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2025 AMC 10A Problems/Problem 15: Difference between revisions

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==Soltion 2 (simpler)==
==Soltion 2 (simpler)==
Draw segment <imath>AE.</imath> <imath>\segment{AE}</imath>

Revision as of 14:03, 6 November 2025

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

Soltion 2 (simpler)

Draw segment $AE.$ $\segment{AE}$ (Error compiling LaTeX. Unknown error_msg)

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