Answer:
50 sq. units
Step-by-step explanation:
Join one diagonal of the trapezoid ABCD.
Assume that AC is the diagonal.
So, Area of the trapezoid = Area of Δ ABC + Area of Δ CAD.
Now, coordinates of A(-2,2), B(2,5), C(11,-7) and D(1,-2) are given.
We know, that three vertices of a triangle are [tex](x_{1},y_{1} )[/tex], [tex](x_{2},y_{2} )[/tex] and [tex](x_{3},y_{3} )[/tex] respectively, then the area of the triangle will be given by
[tex]\frac{1}{2} |x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+ x_{3}(y_{1}-y_{2} ) |[/tex]
Hence, area of Δ ABC will be
[tex]\frac{1}{2} |-2(5-(-7))+2(-7-2)+11(2-5)|= 37.5[/tex] sq units.
Again, area of Δ CAD will be
[tex]\frac{1}{2} |11(2-(-2))-2(-2-(-7))+1(-7-2)|= 12.5[/tex] sq units.
Therefore, the area of the trapezoid ABCD will be (37.5 + 12.5) = 50 sq, units. (Answer)