Fall 2021 AMC 10 B

Label the vertices as shown in the diagram. Then $△ABC$, $△CDE$, and $△EFG$ are similar. Because $CD=2$ and $DE=\frac{3-2}{2}=\frac{1}{2}$, the lengths of the legs of each of these triangles are in the ratio of $4$ to $1$. It follows that $FG=\frac{3}{4}$, so the base of the isosceles triangle has length $\frac{3}{4}+3+\frac{3}{4}=\frac{9}{2}$. Similarly, because $BC=1$, similar triangles give $AB=4$. It follows that the altitude of the isosceles triangle is $4+2+3=9$. The area of the triangle is then given by $\frac{1}{2}\cdot \frac{9}{2}\cdot 9=20\frac{1}{4}$.

Place the triangle in a coordinate plane with the base on the x-axis and the apex on the positive y-axis. The upper right vertex of the large square is $(\frac{3}{2},3)$, and the upper right vertex of the small square is $(1,5)$. The side of the triangle in the first quadrant has slope $$\frac{3-5}{\frac{3}{2}-1}=-4,$$ and its equation is $y=9-4x$. Thus the side intersects the x-axis at $(\frac{9}{4},0)$ and the y-axis at $(0,9)$. Therefore the triangle has base $2\cdot \frac{9}{4}=\frac{9}{2}$ and altitude $9$, so its area is $\frac{1}{2}\cdot \frac{9}{2}\cdot 9=20\frac{1}{4}$.