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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 14"

(→‎Solution (Law of Cosines and Equilateral Triangle Area))
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Let the side length of the hexagon be <math>s</math>.  
 
Let the side length of the hexagon be <math>s</math>.  
  
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The area of each isosceles triangle is <math>\frac{1}{2}s^2\sin{30}=\frac{1}{4}s^2</math>.  
+
The area of each isosceles triangle is <math>\frac{1}{2}s^2\sin{30}=\frac{1}{4}s^2</math> by the fourth formula [[Area#Area_of_Triangle|here]].
  
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By the [[Law of Cosines]], the square of the side length of equilateral triangle BDF is <math>2s^2-2x^2\cos{30}=(2-\sqrt{3})s^2</math>. Hence, the area of the triangle is <math>\frac{\sqrt{3}}{4}(2-\sqrt{3})s^2=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2</math>.  
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By the [[Law of Cosines]], the square of the side length of equilateral triangle BDF is <math>2s^2-2x^2\cos{30}=(2-\sqrt{3})s^2</math>. Hence, the [[Area_of_an_equilateral_triangle|area of the equilateral triangle]] is <math>\frac{\sqrt{3}}{4}(2-\sqrt{3})s^2=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2</math>.  
  
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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: <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>.

Revision as of 19:10, 23 November 2021

Solution (Law of Cosines and Equilateral Triangle Area)

Isosceles triangles $ABE$, $CBD$, and $EDF$ are identical by SAS similarity. $BF=BD=DF$ by CPCTC, and triangle $BDF$ is equilateral.

Let the side length of the hexagon be $s$.

The area of each isosceles triangle is $\frac{1}{2}s^2\sin{30}=\frac{1}{4}s^2$ by the fourth formula here.

By the Law of Cosines, the square of the side length of equilateral triangle BDF is $2s^2-2x^2\cos{30}=(2-\sqrt{3})s^2$. Hence, the area of the equilateral triangle is $\frac{\sqrt{3}}{4}(2-\sqrt{3})s^2=\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2$.

So, the total area of the hexagon is the area of the equilateral triangle plus thrice the area of each isosceles triangle or $\frac{\sqrt{3}}{2}s^2-\frac{3}{4}s^2+3(\frac{1}{4}s^2)=\frac{\sqrt{3}}{2}s^2=6\sqrt{3}$. The perimeter is $6s=\boxed{12\sqrt{3}}$.

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