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 May 29th, 2015, 05:13 AM #1 Newbie   Joined: May 2015 From: UK Posts: 1 Thanks: 0 Incircle of a regular polygon Hi Guys, Can any of you work this out? I cant find anything online to help me. I have a circle (well actually its a kids swimming pool) 3m in diameter. I want to cover that circle (ok - pool) with a regular 6 sided polygon (ok! A pool cover) that I am going to make. How do I work out the length of each of the sides of the polygon that will cover the circle (pool!!!) Anyone, anyone???? Thanks Mark
 May 29th, 2015, 06:20 AM #2 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,133 Thanks: 719 Math Focus: Physics, mathematical modelling, numerical and computational solutions If you draw a hexagon, you can split the hexagon into 6 triangles by drawing lines from each of the vertices to the center. You can then divide each of these triangles into 2 triangles by drawing lines from the midpoint of each side to the center. What you should end up with is a hexagon split into 12 identical triangles. In order for the swimming pool to fit, the shortest side of each triangle must be 3m or more. Furthermore, the inner angle (at the center) must be $\displaystyle 360^{\circ}/12 = 30 ^{\circ}$. So, you can get the length of the opposite side of the angle using trigonometry: $\displaystyle \tan\theta = \frac{opp.}{adj.}$ $\displaystyle \tan 30^{\circ} = \frac{opp.}{3}$ The left-hand side can be evaluated to be: $\displaystyle \tan 30^{\circ} = \frac{1}{\sqrt{3}}$ So... with a little algebraic rearrangement... $\displaystyle opp. = \frac{3}{\sqrt{3}} = \sqrt{3}$ Since the length of a side of the hexagon is twice what we've just found, the length of each side must be greater than *6 $\displaystyle 2\sqrt{3} = 3.464$ So I'd recommend getting wooden/metal/whatever beams with length a bit larger than this value (3.6m maybe?) and then cut them down to size as appropriate.

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### incircle of a hexagon

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