Difference between revisions of "2021 Fall AMC 12A Problems/Problem 14"
Ehuang0531 (talk | contribs) (→Solution (Law of Cosines and Equilateral Triangle Area)) |
Ehuang0531 (talk | contribs) (→Solution (Law of Cosines and Equilateral Triangle Area)) |
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Isosceles triangles <math>ABE</math>, <math>CBD</math>, and <math>EDF</math> are identical by SAS similarity. <math>BF=BD=DF</math> by CPCTC, and triangle <math>BDF</math> is equilateral. | Isosceles triangles <math>ABE</math>, <math>CBD</math>, and <math>EDF</math> are identical by SAS similarity. <math>BF=BD=DF</math> by CPCTC, and triangle <math>BDF</math> is equilateral. | ||
− | Let the side length of the hexagon be <math>s</math>. The area of each | + | Let the side length of the hexagon be <math>s</math>. |
+ | |||
+ | The area of each isosceles triangle is <math>\frac{1}{2}s^2\sin{30}=\frac{1}{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 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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+ | 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>. |
Revision as of 19:06, 23 November 2021
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 Law of Cosines, the square of the side length of equilateral triangle BDF is . Hence, the area of the triangle is
.
So, the total area of the hexagon is the area of the equilateral triangle plus thrice the area of each isosceles triangle: . The perimeter is
.