2021 AMC 10A Problems/Problem 21

Revision as of 23:04, 11 February 2021 by Pi is 3.14 (talk | contribs) (Solution (Misplaced problem?))

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 $16+3+36=\boxed{55~\textbf{(C)}}$


Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles)

https://youtu.be/ptBwDcmDaLA

~ pi_is_3.14