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

Revision as of 02:26, 9 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 $\triangle ABC$ ?

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

Video Solution

https://youtu.be/CCYoHk2Af34

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 $7\sqrt{x^2-1}=5x-1$ . We square both sides, and this is simply a quadratic in $x$ : $24x^2+10x-50=0$ , which has positive root $x=\frac{5}{4}$ . Since $AB=1$ , we can plug this into the Pythagorean Theorem, with $\frac{5}{4}$ being the hypotenuse, to get $BC=\frac{3}{4}$ , and ${1}*\frac{3}{4}/2$ to equal $[ABC]= \boxed{\textbf{(A)} \frac{3}{8}}$

Solution by HumblePotato, written by lhfriend,

~Corrected all incorrect side length labels and fixed typos and major errors ~Neo

Minor edit by SixthGradeBookWorm927

Minor edit by aldzandrtc

Minor edit by rcll (I lowkey fixed this shit, as aldzandrtc would have said)

Minor edit by pyraminx

Solution 2 (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 3 (Trigonometry)

Using the Pythagorean theorem, I can get $AE=5 \sqrt{2}$ . 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:

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

$=\frac{7}{10}+ \frac{1}{10}$

$=\frac{4}{5}$

Thus, $\sin(a+b)=\frac{4}{5}$ , so $\sin(180-a-b)=\frac{4}{5}$ . This indicates that if a perpendicular is dropped from point $F$ to the extension of line $DE$ , and the foot of the latitude is point $G$ , $\triangle EGF$ is a 3-4-5 triangle. Because $\triangle EGF\sim\triangle CDE$ , $\triangle CDE$ is also a 3-4-5 triangle. Using ratios,:

$CE=\frac{5}{4}\times 5$

$=\frac{25}{4}$

Therefore, $CE=\frac{25}{4}$ , so $BC=\frac{3}{4}$ , so $\triangle ABC$ has area $\frac{3}{8}$ , or $\boxed A$ .

~Lollipop316

P.S. Thank you to eqb5000 and i_am_not_suk_at_math for pointing out and helping me fix typos.

Solution 4 (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

Remark - You could also redraw the figure to scale. That way, you won't have to assume it's drawn to scale in the provided figure. You would have to draw the diagonal AE, measure it as $\sqrt{7^2+1^2}$ and use pythagorean theorem to find ED as 5. Once you have that information you could sketch the whole figure to scale.

~wisegod62 (Remark and LaTeX formatting)

Solution 5 (another cheese)

Inspired by 2022 AMC 10B Problem 16, another right-triangle and rectangle mashup, we can pretend that the problem uses $3-4-5$ similarity, because AMC problems do that a lot. In this case, we can verify it works by setting $BC = \frac{3}{4}, AC = \frac{5}{4}$ by $3-4-5$ similarity, so $CD = \frac{15}{4}$ and $CE = \frac{25}{4}$ . Then, we see that $BC + CE = \frac{25+3}{4} = 7 = BE = AF$ , which is our only other condition, so this setup works and we can bubble in $\frac{1}{2} \cdot \frac{3}{4} \cdot 1 = \boxed{\textbf{(A)} \frac{3}{8}}$ . of course, if it weren't $3-4-5$ , we can use any other technique.

~tiguhbabehwo

Solution 6 (area)

Connect $AE$ , which equals $\sqrt{50}$ . Then calculate $ED$ , which equals $5$ . $\triangle ABC \sim \triangle EDC$ (AA). $\dfrac{AB}{ED} = \dfrac{1}{5}$ .

Therefore, $\dfrac{\text{Area of } \triangle ABC}{\text{Area of } \triangle EDC} = \dfrac{1}{25}$ .

Also, let the area of $\triangle ACE = x$ . The area of $\triangle ABC = \dfrac{7}{2} - x$ . The area of $\triangle EDC = \dfrac{25}{2} - x$ .

Solve: $x = \dfrac{25}{8}$

So, $\text{Area of } \triangle ABC = \dfrac{7}{2} - \dfrac{25}{8} = \dfrac{3}{8}$ .

~sandpiper357

Chinese Video Solution

https://www.bilibili.com/video/BV1nhkUByE3V/

~metrixgo

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

Revision as of 02:26, 9 November 2025

Contents

Problem

Video Solution

Solution 1

Solution 2 (thorough)

Solution 3 (Trigonometry)

Solution 4 (risky, but it works!)

Solution 5 (another cheese)

Solution 6 (area)

Chinese Video Solution

Video Solution (Fast and Easy)

Video Solution by SpreadTheMathLove

Video Solution

Video Solution by Daily Dose of Math

See Also

@@ Line 148: / Line 148: @@
 ~sandpiper357
+==Chinese Video Solution==
+https://www.bilibili.com/video/BV1nhkUByE3V/
+~metrixgo
 == Video Solution (Fast and Easy) ==