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2008 AMC 10B Problems/Problem 7

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Problem

An equilateral triangle of side length 10 is completely filled in by non-overlapping equilateral triangles of side length 1. How many small triangles are required?

\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000

Solution

The area of the large triangle is \frac{10^2\sqrt3}{4}, while the area each small triangle is \frac{1^2\sqrt3}{4}. Dividing these two quantities, we get 100, therefore \boxed{100} small triangles can fit in the large one.


Another Solution: unitsize(0.5cm);defaultpen(0.8);for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); }for (int i=0...

The number of triangles is 1+3+\dots+19 = \boxed{100}.

See also

2008 AMC 10B (ProblemsResources)
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
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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