Difference between revisions of "2021 Fall AMC 12A Problems/Problem 14"
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==Solution (Law of Cosines and Equilateral Triangle Area)== | ==Solution (Law of Cosines and Equilateral Triangle Area)== | ||
− | Isosceles triangles <math>ABF</math>, <math>CBD</math>, and <math>EDF</math> are congruent by [[ | + | Isosceles triangles <math>ABF</math>, <math>CBD</math>, and <math>EDF</math> are congruent by [[Congruent_(geometry)#SAS_Congruence|SAS congruence]]. By [[CPCTC]], <math>BF=BD=DF</math>, so triangle <math>BDF</math> is equilateral. |
Let the side length of the hexagon be <math>s</math>. | Let the side length of the hexagon be <math>s</math>. | ||
− | The area of each isosceles triangle is <math>\frac{1}{2}s | + | The area of each isosceles triangle is <math>\frac{1}{2}s \cdot s \cdot \sin{30}=\frac{1}{4}s^2</math> by the fourth formula [[Area#Area_of_Triangle|here]]. |
− | By the [[Law of Cosines]] on triangle <math>ABF</math>, <math>BF^2= | + | By the [[Law of Cosines]] on triangle <math>ABF</math>, <math>BF^2=s^2+s^2-2s^2\cos{30^\circ}=2s^2-\sqrt{3}s^2</math>. Hence, the [[Area_of_an_equilateral_triangle|area of the equilateral triangle]] <math>BDF</math> is <math>\frac{\sqrt{3}}{4}\left(2s^2-\sqrt{3}s^2\right)=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2</math>. |
The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or <math>3\left(\frac{1}{4}s^2\right)+\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}</math>. Hence, <math>s=2\sqrt{3}</math> and the perimeter is <math>6s=\boxed{\textbf{(E)} \: 12\sqrt{3}}</math>. | The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or <math>3\left(\frac{1}{4}s^2\right)+\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}</math>. Hence, <math>s=2\sqrt{3}</math> and the perimeter is <math>6s=\boxed{\textbf{(E)} \: 12\sqrt{3}}</math>. |
Revision as of 21:11, 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 congruent by SAS congruence. By CPCTC,
, so 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 on triangle ,
. Hence, the area of the equilateral triangle
is
.
The total area of the hexagon is thrice the area of each isosceles triangle plus the area of the equilateral triangle, or . Hence,
and the perimeter is
.
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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