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

Revision as of 15:36, 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, $\angle ABC=90°$ . We are given that $\angle ADE=90°$ , 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 $x$ : $24x^2+10x-50=0$ , 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, mistake edited by neo changed $24x^2+10x-56=0$ to $24x^2+10x-50=0$

Soltion 2 (less algebra)

Draw segment $AE.$ Segment $AE$ is the diagonal of rectangle $ABEF,$ so its diagonals have length $\sqrt{7^2+1^2}=\sqrt{50}=5\sqrt2.$ From right triangle $AED,$ we use pythagorean theorem to find $DE = 5.$

Now, we see similar triangles $CDE$ and $ABC$ . Let $CE = a,$ and $CD = b.$ We can find that $AC = 5-b,$ and $CB = 7-a.$ These triangles have a ratio of $\frac {AB}{DE} = \frac{1}{5}.$ So we get that $\frac {5-b}{a} = \frac{1}{5}.$ Cross multplying, we get $a =25-5b.$ And also $\frac{CB}{CD} = \frac{1}{5} = \frac{7-a}{b}.$ Cross multiplying gives $35-5a=b.$ Solving the system of equations, we find $a = 25/4,$ which means $CB = 7-25/4 = 3/4.$ $[ABC] = CB/2,$ which gives $\boxed{[ABC] = 3/8}.$

~ eqb5000/Esteban Q.

@@ Line 38: / Line 38: @@
 Let <imath>AB=x</imath>. By the Pythagorean Theorem, <imath>AC=\sqrt{x^2-1}</imath>. Since <imath>FB=EC=7</imath>, <imath>EA=7=\sqrt{x^2-1}</imath>. Because <imath>AB=x</imath> and <imath>BD=5</imath>, <imath>AD=5-x</imath>. By similar triangles, <cmath>\frac{7-\sqrt{x^2-1}}{x}=\frac{5-x}{\sqrt{x^2-1}}</cmath>. Cross-multiplying, we get that <cmath>7\sqrt{x^2-1}-x^2+1=5x-x^2</cmath>, so <cmath>7\sqrt{x^2-1}=5x-1</cmath>. This is simply a quadratic in <imath>x</imath>: <cmath>24x^2+10x-50=0</cmath>, which has positive root <imath>x=\frac{5}{4}</imath>. Since <imath>BC=1</imath>, <imath>AC=\frac{3}{4}</imath>, so <imath>[ABC]=\textbf{(A)} \frac{3}{8}</imath>
-Solution by HumblePotato, written by lhfriend, mistake edited by neo changed <imath>24x^2+10x-56=0</imath> to <imath>24x^2+10x-50=0
+Solution by HumblePotato, written by lhfriend, mistake edited by neo changed <imath>24x^2+10x-56=0</imath> to <imath>24x^2+10x-50=0</imath>
 ==Soltion 2 (less algebra) ==

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

Revision as of 15:36, 6 November 2025

Problem

Solution 1

Soltion 2 (less algebra)