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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== Video Solution By ThePuzzlr ==
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https://youtu.be/br11LJJD4-c
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~ MathIsChess
  
 
==Solution 1 by OmegaLearn using Equiangular Hexagon Properties==
 
==Solution 1 by OmegaLearn using Equiangular Hexagon Properties==

Revision as of 10:41, 12 November 2022

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?

2022 AMC10A 21 Figure.png

$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9$

Video Solution By ThePuzzlr

https://youtu.be/br11LJJD4-c

~ MathIsChess

Solution 1 by OmegaLearn using Equiangular Hexagon Properties

https://youtu.be/-QHhR2r9HgQ

~ pi_is_3.14

See Also

2022 AMC 10A (ProblemsAnswer KeyResources)
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

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