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Difference between revisions of "2008 AMC 10B Problems/Problem 7"
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==Solution==
 
==Solution==
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The area of the large triangle is <math>\frac{10^2\sqrt3}{4}</math>, while the area each small triangle is <math>\frac{1^2\sqrt3}{4}</math>. Dividing these two quantities, we get 100, therefore 100 small triangles can fit in the large one.
  
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Another Solution:
 
<asy>
 
<asy>
 
unitsize(0.5cm);
 
unitsize(0.5cm);
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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 11:31, 14 February 2009

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 100 small triangles can fit in the large one.


Another Solution: [asy] 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; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); } for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); } [/asy]

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

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
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