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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, $\angle ABC=90°$. We are given that $\angle ADE=90°$, and since $\angle EAD=\angle BAC$ by vertical angles, $\triangle ECD \sim \triangle BAC$. 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 $24x^2+10x-50=0$ Minor edit (EAD->ECD, added \sim) by SixthGradeBookWorm927

Solution 2 (less algebra, simpler)

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.

Solution 3 (10 second solution🔥)

From the answer choices, we can deduce the following. If the answer is $A$, then since $AB = 1$, we have $CB = \tfrac{3}{4}$ and $AC = \tfrac{5}{4}.$ Now, because $AB = 5$, it follows that $CB = \tfrac{15}{4}.$ Since $CB = \tfrac{3}{4}$ and $EB = 7$, we can find $EC = \tfrac{25}{4}.$ We already know that $\triangle EDC \sim \triangle ABC,$ and these calculated side lengths are consistent with the given ratio of similarity. Moreover, we can observe that both triangles are $3$–$4$–$5$ right triangles. Therefore, the answer is $\boxed{A}.$ Remark: This solution only works here because our answer is A. In a real test it is not ideal to do this ~ WildSealVM / Vincent M.

Solution 4 (thorough)

From the diagram, $\angle BCA$ and $\angle DCE$ are vertical angles and hence congruent. Additionally, $\angle B = \angle D = 90^\circ$, so we have by AA Similarity that $\triangle BCA \sim \triangle DCE$.

Let $BC = x$ so $EC = 7 - x$ and $AC = y$ so $CD = 5 - y$. Since the two triangles are similar, we have $\frac{BC}{AC} = \frac{CD}{EC}$. Plugging in the variables gives $\frac{x}{y} = \frac{5 - y}{7 - x}$.

Cross multiplying yields $(7 - x)(x) = (5 - y)(y) \implies 7x - x^2 = 5y - y^2 \implies 7x + (y^2 - x^2) = 5y$.

By applying the Pythagorean Theorem on $\triangle BCA$, we get $x^2 + 1 = y^2 \implies 1 = y^2 - x^2$.

Therefore, $y = \frac{7x + 1}{5}$, and plugging this back into $x^2 + 1 = y^2$:

$x^2 + 1= (\frac{7x+1}{5})^2$

$25(x^2 + 1) = (7x+1)^2$

$25x^2 + 25 = 49x^2 + 14x + 1$

$0 = 49x^2 + 14x + 1 - 25x^2 - 25$

$0 = 24x^2 + 14x - 24$

$0 = 12x^2 + 7x - 12$

$x = \frac{-7 \pm \sqrt{7^2 - 4(12)(-12)}}{2\cdot 12}$

$= \frac{-7 \pm \sqrt{49 + 576}}{24}$

$= \frac{-7 \pm \sqrt{625}}{24}$

$= \frac{-7 \pm 25}{24}$

Therefore, $x = \frac{-7+25}{24}=\frac{18}{24}=\frac{3}{4}$.

The area of $\triangle BCA$ is therefore $\frac{x \cdot 1}{2}=\frac{3}{8}=\boxed{A}$. ~hxve

Video Solution (Fast and Easy)

https://youtu.be/RvU1P9qRu84?si=Ynf6wWPNB1EuF_mq ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

See Also

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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