+100 The convex pentagon $ABCDE$ has $\angle A = \angle B = 120^\circ$, $EA = AB = BC = 2$ and $CD = DE = 4$. What is the area of $ABCDE$?
+100 I got 11sqrt3 but it wasn't correct
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half hexagon at the bottom = 3sqrt3
((3sqrt3/2)*2^2)/2
top triangle = 8sqrt3
edge length 2= long diagonal 4 << length from E to C
draw point halfway E to C to create point F
EF=CF=2
pythagorean theorem: DF= 2sqrt3
area of top triangle = DF*4 = 8sqrt3
add up= 10sqrt3
+129899
I get something a little different, atroshus.....
Refer to the following image :
Note that the area of triangle EAB = (1/2)(2)(2)sin120 = 2 * √3 / 2 = √3
And this triangle is isosceles......so angle ABE = 30°
But since angle ABC = 120°.....then triangle EBC is a right triangle with angle EBC = 90°
And since EB is a transversal cutting parallel segments AB and EC....then angle ABE = angle BEC = 30°....then triangle EBC is a 30 - 60 - 90 right triangle with angle BCE = 60°
And since BC = 2 and is opposite the 30° angle......then the hypotenuse EC is twice this = 4
So......the area of triangle EBC = (1/2)(2)(4)sin 60 = 4 * √3/2 = 2 √3
And triangle CDE is equilateral with an area = (1/2)(4)(4) sin60 = 8 √3 / 2 = 4√3
So the total area of ABCDE = [ √3 + 2√3 + 4√3 ] = 7√3 units^2
