Difference between revisions of "2021 AMC 10A Problems/Problem 21"
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==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 <math>192\sqrt{3}</math>, so the side length is <math>\sqrt{192\cdot 4}=16\sqrt{3}</math>. The area of the second triangle is <math>324\sqrt{3}</math>, so the side length is <math>\sqrt{4\cdot 324}=36</math>. We can set the first value equal to <math>AB+CD+EF</math> and the second equal to <math>BC+DE+FA</math> 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 <math>16\sqrt{3}+36</math> and <math>16+3+36=\boxed{55~\textbf{(C)}}</math> | 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 <math>192\sqrt{3}</math>, so the side length is <math>\sqrt{192\cdot 4}=16\sqrt{3}</math>. The area of the second triangle is <math>324\sqrt{3}</math>, so the side length is <math>\sqrt{4\cdot 324}=36</math>. We can set the first value equal to <math>AB+CD+EF</math> and the second equal to <math>BC+DE+FA</math> 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 <math>16\sqrt{3}+36</math> and <math>16+3+36=\boxed{55~\textbf{(C)}}</math> | ||
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| + | == Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles) == | ||
| + | https://youtu.be/ptBwDcmDaLA | ||
| + | |||
| + | ~ pi_is_3.14 | ||
Revision as of 22:04, 11 February 2021
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 
, so the side length is 
. The area of the second triangle is 
, so the side length is 
. We can set the first value equal to 
and the second equal to 
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 
and 
Video Solution by OmegaLearn (Angle Chasing and Equilateral Triangles)
~ pi_is_3.14
