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> | ||
| − | ==Solution | + | == Video Solution By ThePuzzlr == |
| + | https://youtu.be/br11LJJD4-c | ||
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| + | ~ MathIsChess | ||
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| + | ==Video Solution by OmegaLearn using Equiangular Hexagon Properties== | ||
https://youtu.be/-QHhR2r9HgQ | https://youtu.be/-QHhR2r9HgQ | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
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== See Also == | == See Also == | ||
Revision as of 09:43, 12 November 2022
Contents
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?
Video Solution By ThePuzzlr
~ MathIsChess
Video Solution by OmegaLearn using Equiangular Hexagon Properties
~ pi_is_3.14
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. 
