Question

What is the area of trapezoid ABCD ? Enter your answer as a decimal or whole number in the box. Do not round at any steps. units²

Answer 1

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 = $\frac{1}{2}\times AB \times (AD + BC)$ {Since, AD ║ BC and AB is the perpendicular distance between the parallel lines }

Now, AB = $\sqrt{(- 3 - 1)^{2}+ (2 - 5)^{2} } = 5$ units

AD = $\sqrt{(- 3 - 0)^{2}+ (2 - (-2))^{2} } = 5$ units

Again, BC = $\sqrt{(1-7)^{2}+ (5 - (-3))^{2} } = 10$ units

Therefore, the area of trapezoid ABCD = $\frac{1}{2} (10 + 5) \times 5 = 37.5$ sq. units. (Answer)

We know, that the distance formula between two points $(x_{1}, y_{1})$ and  $(x_{2}, y_{2})$ is  

$\sqrt{(x_{1} - y_{1} )^{2} + (x_{2} - y_{2} )^{2} }$.