2025 AMC 10A Problems/Problem 15

Problem

In the figure below, $ABEF$ is a rectangle, $\overline{AD}\perp\overline{DE}$ , $AF=7$ , $AB=1$ , and $AD=5$ . $[asy] unitsize(1cm); pair A, B, C, D, E, F; A = (5, 5); B = (5.6, 4.2); C = (5, 3.75); D = (5, 0); E = (0, 0); F = (-0.6, 0.8); fill(A--B--C--cycle, gray); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); label("$A$", A, N); label("$B$", B, (1,0)); label("$C$", C, SE); label("$D$", D, (1,0)); label("$E$", E, S); label("$F$", F, W); draw(A--D--E); draw(A--B--E--F--A); draw(rightanglemark(C, D, E)); [/asy]$ What is the area of $\triangle ABC$ ?

$\textbf{(A) } \frac{3}{8} \qquad\textbf{(B) } \frac{4}{9} \qquad\textbf{(C) } \frac{1}{8}\sqrt{13} \qquad\textbf{(D) } \frac{7}{15} \qquad\textbf{(E) } \frac{1}{8}\sqrt{15}$

Solution

Solution 1

Because $ABEF$ is a rectangle, $\angle BCA=90°$ . We are given that $\angle BDE=90°$ , and since $\angle EAD=\angle BAC$ by vertical angles, $\triangle EAD~\triangle BAC$ . Let $AB=x$ . By the Pythagorean Theorem, $AC=\sqrt{x^2-1}$ . Since $FB=EC=7$ , $EA=7=\sqrt{x^2-1}$ . Because $AB=x$ and $BD=5$ , $AD=5-x$ . By similar triangles, $\frac{7-\sqrt{x^2-1}}{x}=\frac{5-x}{\sqrt{x^2-1}}$ . Cross-multiplying, we get that $7\sqrt{x^2-1}-x^2+1=5x-x^2$ , so $7\sqrt{x^2-1}=5x-1$ . This is simply a quadratic in $x$ : $24x^2+10x-56=0$ , which has positive root $x=\frac{5}{4}$ . Since $BC=1$ , $AC=\frac{3}{4}$ , so $[ABC]=\textbf{(A)} \frac{3}{8}$

Solution by HumblePotato, written by lhfriend

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2025 AMC 10A Problems/Problem 15

Problem

Solution 1