Difference between revisions of "2022 AMC 12B Problems/Problem 10"
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==Solution== | ==Solution== | ||
− | Consider triangle <math> | + | Consider triangle <math>AFG</math>. <math>AG = 1</math> and <math>AF = 2</math>. <math>\angle GAF = 120 ^{\circ}</math> because it is an interior angle of a regular hexagon. By the [[Law of Cosines]], we have: |
− | <cmath> | + | <cmath>\begin{align*} |
+ | FG^2 &= AG^2 + AF^2 - 2 \cdot AG \cdot AF \cdot \cos \angle GAF \\ | ||
+ | FG^2 &= 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos 120 ^{\circ} \\ | ||
+ | FG^2 &= 1^2 + 2^2 - 4 \cdot \left( \frac 12 \right) \\ | ||
+ | FG^2 &= 7 \\ | ||
+ | FG &= \sqrt 7. | ||
+ | \end{align*}</cmath> | ||
− | + | By [[SAS Congruence]], triangles <math>AFG</math>, <math>BCG</math>, <math>CDH</math>, and <math>EFH</math> are congruent, and by [[CPCTC]], quadrilateral <math>GCHF</math> is a rhombus. Therefore, its perimeter is <math>4 \cdot FG = \boxed{\textbf{(D)} \ 4 \sqrt 7}</math>. | |
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− | By [[SAS Congruence]], triangles <math> | ||
== See Also == | == See Also == |
Revision as of 19:21, 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 .
and
.
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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