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dleete

 Jun 26, 2018
edited by yasbib555  Jun 27, 2018
 #1
avatar+98182 
+2

I'm assuming that we have the following :

 

 

Let A  = (0,0)

Since BE   = 2, then AB  =  3BE  = 6...so....AE  = 4

And since angle BAC  = 60°, the point B  is  (6 cos 60°, 6 sin 60°)  = (3, 3√3)

And point  E  is   ( 4cos 60°, 4 sin 60°)   = (2, 2√3)

And  since  AD  = 3, then AC  = 3AD = 9

So  D  is  (3, 0)

 

So.....the  area  of BCDE  =  area  of triangle ABC  -  area of triangle  AED

 

Area of triangle ABC  =  (1/2)DB * AC  = (1/2) (3√3) (9)  = (13.5)√3  units^2

 

And area  of triangle AED  = (1/2) (AD) (2√3)  = (1/2)(3)(2√3)  = 3√3 units^2

 

So...the area  of  quadrilateral BCDE  = [ (13.5)√3  - 3√3] units^2  = (10.5√3) units^2

 

cool cool cool

 Jun 26, 2018

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