Difference between revisions of "2008 AMC 10B Problems/Problem 7"
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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 <math>\boxed{100}</math> small triangles can fit in the large one. | + | '''(C)''' 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 <math>\boxed{100}</math> small triangles can fit in the large one. |
Revision as of 14:41, 6 February 2011
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
(C) The area of the large triangle is 
, while the area each small triangle is 
. Dividing these two quantities, we get 100, therefore 
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]](http://latex.artofproblemsolving.com/2/b/6/2b6049d158759587caa572b1d8911fc0deb0b10b.png)
The number of triangles is 
.
Also, another way to do it is to notice that as you go row by row (from the bottom), the number of triangles decrease by 2 from 19, so we have:
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 | ||
