Melody

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

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

This was a really good question Alan, thanks for posting it :)   

Nice answer Chris  

Yours looks a bit easier than mine.

I suppose I should include my approx answers so that the answersc an be compared.

Yes I forgot.  You might get a cookie too:)

Your question doesn't make sense kes. :/

All th angles must be less than 180 degrees.

and little letters stand for sides.

2nd. Then.    asin 

Congratulations on that award you are getting.  That is a great achievement!

Try putting the first 'array' in parentheses.  :/

also you forget the +6 on the second line.  So your answer is not going to be correct. :(

Please note that Einstein already knows the answer so you will not be helping him and he will not thank you.  He may, however, give you points.

This is a test. He wants to know if you are half as bright as him.

Don't let that put you off. He does ask good question and you might pass his test.

@@ End of Day Wrap   Wed 13/5/15   Sydney, Australia Time   7:40pm  ♪ ♫

Good day to you all,

Great answers today from GirlsNightOut, Civonamzuk, DarkBlaze347, Alan, MathGod1, foodisloveandlife, EinsteinJr, Brodudedoodebrodude, CPhill, Calicalilac, MaddiB926, Geno3141, Headingnorth, Zacismyname and Feliciabaugh110722.  Thanks everyone