Difference between revisions of "2021 Fall AMC 12A Problems/Problem 14"
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So, the total area of the hexagon is the area of the equilateral triangle plus thrice the area of each isosceles triangle or <math>\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2+3(\frac{1}{4}s^2)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}</math>. The perimeter is <math>6s=\boxed{12\sqrt{3}}</math>. | So, the total area of the hexagon is the area of the equilateral triangle plus thrice the area of each isosceles triangle or <math>\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2+3(\frac{1}{4}s^2)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}</math>. The perimeter is <math>6s=\boxed{12\sqrt{3}}</math>. | ||
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{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:05, 23 November 2021
Problem
In the figure, equilateral hexagon has three nonadjacent acute interior angles that each measure . The enclosed area of the hexagon is . What is the perimeter of the hexagon?
Solution (Law of Cosines and Equilateral Triangle Area)
Isosceles triangles , , and are identical by SAS similarity. by CPCTC, and triangle is equilateral.
Let the side length of the hexagon be .
The area of each isosceles triangle is by the fourth formula here.
By the Law of Cosines, the square of the side length of equilateral triangle BDF is . Hence, the area of the equilateral triangle is .
So, the total area of the hexagon is the area of the equilateral triangle plus thrice the area of each isosceles triangle or . The perimeter is .
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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