# Difference between revisions of "2008 AMC 10B Problems/Problem 7"

## 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

$[asy] unitsize(0.5cm); defaultpen(0.8); for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); } for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); } for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); } [/asy]$

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

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.