2021 AMC Fall 10B Problem 13

Measure height from the base of the triangle. The two slanted sides are straight, so the triangle gets narrower at a steady rate as you go up. Two widths are known, and that is enough to find everything.

1. Width at height 3. The big square has side $3$ and sits on the base. Its top corners touch the triangle's sides, so at height $3$ the triangle is $3$ wide.
2. Width at height 5. The small square sits on top of the big one, so its top is at height $3 + 2 = 5$. Its top corners touch the triangle's sides, so at height $5$ the triangle is $2$ wide.
3. The rate. Going up $2$ units, the width dropped by $3 - 2 = 1$. So the width drops $\frac12$ for every $1$ unit up.
4. The base. Going down $3$ units from height $3$ to the base, the width grows by $3 \times \frac12 = \frac32$. So the base is $3 + \frac32 = \frac92$.
5. The top point. Starting from width $3$ at height $3$, losing $\frac12$ per unit, the width reaches $0$ after $6$ more units. So the triangle's height is $3 + 6 = 9$.
6. Area.
$$\frac12 \times \frac92 \times 9 = \frac{81}{4} = 20\tfrac14.$$

The answer is $\boxed{\textbf{(B)}\ 20\frac14}$.

Cutting a triangle straight across at any height gives a smaller copy of the triangle, so the width changes at a constant rate as you climb. Two measured widths fix that rate and pin down the whole triangle. This is the same as a similar-triangles argument, just packed into one rate calculation.

Use similar triangles directly. Above height $5$ there is a small triangle with base $2$. Above height $3$ there is a bigger triangle with base $3$. They are the same shape, so their heights are in the ratio $2 : 3$. Their heights also differ by $2$ (the small square). Heights $4$ and $6$ fit both facts. So the top point is at height $3 + 6 = 9$, and the whole triangle, with height $9$, is $\frac96$ times the base-$3$ triangle, giving base $3 \times \frac96 = \frac92$.