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The convex pentagon $ABCDE$ has $\angle A = \angle B = 120^\circ$, $EA = AB = BC = 2$ and $CD = DE = 4$. What is the area of $ABCDE$?

Atroshus  Aug 29, 2017
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 #1
avatar+38 
0

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

Atroshus  Aug 29, 2017
 #2
avatar+76038 
+2

 

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

 

 

 

cool cool cool

CPhill  Aug 29, 2017

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