Math CentralQuandaries & Queries


Question from pat:

I want to make raised-bed gardens - usually 4' square, but want to be more creative by creating octagonal boxes. The ideal size for square beds is 4'x4'. The boards are 8' long, so I want to get as much out of a board as I can. If I want an octagon with a diameter of 4', I think the sides would be (rounded off) about 1.5' each using 2 full boards with a little waste and the growing space would be less than a 4' square (16 square feet). If I want to increase the length of each side to 2' each (getting the max from a board), what will the diameter of this octagon be so I can determine the growing space of the box. (The reason a 4' square is important in gardening, besides maximizing the wood, is that a person can reach into the center to plant and pick veggies without stepping on the soil, so diameter is important). My math days are long over and I'm having trouble working with octagons! Thank you!

Hi Pat,

I did something similar. I created raised beds in the shape of regular hexagons but with a diameter 6 feet. As you said I wanted to be able to weed and pick the garden while sitting on the frame.

You can find an expression for the area of a regular polygon ($A_{RP}$) in the response that Stephen gave to a question about a regular polygon with 14 sides. His expression is

\[A_{RP} = \frac{a^2 n}{4 \tan\left(\frac{360^o}{2 n}\right)}\]

where $n$ is the number of sides of the polygon and $a$ is the side length. Thus for your octagon with side length 2 feet I get

\[A_{RP} = \frac{2^2 \times 8}{4 \tan\left(\frac{360^o}{16}\right)} = 19.3 \mbox{ square feet}\]


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