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(a) Prove that one regular hexagon, six squares, and six equilateral triangles, all with the same side length, can be assembled to form a regular dodecagon. (Begin with a diagram of course, but you must also show that at all points where two or more polygons "fit" together, the angles add up to the correct amount.)

(b) The distance between two opposite vertices of the dodecagon is 2. Find the area of the dodecagon.

 

 

Thanks so much!  (Detailed answers are greatly appreciated!)

 #1
avatar+1442 
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Don't really need help with (a) anymore but if you want...

 #2
avatar+87294 
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Second one.....a (regular)  dodecagon is a 12-sided figure

 

We have twelve congruent triangles......if the distance between two opposite vertices of the dodecagon is 2...then  the side of one of these triangles  = 1

 

And the apex angle of each  of these triangles   =  360° / 12  =  30°....so...the area is just

 

12 (1/2) (side length of a triangle)^2 *  sin (apex angle)  = 

 

12 (1/2) (1^2)sin(30°)  =

 

12 (1/2) (1/2)  =  

 

12/ 4   =

 

3 units^2

 

 

 

cool cool cool

CPhill  Apr 3, 2018
 #3
avatar+1442 
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Thank you!


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