Question

In the figure below, $ABEF$ is a rectangle, $AD \perp DE,$ $AF = 7,$ $AB = 1,$ and $AD = 5.$ What is the area of $\triangle ABC?$

$\dfrac{3}{8}$

$\dfrac{4}{9}$

$\dfrac{1}{8}\sqrt{13}$

$\dfrac{7}{15}$

$\dfrac{1}{8}\sqrt{15}$

Solution:

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.

2025 AMC 10A Problem 15

Solution:

Problem 15 in Other Years