Difference between revisions of "2022 AMC 12B Problems/Problem 10"
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Note: The sum of the interior angles of any polygon with <math>n</math> sides is given by <math>180 ^{\circ} (n - 2)</math>. Therefore, the sum of the interior angles of a hexagon is <math>720 ^{\circ}</math>, and each interior angle of a regular hexagon measures <math>\frac{720 ^{\circ}}{6} = 120 ^{\circ}</math>. | Note: The sum of the interior angles of any polygon with <math>n</math> sides is given by <math>180 ^{\circ} (n - 2)</math>. Therefore, the sum of the interior angles of a hexagon is <math>720 ^{\circ}</math>, and each interior angle of a regular hexagon measures <math>\frac{720 ^{\circ}}{6} = 120 ^{\circ}</math>. | ||
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+ | ==Video Solution 1== | ||
+ | https://youtu.be/pIdM5l3CyUY | ||
+ | |||
+ | ~Education, the Study of Everything | ||
== See Also == | == See Also == |
Revision as of 19:48, 19 November 2022
Contents
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 . Note that , , and because it is an interior angle of a regular hexagon. (See note for details.)
By the Law of Cosines, we have:
By SAS Congruence, triangles , , , and are congruent, and by CPCTC, quadrilateral is a rhombus. Therefore, the perimeter of is .
Note: The sum of the interior angles of any polygon with sides is given by . Therefore, the sum of the interior angles of a hexagon is , and each interior angle of a regular hexagon measures .
Video Solution 1
~Education, the Study of Everything
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.