Difference between revisions of "2022 AMC 12B Problems/Problem 10"
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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>. | 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>. | ||
− | Note: The sum of the | + | 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>. |
== See Also == | == See Also == |
Revision as of 19:28, 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.See note. 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
.
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
.
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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