Difference between revisions of "2008 AMC 8 Problems/Problem 6"
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− | ==Problem== | + | == Problem == |
In the figure, what is the ratio of the area of the gray squares to the area of the white squares? | In the figure, what is the ratio of the area of the gray squares to the area of the white squares? | ||
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<asy> | <asy> | ||
size((70)); | size((70)); | ||
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fill((5,0)--(7.5,2.5)--(10,0)--(7.5,-2.5)--cycle, gray); | fill((5,0)--(7.5,2.5)--(10,0)--(7.5,-2.5)--cycle, gray); | ||
</asy> | </asy> | ||
+ | |||
<math> \textbf{(A)}\ 3:10 \qquad\textbf{(B)}\ 3:8 \qquad\textbf{(C)}\ 3:7 \qquad\textbf{(D)}\ 3:5 \qquad\textbf{(E)}\ 1:1 </math> | <math> \textbf{(A)}\ 3:10 \qquad\textbf{(B)}\ 3:8 \qquad\textbf{(C)}\ 3:7 \qquad\textbf{(D)}\ 3:5 \qquad\textbf{(E)}\ 1:1 </math> | ||
− | ==Solution== | + | == Solution == |
Dividing the gray square into four smaller squares, there are <math>6</math> gray tiles and <math>10</math> white tiles, giving a ratio of <math>\boxed{\textbf{(D)}\ 3:5}</math>. | Dividing the gray square into four smaller squares, there are <math>6</math> gray tiles and <math>10</math> white tiles, giving a ratio of <math>\boxed{\textbf{(D)}\ 3:5}</math>. | ||
− | ==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}} |
Latest revision as of 12:36, 19 October 2020
Problem
In the figure, what is the ratio of the area of the gray squares to the area of the white squares?
Solution
Dividing the gray square into four smaller squares, there are gray tiles and white tiles, giving a ratio of .
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.