What is the area of the trapezoid shown? Express your answer in simplest exact form.

It seems from Chris's answer that maybe this is a continuation of an earlier problem.

If that is the case can you please just continue on the same thread next time.

If you think it will help you can draw our attention to it with a new post that just has the address of the old post pasted into it  asking us to go back and take another look. 

It is much better if all questions and answers to it stay on the same thread. :)

Based on my previous answer......the altitude (height) = √3 ...the top base is 2 and the bottom base is 4

So...the area  =

(1/2)h(b₁ + b₂) =

(1/2)√3 (2 + 4)  = 3√3  sq units

Well, if you draw a right angled triangle in the corner then  you will have

$$\\sin60=\frac{h}{2}\\\\
\frac{\sqrt{3}}{2}=\frac{h}{2}\\\\
h=\sqrt3\\
$Now the area of a trapezium is the average of the 2 paralel sides times the height$\\
$So we need the other parallel side length$\\
$The bit hanging out the end I'll call it x is given by $\\
cos60=x/2\\
1/2=x/2\\
x=1\\
$so the bottom is $1+2+1=4\; units \\
$So the area is \\
A=[(4+2)/2]*\sqrt3\\
A=3\sqrt3 \;\;unit\; thingies$$