Difference between revisions of "2021 AMC 10A Problems/Problem 21"
Alancenwang (talk | contribs) (Created page with "The amc 10 has not begun yet.") |
Sugar rush (talk | contribs) |
||
| Line 1: | Line 1: | ||
| − | The | + | ==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> | ||
Revision as of 17:06, 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 
