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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.
Please help. I don't know how to start!
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hi Enshrouded_
From the shapes you are given, you're going to have to rely on rotational symmetry, order six.
I have a diagram, but let's see if you can draw it yourself.
Start with the hexagon.
Add six squares around the outside.
You can finish it from there.
Good luck.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I got the first part but how do I do part (b)?
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Now forget about the other shapes and just consider a regular dodecahedron. Imagine it split into 12 congruent isosceles triangles meeting at the centre. Each triangle has two sides each 1 in length and the angle between them is 360/12.
So you can calculate the area of each one and multiply by 12.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I got it! Thanks bob!
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