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

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<imath>\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}</imath>
<imath>\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}</imath>


[[2025 AMC 10A Problems/Problem 15|Solution]]
==Video Solution==
https://youtu.be/CCYoHk2Af34


==Solution 1==
==Solution 1==


Because <imath>ABEF</imath> is a rectangle, <imath>\angle ABC=90°</imath>. We are given that <imath>\angle ADE=90°</imath>, and since <imath>\angle EAD=\angle BAC</imath> by vertical angles, <imath>\triangle ECD \sim \triangle BAC</imath>.  
Because <imath>ABEF</imath> is a rectangle, <imath>\angle ABC=90°</imath>. We are given that <imath>\angle ADE=90°</imath>, and since <imath>\angle ECD=\angle ACB</imath> by vertical angles, <imath>\triangle ECD \sim \triangle ACB</imath>.  
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>
Let <imath>AC=x</imath>. By the Pythagorean Theorem, <imath>CB=\sqrt{x^2-1}</imath>. Since <imath>AF=BE=7</imath>, <imath>EC=7-\sqrt{x^2-1}</imath>. Because <imath>AC=x</imath> and <imath>AD=5</imath>, <imath>CD=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>. We square both sides, and this is simply a quadratic in <imath>x</imath>: <cmath>24x^2+10x-50=0</cmath>, which has a positive root <imath>x=\frac{5}{4}</imath>. Since <imath>AB=1</imath>, we can plug this into the Pythagorean Theorem, with <imath>\frac{5}{4}</imath> being the hypotenuse, to get <imath>BC=\frac{3}{4}</imath>, and <imath>{1}\cdot \frac{\frac{3}{4}}{2}</imath> to equal <imath>[ABC]= \boxed{\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</imath>
Solution by HumblePotato, written by lhfriend,  
Minor edit (EAD->ECD, added \sim) by SixthGradeBookWorm927


==Solution 2 (less algebra, simpler) ==
~Corrected all incorrect side length labels and fixed typos and major errors ~Neo


Draw segment <imath>AE.</imath> Segment <imath>AE</imath> is the diagonal of rectangle <imath>ABEF,</imath> so its diagonals have length <imath>\sqrt{7^2+1^2}=\sqrt{50}=5\sqrt2.</imath> From right triangle <imath>AED,</imath> we use pythagorean theorem to find <imath>DE = 5.</imath>
Minor edit by SixthGradeBookWorm927


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


~ eqb5000/Esteban Q.
Minor edit by rcll (I lowkey fixed this shi, as aldzandrtc would have said)
 
Minor edit by pyraminx
 
Minor edit by wisewigglyjaguar
 
Minor edit by jeffyang2025
 
==Solution 2 (simpler)==


==Solution 3 (10 second solution🔥) ==
Draw segment <imath>AE.</imath> Segment <imath>AE</imath> is the diagonal of rectangle <imath>ABEF,</imath> and its diagonals have length <imath>\sqrt{7^2+1^2}=\sqrt{50}=5\sqrt2.</imath> From right triangle <imath>AED,</imath> we use pythagorean theorem to find <imath>DE = 5.</imath>


From the answer choices, we can deduce the following.
Now, we see similar triangles <imath>\triangle CDE</imath> and <imath>\triangle CBA</imath>. Let <imath>CE = a,</imath> and <imath>CD = b.</imath> We can find that <imath>AC = 5-b,</imath> and <imath>CB = 7-a.</imath> These triangles have a ratio of <imath>\frac {AB}{DE} = \frac{1}{5}.</imath> So we get that <imath>\frac {5-b}{a} = \frac{1}{5}.</imath> Cross multplying, we get <imath>a =25-5b.</imath> And also <imath>\frac{CB}{CD} = \frac{1}{5} = \frac{7-a}{b}.</imath> Cross multiplying gives <imath>35-5a=b.</imath> Solving the system of equations, we find <imath>a = 25/4,</imath> which means <imath>CB = 7-25/4 = 3/4.</imath> <imath>[ABC] = CB/2,</imath> which gives <imath>\boxed{[ABC] = 3/8}.</imath>
If the answer is <imath>A</imath>, then since <imath>AB = 1</imath>, we have <imath>CB = \tfrac{3}{4}</imath> and <imath>AC = \tfrac{5}{4}.</imath>
Now, because <imath>AD = 5</imath>, it follows that <imath>CB = \tfrac{15}{4}.</imath>
Since <imath>CB = \tfrac{3}{4}</imath> and <imath>EB = 7</imath>, we can find <imath>EC = \tfrac{25}{4}.</imath>
We already know that <imath>\triangle EDC \sim \triangle ABC,</imath> and these calculated side lengths are consistent with the given ratio of similarity.
Moreover, we can observe that both triangles are <imath>3</imath>–<imath>4</imath>–<imath>5</imath> right triangles.
Therefore, the answer is <imath>\boxed{A}.</imath> 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) ==
~ eqb5000/Esteban Q.
 
