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 1

(C) 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.


Solution 2

[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 $1+3+\dots+19 = \boxed{100}$.

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: $19+17+15...+3+1 = \frac{19+1}{2}\cdot 10 = \boxed{100}$

A fourth solution is to notice that the small triangles are similar to the large triangle as they are both equilateral. Therefore, the ratio of their areas is the square of the ratios of their side lengths. Hence the ratio of their areas is $(1/10)^2=1/100$, so the answer is $\boxed{100}$.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
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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