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.
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!