Difference between revisions of "2022 AMC 10A Problems/Problem 21"
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<math>\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math> | <math>\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9</math> | ||
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==Solution 1 by OmegaLearn using Equiangular Hexagon Properties== | ==Solution 1 by OmegaLearn using Equiangular Hexagon Properties== | ||
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~ pi_is_3.14 | ~ pi_is_3.14 | ||
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+ | == Video Solution By ThePuzzlr == | ||
+ | https://youtu.be/br11LJJD4-c | ||
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+ | ~ MathIsChess | ||
== See Also == | == See Also == |
Revision as of 09:42, 12 November 2022
Contents
[hide]Problem
A bowl is formed by attaching four regular hexagons of side 1 to a square of side 1. The edges of adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl?
Solution 1 by OmegaLearn using Equiangular Hexagon Properties
~ pi_is_3.14
Video Solution By ThePuzzlr
~ MathIsChess
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.