)B(-4-5)C(-1,-6)D(4,5)

Answer 1

In Quadrilateral ABCD

Area of triangle ABC =17 sq units

Area of triangle ADC = 64.5 sq units

since area of triangle ABC +ADC= 81.5 sq units

Answer 2

Step-by-step explanation:

area of quadilateral = sum of two triangle

                                  = area of first triangle + area of second triangle

                                  = 1/2   $\left[\begin{array}{ccc}-5&7&1\\-4&-5&1\\-1&-6&1\end{array}\right]$   +1/2$\left[\begin{array}{ccc}-5&7&1\\-1&-6&1\\4&5&1\end{array}\right]$

                                  = 1/2 (35)  + 109/2    

                                  = 144/2

                                 = 72 square units

open above bracket  -5 (-5+6) -7 (-4+1) +1( 24-5)  and -5(-6-5)-7(-1-4)+1(-5+24)