Difference between revisions of "2022 AMC 10A Problems/Problem 21"
Pi is 3.14 (talk | contribs) (Created page with "==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 a...") |
Pi is 3.14 (talk | contribs) (→Solution 1 using Hexagon Properties) |
||
| Line 6: | Line 6: | ||
<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 1 using Hexagon Properties== | + | ==Solution 1 using Equiangular Hexagon Properties== |
https://www.youtube.com/watch?v=-QHhR2r9HgQ&list=PLT9bNzqjDoMl3jNviYrczw7Ck_ArS54Xn&index=5 | https://www.youtube.com/watch?v=-QHhR2r9HgQ&list=PLT9bNzqjDoMl3jNviYrczw7Ck_ArS54Xn&index=5 | ||
Revision as of 03:16, 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?
https://i.ibb.co/hRkhJcs/pic.png
Solution 1 using Equiangular Hexagon Properties
https://www.youtube.com/watch?v=-QHhR2r9HgQ&list=PLT9bNzqjDoMl3jNviYrczw7Ck_ArS54Xn&index=5
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. 
