Difference between revisions of "2008 AMC 8 Problems/Problem 6"
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== Solution == | == Solution == | ||
| − | Dividing the gray square into four smaller squares, there are | + | Dividing the gray square into four smaller squares, there are 6 gray tiles and 10 white tiles, giving a ratio of D 3:5 |
== See Also == | == See Also == | ||
{{AMC8 box|year=2008|num-b=5|num-a=7}} | {{AMC8 box|year=2008|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:56, 8 August 2021
Problem
In the figure, what is the ratio of the area of the gray squares to the area of the white squares?
![[asy] size((70)); draw((10,0)--(0,10)--(-10,0)--(0,-10)--(10,0)); draw((-2.5,-7.5)--(7.5,2.5)); draw((-5,-5)--(5,5)); draw((-7.5,-2.5)--(2.5,7.5)); draw((-7.5,2.5)--(2.5,-7.5)); draw((-5,5)--(5,-5)); draw((-2.5,7.5)--(7.5,-2.5)); fill((-10,0)--(-7.5,2.5)--(-5,0)--(-7.5,-2.5)--cycle, gray); fill((-5,0)--(0,5)--(5,0)--(0,-5)--cycle, gray); fill((5,0)--(7.5,2.5)--(10,0)--(7.5,-2.5)--cycle, gray); [/asy]](http://latex.artofproblemsolving.com/3/c/4/3c4476181cee26306aeb555b3e4b864d592dca35.png)
Solution
Dividing the gray square into four smaller squares, there are 6 gray tiles and 10 white tiles, giving a ratio of D 3:5
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
