2008 AMC 10B Problem 7

Start small to see the pattern, then count row by row.

1. A triangle of side $2$ splits into $4$ small triangles: $1$ pointing up on top, and $3$ in the bottom row (two pointing up, one pointing down between them).
2. In general, cut the big triangle into horizontal strips of height one small triangle. The top strip holds $1$ triangle. Each strip below holds $2$ more than the one above it: $1, 3, 5, 7, \ldots$
3. With side $10$ there are $10$ strips, so the counts are
$$1, 3, 5, 7, 9, 11, 13, 15, 17, 19.$$
4. Add them by pairing the ends: $1 + 19 = 20$, $3 + 17 = 20$, $5 + 15 = 20$, $7 + 13 = 20$, $9 + 11 = 20$. Five pairs of $20$ make $100$.

The answer is $\boxed{\textbf{(C)}\ 100}$.

The row count shows the pattern: side $1$ needs $1$ triangle, side $2$ needs $4$, side $3$ needs $9$, and side $10$ needs $100$. The number of small pieces grows with the square of the side, exactly like a $10 \times 10$ square holding $100$ unit squares. Next time a shape is filled with smaller copies of itself, think "side ratio squared".

Use area. Every equilateral triangle has the same shape, and making the sides $10$ times as long makes the area $10 \times 10 = 100$ times as large. The small triangles fill the big one with no gaps or overlaps, so it takes $100$ of them.

All equilateral triangles are similar, and the side ratio here is $10 : 1$. Areas of similar figures scale by the square of the side ratio, so the big triangle has $10^2 = 100$ times the area of a unit triangle. Since the small triangles tile it without gaps or overlaps, exactly $100$ of them are needed.

The answer is $\boxed{\textbf{(C)}\ 100}$.