Difference between revisions of "2022 AMC 12B Problems/Problem 10"
Ehuang0531 (talk | contribs) (Created page with "==Problem== Regular hexagon <math>ABCDEF</math> has side length <math>2</math>. Let <math>G</math> be the midpoint of <math>\overline{AB}</math>, and let <math>H</math> be th...") |
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==Solution== | ==Solution== | ||
+ | Consider triangle <math>BCG</math>. <math>BG</math> = 1 and <math>BC</math> = 2. <math>\angle CBG = 120 ^{\circ}</math> because it is an interior angle of a regular hexagon. By the [[Law of Cosines]], we have: | ||
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+ | <cmath> CG^2 = BG^2 + BC^2 - 2 \cdot BG \cdot BC \cdot \cos \angle CBG \\ CG^2 = 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos 120 ^{\circ} \\ CG = \sqrt 7.</cmath> | ||
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+ | By [[SAS Congruence]], triangles <math>BCG</math>, <math>CDH</math>, <math>HEF</math>, and <math>AFG</math> are congruent, and by [[CPCTC]], quadrilateral <math>GCHF</math> is a rhombus. Therefore, its perimeter is <math>4 \cdot CG = \boxed{\textbf{(D)} \ 4 \sqrt 7}</math>. | ||
== See Also == | == See Also == |
Revision as of 17:45, 17 November 2022
Problem
Regular hexagon has side length . Let be the midpoint of , and let be the midpoint of . What is the perimeter of ?
Solution
Consider triangle . = 1 and = 2. because it is an interior angle of a regular hexagon. By the Law of Cosines, we have:
By SAS Congruence, triangles , , , and are congruent, and by CPCTC, quadrilateral is a rhombus. Therefore, its perimeter is .
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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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