2021 AMC 10A Problems/Problem 21

Revision as of 17:06, 11 February 2021 by Sugar rush (talk | contribs)

Solution (Misplaced problem?)

Note that the extensions of the given lines will determine an equilateral triangle because the hexagon is equiangular. The area of the first triangle is $192\sqrt{3}$, so the side length is $\sqrt{192\cdot 4}=16\sqrt{3}$. The area of the second triangle is $324\sqrt{3}$, so the side length is $\sqrt{4\cdot 324}=36$. We can set the first value equal to $AB+CD+EF$ and the second equal to $BC+DE+FA$ by substituting some lengths in with different sides of the same equilateral triangle. The perimeter of the hexagon is just the sum of these two, which is $16\sqrt{3}+36$ and $16+3+36=\boxed{55~\textbf{(C)}}$