2022 AMC 12B Problems/Problem 10
Problem
Regular hexagon has side length . Let be the midpoint of , and let be the midpoint of . What is the perimeter of ?
Solution 1 (No trig)
Let the center of the hexagon be . , , , , , and are all equilateral triangles with side length . Thus, , and . By symmetry, . Thus, by the Pythagorean theorem, . Because and , . Thus, the solution to our problem is
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Solution 2
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
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See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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All AMC 12 Problems and Solutions |
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