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According to Wikipedia, for a uniform n-gonal antiprism with edge length $a$, SA = $\frac{n}{2} \left( \cot\frac{\pi}{n} + \sqrt{3} \right) a^2$ . This formula seems unnecessarily complex to me, as I deduced the surface area should just be the area of each base plus the areas of the 2n equilateral triangles, which means the formula should be SA = 2B + 2n$\Bigl(\!\frac{a^2\sqrt{3}}{4}\Bigr)$. Can anyone show these formulas are equivalent? (Pictures are always appreciated.) I'm also wondering whether there is a specific reason to use the first formula rather than the second.

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    $\begingroup$ What's your formula for $B$? ... Since you've mentioned that the area is the two bases, plus those triangles, you could say that the area is simply $2B+2nT$ (with $B$ the base area and $T$ the triangle area). But that's not terribly helpful. The reason to use the first formula —and, presumably, the reason you incorporated the formula for $T$ in the second— is because it expresses the full area in terms of just $n$ and the side-length $a$. $\endgroup$
    – Blue
    Commented Feb 12 at 7:44
  • $\begingroup$ ^Depends on which type of regular polygon forms the base of a given antiprism. $\endgroup$
    – John
    Commented Feb 12 at 7:50
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    $\begingroup$ "Depends on which type of regular polygon..." Correct, which means that just calling it $B$ leaves more work for the reader. By incorporating the formula for $B$, the first formula isn't unnecessarily complex; it's as complex as needed to be fully informative. (BTW: "^" isn't useful here, since comments can be presented out of order. If you're wanting to direct a comment at someone (and ensure that the someone gets a notification of that comment), use @name .) Cheers! $\endgroup$
    – Blue
    Commented Feb 12 at 8:19

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Obviously, you have figured out the areas of the triangles: That's your $2n(a^2\frac{\sqrt{3}}4)$. However, the area of the base is not always immediately clear. Imagine if I have an octagon for example, we should have a formula for calculating the base area!

We do that by splitting the base into n equal isosceles triangles. At the centre, obviously the angle will be $\frac{2\pi}{n}$, and the outer side will be a! So using the cosine rule, we can find the length of the inner sides of the triangle to be $\frac{a}{2*sin(pi/n)}$. Using the formula for the area of a triangle, we get each triangle has an area of $\frac{a^2cot(\frac{\pi}{n})}{4}$ Now there are n many of these triangles on each base, and two base sides, so multiply that by 2n to get to that required area!

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    $\begingroup$ Perfect! Thank you so much :-) $\endgroup$
    – John
    Commented Feb 12 at 8:08

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