2025 AMC 10A Problem 15

Let $x = BC.$ Then $AC = \sqrt{1 + x^2},$ $CE = 7 - x,$ and $CD = 5 - \sqrt{1 + x^2}$

Use $\triangle ABC \sim \triangle EDC$ to set up a proportion, then solve for $x$

Let $x = BC.$ Since $ABEF$ is a rectangle with $AB = 1$ and $AF = 7,$ and $AD = 5,$ we get $AC = \sqrt{1 + x^2},$ $CE = 7 - x,$ and $CD = 5 - \sqrt{1 + x^2}.$ The triangles $\triangle ABC$ and $\triangle EDC$ are similar, so $\frac{7 - x}{\sqrt{1 + x^2}} = \frac{5 - \sqrt{1 + x^2}}{x}.$ Clear denominators and square to get $24x^2 + 14x - 24 = 0,$ which factors as $(4x - 3)(3x + 4) = 0.$ The positive root is $x = \tfrac{3}{4}.$ So the area is $\tfrac12 \cdot \tfrac34 \cdot 1 = \tfrac{3}{8}.$ Thus, A is the correct answer.