Difference between revisions of "2021 Fall AMC 12A Problems/Problem 14"
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− | <math>\textbf{(A)} 4 \qquad \textbf{(B)} 4\sqrt3 \qquad \textbf{(C)} 12 \qquad \textbf{(D)} 18 \qquad \textbf{(E)} 12\sqrt3</math> | + | <math>\textbf{(A)} \: 4 \qquad \textbf{(B)} \: 4\sqrt3 \qquad \textbf{(C)} \: 12 \qquad \textbf{(D)} \: 18 \qquad \textbf{(E)} \: 12\sqrt3</math> |
==Solution (Law of Cosines and Equilateral Triangle Area)== | ==Solution (Law of Cosines and Equilateral Triangle Area)== |
Revision as of 19:59, 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 35 |
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All AMC 12 Problems and Solutions |
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