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 $\angle ECD=\angle ACB$ by vertical angles, $\triangle ECD \sim \triangle ACB$ . Let $AC=x$ . By the Pythagorean Theorem, $CB=\sqrt{x^2-1}$ . Since $AF=BE=7$ , $EC=7-\sqrt{x^2-1}$ . Because $AC=x$ and $AD=5$ , $CD=5-x$ . 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 $AB=1$ , $BC=\frac{3}{4}$ , $[ABC]= \boxed{\textbf{(A)} \frac{3}{8}}$

Solution by HumblePotato, written by lhfriend,

~Corrected all incorrect side length labels and fixed typos and major errors (changed $24x^2+10x-56=0$ to $24x^2+10x-50=0$ , $\triangle ECD \sim \triangle ACB$ etc.)~Neo

Minor edit (Changed $EAD$ to $ECD$ , added $\sim$ ) by SixthGradeBookWorm927

Solution 2 (less algebra, simpler)

Draw segment $AE.$ Segment $AE$ is the diagonal of rectangle $ABEF,$ and 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 $\triangle CDE$ and $\triangle CBA$ . 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 $AD = 5$ , it follows that $CD = \tfrac{15}{4}.$ 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

Solution 5 (Trigonometry)

Using the Pythagorean theorem, I can get $AE=\frac{5}{\sqrt{2}}$ . Then, because $AD=5$ , $ED=5$ . Now, let $\angle FEA=a$ and $\angle AED=b$ . $\sin a=\frac{7}{5\sqrt{2}}, \sin b=\frac{1}{\sqrt{2}}, \cos a=\frac{1}{5\sqrt{2}},$ and $\cos b=\frac{1}{\sqrt{2}}$ . Then, applying the sine addition formula, I get:

$\sin(a+b)=\frac{7}{5\sqrt{2}}\times\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\times\frac{1}{5\sqrt{2}}$

To be continued and finished by end of Nov 8.

$\textbf{Note}$: $AE$ is equal to $\sqrt{7^2+1^2} = \sqrt{50} = 5\sqrt{2},$ not $\frac {5}{\sqrt2}.$

Yeah, AE doesn’t equal what you said

~eqb5000 (note)

~Lollipop316

~minor $\LaTeX$ edits by i_am_not_suk_at_math (saharshdevaraju 15:12, 7 November 2025 (EST)saharshdevaraju)

Solution 6 (risky, but it works!)

Using a ruler (which is permitted during the exam), and assuming the diagram is to-scale, we can measure the physical lengths of $AB$ and $BC$ , and determine the scale factor in order to calculate $BC$ 's actual math length. In my specific case (potentially could vary), $AB$ was $1$ cm and $BC$ was between $0.7$ and $0.8$ cm. So, the scale with cm is $1:1$ , and the length of $BC$ is around $0.75$ , so the area is $\frac{1}{2} \cdot 0.75 \cdot 1 = \frac{3}{8}$ . To assure ourselves that $\frac{3}{8}$ is the most accurate estimation, we know that $\frac{4}{9}$ is around $0.44$ (too big), $\frac{\sqrt(13)}{8}$ even bigger (so also too big), $\frac{7}{15}$ is just under $0.5$ , and $\frac{\sqrt(15)}{8}$ is even bigger than $\frac{\sqrt(13)}{8}$ , so most likely the answer is $\boxed{\textbf{(A)} \frac{3}{8}}$ .

~vaishnav

Video Solution (Fast and Easy)

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

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

Video Solution by Daily Dose of Math

https://youtu.be/5Fjos1vBt0A

~Thesmartgreekmathdude

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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