, 2021

Label the vertices as shown in the diagram.

Then $\triangle ABC$, $\triangle CDE$, and $\triangle EFG$ 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 $\frac{1}{2} \cdot \frac{9}{2} \cdot 9 = 20\frac{1}{4}$.