Answer:
37.5 sq. units.
Step-by-step explanation:
The coordinate of vertices of trapezoid ABCD are given,
From the diagram it is clear the area of ABCD = [tex]\frac{1}{2}\times AB \times (AD + BC)[/tex] {Since, AD ║ BC and AB is the perpendicular distance between the parallel lines }
Now, AB = [tex]\sqrt{(- 3 - 1)^{2}+ (2 - 5)^{2} } = 5[/tex] units
AD = [tex]\sqrt{(- 3 - 0)^{2}+ (2 - (-2))^{2} } = 5[/tex] units
Again, BC = [tex]\sqrt{(1-7)^{2}+ (5 - (-3))^{2} } = 10 [/tex] units
Therefore, the area of trapezoid ABCD = [tex]\frac{1}{2} (10 + 5) \times 5 = 37.5[/tex] sq. units. (Answer)
We know, that the distance formula between two points [tex](x_{1}, y_{1})[/tex] and [tex](x_{2}, y_{2})[/tex] is
[tex]\sqrt{(x_{1} - y_{1} )^{2} + (x_{2} - y_{2} )^{2} }[/tex].
