a landscape timber is 3 and a half inches wide and to make a polygon how long does each one need to be and what angle do they need to be cut to get a 36 inch inside dia

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- Nov 13th 2013, 08:13 AM #1

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- Nov 13th 2013, 08:40 AM #2

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## Re: well box

I am not certain what you are asking. To have an 'inside diameter" you have to have one circle insides another (or a circle inside

**something**) but see no mention of a circle. And you certainly have no "36 inch diameter" in a 3 and half inch region! It is possible that you want to make a cut from one side of the timber that will exit the other side of the timber at a distance of 36 inches. In that case, looking from the side, there is a right triangle with one leg, the end of the timber, of length 3.5 inches, another leg, the length to the end of the cut, of 36 inches. That is a right triangle in which we are given the "opposite side" and the "near side". The angle is given by $\displaystyle tan^{-1}(3.5/36)= tan^{-1}(0.09722)$.

- Nov 13th 2013, 09:00 AM #3

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- Nov 13th 2013, 06:46 PM #4

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- Nov 14th 2013, 04:00 AM #5

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- Nov 14th 2013, 04:04 AM #6

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- Nov 15th 2013, 07:28 AM #7

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## Re: well box

Hi again,

The angle for the octagon is the same no matter what length the outer side is. So if you want to cut 16 inch boards (presumably so you can get exactly 6 pieces from an 8 foot 2 by 4), you still want to cut angles of 67.5 degrees. A little more trigonometry shows that the diameter of the inner octagon will be 34.2 inches for 16 inch outer sides. In terms of area, the area of any octagon with radius r is $\displaystyle 2\sqrt2r^2$. Applying this you get for 16 inch outer sides, an area of 5.75 square feet; for your original question, the inner area is 6.36 square feet.

- Nov 15th 2013, 07:52 AM #8

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- Nov 15th 2013, 08:12 AM #9

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