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I have a friend who is trying to build a wooden retaining wall around a trampoline. He wants the wall to be in the shape of a regular polygon with sides between $24"-30"$. He asked me to figure out how many sides he should make it, and then calculate the angles so that he knows how to cut the wood. The trampoline has a diameter of $16'$.

So I attacked the problem at first using Geometry. I found that the area of the trampoline is approximately $201.1$ $ft^2$. Then I used the regular polygon formula $A=\frac{a\cdot p}{2}$, where $a$ is the apothem and $p$ is the perimeter for all sides from $24"-30"$. To make a long story short. I found that a polygon of $24$ sides with $26"$ sides would best do the trick, and another mathematics friend of mine concurred. Since we have 24 sides, using the internal angle formula, I know that the sum of the interir angles is $A(24)=180(24-2)=3960$ which means each of the $24$ angles should be $165$ degrees. Halving this gives me cuts of $82.5$ degrees.

However, using Trig, if the long side of the triangle is $96"$ and the short side is $13"$, then $\tan{\theta}=\frac{96}{13}$ which implies that $\theta=82.288$, not $82.5$

Why the discrepancy? I know these numbers are close, but shouldn't the math agree? I don't see where my error lies if there is an error in my math.

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  • $\begingroup$ As you state, your calculation $\tan\theta = 96/13$ is for a right triangle with "long side" (the polygon's apothem) of $96$ inches and "short side" (half the polygon's side-length) of $13$ inches. The problem is that $96$ inches is the radius of the trampoline, NOT the apothem of the polygonal wall. The apothem is actually about $98.74$ inches, so that $\theta \approx \text{atan}(98.74/13) \approx 82.4996^\circ$, which "agrees" with the true value of $82.5^\circ$. $\endgroup$ – Blue May 1 '14 at 11:52
  • $\begingroup$ Right, but if the circle is inscribed, then the apothem would be the midpoint of the side, which is tangent to the circle. Therefore the apothem should be the radius. In the picture below for the given answer, it's as if the midpoint of LM is tangent to the circle. $\endgroup$ – Iceman May 1 '14 at 12:43
  • $\begingroup$ Since the apothem is perpendicular to the side, which is tangent to the circle, this means that the apothem is the radius, right? $\endgroup$ – Iceman May 1 '14 at 12:50
  • $\begingroup$ If the polygon were tangent to the circle, then yes: the apothem would match the radius. However, a $24$-gon of side-length of $26$ inches is NOT tangent to the circle of radius $96$ inches. As I mentioned in my first comment, the apothem of such a $24$-gon is $98.74$ inches. In other words, the side length you chose ($26$ inches) gives only an approximation of the perfectly-circumscribing $24$-gon; the wall is slightly too large for the trampoline (which is probably a good thing). For a snug (tangential) fit, the side-length should be closer to $25.2773$ inches. $\endgroup$ – Blue May 1 '14 at 13:02
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    $\begingroup$ Okay, i think I get it...it is physically impossible to have a 24-gon with exactly both 26" sides and 8' apothem. There must have been a level of rounding i forgot to take into account. Then the million dollar question is this: knowing the apothem, how do you calculate the actual side length for an $n$-gon? Clearly, if the apothem is 8 ft, then there should be a way to calculate right? $\endgroup$ – Iceman May 1 '14 at 13:16
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EDIT: I had it the wrong way round.

I think it may be due to the fact that you're using the distance to the base of the triangle (the long side) as half the diameter of the trampoline. If you look at this diagram this isn't strictly accurate since there's a little difference.

enter image description here

Which as Blue states the you can work out the circum radius of the polygon with: $$ \frac{26}{2 \sin \left(\frac{\pi }{24}\right)}= 13 \csc \left(\frac{\pi }{24}\right)\approx 99.5969 $$ Then you still need to account for the difference $d$ in the second diagram, you can find the distance with $d$ accounted for using: $$\sqrt{r^2-\frac{c^2}{4}} \rightarrow \sqrt{\left(\frac{26}{2 \sin \left(\frac{\pi }{24}\right)}\right)^2-\frac{26^2}{4}}=\sqrt{169 \csc ^2\left(\frac{\pi }{24}\right)-169} \approx 98.7448$$ where $r$ is the circum radius and $c$ is the length of the side of the polygon. Thus you can then find the angle to be: $$ \tan ^{-1}\left(\frac{1}{13} \sqrt{169 \csc ^2\left(\frac{\pi }{24}\right)-169}\right) \times \frac{180}{\pi}=82.5^\circ $$

Of course you could just use the inverse cosine: $$\cos ^{-1}\left(\frac{26/2}{\left(13 \csc \left(\frac{\pi }{24}\right)\right)}\right) \times \frac{180}{\pi}=82.5^\circ$$ to skip a step from the above working but I thought that since you were using $\tan$ I would too.
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Diagram for comment enter image description here

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  • $\begingroup$ your second diagram is not accurate. This describes a circumscribed circle around a regular polygon, but I need an inscribed circle, since the retaining wall is outside the trampoline, so that $d$ is nonexistent, since the midpoint of a regular polygon's side is tangent to the circle, and thus the apothem is the radius. $\endgroup$ – Iceman May 1 '14 at 12:45
  • $\begingroup$ @Iceman the inner most circle is the trampoline and the outermost circle I drew just so it was clear that if you were to find the angle using the tangent function you need to take into account $d$ (since you don't need to when doing it with cosine). $\endgroup$ – Jay May 1 '14 at 12:50
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    $\begingroup$ @Iceman Sorry for the late reply, I think I understand what you're saying, (I've added another diagram to the bottom of my answer since I can't add a picture here). The distance $d$ you can see is still in the diagram if you use an inscribed circle, it's just shifted place. So to work out the angle you need to account for it whether you use the outer circle like my method or an inner one like yours. In your case you need to find the length of the apothem which inherently is the circum radius minus $d$. $\endgroup$ – Jay May 1 '14 at 13:41
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    $\begingroup$ @Iceman I used Geoalgebra; I recommend it since it's free, cross platform and relatively easy to use I've used it quite a lot since it's much neater than my attempt at drawing anything accurately! $\endgroup$ – Jay May 1 '14 at 16:44
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    $\begingroup$ @Jay Its actually Geo-Gebra! $\endgroup$ – Sawarnik May 1 '14 at 19:20

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