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ABCD is a rectangle with AB=12 and BC=18. Rectangle AEFG is formed by rotating ABCD about A through an angle of \(30^\circ\). What is the total area of the shaded regions?

bambam89  Nov 5, 2018
edited by bambam89  Nov 5, 2018
 #1
avatar+91360 
0

Need to see a pic, bambam.....sad sad sad

 

 

 

cool cool cool

CPhill  Nov 5, 2018
 #2
avatar+32 
0

Fixed it cheeky

bambam89  Nov 5, 2018
 #3
avatar+91360 
+1

Mmmmmm.......this is an interesting one  !!!!

 

Maybe easier ways to do this....but......

 

Perpendicular to   AB.....draw IJ   through E  so that I is a lies on AB  and J  lies on  DC

 

This divides the gray area on the bottom into 3 areas....

Triangle AIE, Triangle HEJ and Quadrilateral IJCB

 

We can find the area of  Tiriangle AIE as follows

AE = AB  = 12

Angle IAE = 30

Angle AIE  = 90

So  IE  = 6  and AI = 6sqrt(3)

So...the area of Triangle AIE  = (1/2)(6)(6sqrt(3))  =   18sqrt (3)   

 

So....IB  =   12 - AI =  12 - 6qrt(3)

So....the area of Quadrilateral IJBC = BC * IB =  18 * [ 12 - 6sqrt(3) ]  =  216 - 108 sqrt (3)  

 

And we can find the area of Triangle HEJ as follows

Angle HJE = 90

Angle JHE  = 60

EJ  = BC - IE  = 18 - 6  = 12

And HJ is opposite the 30 ° angle in Triangle HEJ  =   12 / sqrt (3) =  4sqrt (3)

So...the area of triangle HEJ = (1/2) (12)(4sqrt(3) )  = 24sqrt(3)

 

So...the "bottom" Grey Area is the sum of the areas in red =  216 - 66sqrt(3)  units^2

 

So....the "white" area  =  Area of  Rectangle ABCD  -  bottom Grey Area =

18 * 12  - [ 216 - 66sqrt(3) ]  =

216 - 216 + 66sqrt(3)  =

66sqrt(3)  units^2

 

And the "top" Grey Area is just the area of Rectangle AEFG - white area  =

Area of Rectangle ABCD - white area  =

216 - 66sqrt (3)    units^2

 

So.....the shaded area is just the bottom grey area + top grey area  

 

[ 216 - 66sqrt (3) ] + [ 216 - 66sqrt (3) ]  =

 

 [432 - 132sqrt(3) ]   units^2

 

 

cool cool cool

CPhill  Nov 5, 2018
edited by CPhill  Nov 5, 2018
 #4
avatar+32 
0

Thank you!

bambam89  Nov 7, 2018

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