Difference between revisions of "2014 AMC 10B Problems/Problem 13"
(→Problem: Finished Diagram from Scratch!!!! :) took like half an hour.....) |
(→Problem: added diagram creds) |
||
Line 15: | Line 15: | ||
draw((0,0)--(30, 17.3205081)--(30, -17.3205081)--(0, 0)); | draw((0,0)--(30, 17.3205081)--(30, -17.3205081)--(0, 0)); | ||
</asy> | </asy> | ||
+ | |||
+ | (Diagram Creds-DivideBy0) | ||
==Solution== | ==Solution== |
Revision as of 12:15, 21 February 2014
Problem
Six regular hexagons surround a regular hexagon of side length as shown. What is the area of
?
(Diagram Creds-DivideBy0)
Solution
We note that the triangular sections in
can be put together to form a hexagon congruent to each of the seven other hexagons. By the formula for the area of the hexagon, we get the area for each hexagon as
. The area of
, which is equivalent to two of these hexagons together, is
.
See Also
2014 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.