Difference between revisions of "2008 AMC 10B Problems/Problem 7"
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The number of triangles is <math>1+3+\dots+19 = \boxed{100}</math>. | The number of triangles is <math>1+3+\dots+19 = \boxed{100}</math>. | ||
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+ | Another Solution: | ||
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+ | The area of the large triangle is 10^2 sqrt(3)/4, while the area each small triangle is 1^2 sqrt(3)/4. Dividing these two quantities, we get 100, therefore 100 small triangles can fit in the large one. | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=6|num-a=8}} | {{AMC10 box|year=2008|ab=B|num-b=6|num-a=8}} |
Revision as of 14:07, 31 January 2009
Problem
An equilateral triangle of side length is completely filled in by non-overlapping equilateral triangles of side length . How many small triangles are required?
Solution
The number of triangles is .
Another Solution:
The area of the large triangle is 10^2 sqrt(3)/4, while the area each small triangle is 1^2 sqrt(3)/4. Dividing these two quantities, we get 100, therefore 100 small triangles can fit in the large one.
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 |