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Usernamehectictar
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 #2
avatar+6959 
+5

Here's another way...but it looks a bit longer than CPhill's!

 

First let's extend line  AB .

Then....

Draw a line perpendicular to  AB  that goes through point  C  .

Draw a line perpendicular to AB  that goes through point  E .

Draw a line parallel to  AB  (and perpendicular to the previous two lines) that goes through point  D .

 

Now let's label these new points of intersection W, X, Y, and Z , like the picture,

and put the information from the problem.

 

∠EAZ  =  ∠CBY  =  180° - 120°  =  60°

 

Since triangles AEZ and BCY are 30-60-90 triangles where the side across from the 90° angle is 2....

the side across from the 30° angle  =  2/2  =  1

the side across from the 60° angle  =  √3

 

And...

WD + DX  =  2 + 1 + 1             And  WD = DX , so we can substitute  WD  in for  DX .

WD + WD  =  2 + 1 + 1

2WD  =  4

WD  =  2

 

Now we can use the Pythagorean theorem to find  CX .

22 + CX2  =  42

4 + CX2  =  16             Subtract  4  from both sides of the equation.

CX2  =  12                   Take the positive square root of both sides.

CX  =  2√3

 

 

And.....

area of ABCDE   =   area of WXYZ  -  2(area of △EDW)  -  2(area of △AEZ)

                           =   (width)(length)  -  2(1/2)(base)(height)  -  2(1/2)(base)(height)

                           =   (2 + 2)(2√3 + √3)  -  2(1/2)(2)(2√3)  -  2(1/2)(1)(√3)

                           =   12√3  -  4√3  -  √3

                           =   7√3  square units

hectictar Aug 24, 2017