Answer:
The area of the trapezoid ABCD = 37.5 sq. units.
Step-by-step explanation:
If we join points B and D on the figure given with the question.
Then area of the trapezoid ABCD = area of Δ ABD + area of Δ BDC
Now, area of Δ ABD = [tex]\frac{1}{2} |- 3(5 - (- 2)) + 1(- 2 - 2) + 0(2 - 5)| = 12.5[/tex] sq, units.
Again area of Δ BDC = [tex]\frac{1}{2} |1(- 2 -(- 3)) +0(- 3 - 5)+ 7(5 - (- 2))|[/tex] = 25 sq, units
Therefore, the area of the trapezoid ABCD = 12.5 + 25 = 37.5 sq. units.
We know the formula of area of a triangle having vertices [tex](x_{1}, y_{1})[/tex], [tex](x_{2}, y_{2})[/tex], and [tex](x_{3}, y_{3})[/tex] will be equal to
Δ = [tex]\frac{1}{2} |x_{1} (y_{2} - y_{3})+ x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |[/tex] (Answer)