==Solution 3 (thorough) ==


From the diagram, <imath>\angle BCA</imath> and <imath>\angle DCE</imath> are vertical angles and hence congruent. Additionally, <imath>\angle B = \angle D = 90^\circ</imath>, so we have by AA Similarity that <imath>\triangle BCA \sim \triangle DCE</imath>.   
From the diagram, <imath>\angle BCA</imath> and <imath>\angle DCE</imath> are vertical angles and hence congruent. Additionally, <imath>\angle B = \angle D = 90^\circ</imath>, so we have by AA Similarity that <imath>\triangle BCA \sim \triangle DCE</imath>.   
Line 98: Line 100:
The area of <imath>\triangle BCA</imath> is therefore <imath>\frac{x \cdot 1}{2}=\frac{3}{8}=\boxed{A}</imath>. ~hxve
The area of <imath>\triangle BCA</imath> is therefore <imath>\frac{x \cdot 1}{2}=\frac{3}{8}=\boxed{A}</imath>. ~hxve


== Video Solution (Fast and Easy) ==
==Solution 4 (Trigonometry) ==
 
Using the Pythagorean theorem, I can get <imath>AE=5 \sqrt{2}</imath>. Then, because <imath>AD=5</imath>, <imath>ED=5</imath>. Now, let <imath>\angle FEA=a</imath> and <imath>\angle AED=b</imath>. <imath>\sin a=\frac{7}{5\sqrt{2}}, \sin b=\frac{1}{\sqrt{2}}, \cos a=\frac{1}{5\sqrt{2}},</imath> and <imath>\cos b=\frac{1}{\sqrt{2}}</imath>. Then, applying the sine addition formula, I get:
 
<imath>\sin(a+b)=\frac{7}{5\sqrt{2}}\times\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\times\frac{1}{5\sqrt{2}}</imath>
 
<imath>=\frac{7}{10}+ \frac{1}{10}</imath>
 
<imath>=\frac{4}{5}</imath>
 
Thus, <imath>\sin(a+b)=\frac{4}{5}</imath>, so <imath>\sin(180-a-b)=\frac{4}{5}</imath>. This indicates that if a perpendicular is dropped from point <imath>F</imath> to the extension of line <imath>DE</imath>, and the foot of the latitude is point <imath>G</imath>, <imath>\triangle EGF</imath> is a 3-4-5 triangle. Because <imath>\triangle EGF\sim\triangle CDE</imath>, <imath>\triangle CDE</imath> is also a 3-4-5 triangle. Using ratios,:
 
<imath>CE=\frac{5}{4}\times 5</imath>
 
<imath>=\frac{25}{4}</imath>
 
Therefore, <imath>CE=\frac{25}{4}</imath>, so <imath>BC=\frac{3}{4}</imath>, so <imath>\triangle ABC</imath> has area <imath>\frac{3}{8}</imath>, or <imath>\boxed A</imath>.
 
~Lollipop316
 
P.S. Thank you to eqb5000 and i_am_not_suk_at_math for pointing out and helping me fix typos.
 
== Solution 5 (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 <imath>AB</imath> and <imath>BC</imath>, and determine the scale factor in order to calculate <imath>BC</imath>'s actual math length. In my specific case (potentially could vary), <imath>AB</imath> was <imath>1</imath> cm and <imath>BC</imath> was between <imath>0.7</imath> and <imath>0.8</imath> cm. So, the scale with cm is <imath>1:1</imath>, and the length of <imath>BC</imath> is around <imath>0.75</imath>, so the area is <imath>\frac{1}{2} \cdot 0.75 \cdot 1 = \frac{3}{8}</imath>. To assure ourselves that <imath>\frac{3}{8}</imath> is the most accurate estimation, we know that <imath>\frac{4}{9}</imath> is around <imath>0.44</imath> (too big), <imath>\frac{\sqrt{13}}{8}</imath> even bigger (so also too big), <imath>\frac{7}{15}</imath> is just under <imath>0.5</imath>, and <imath>\frac{\sqrt{15}}{8}</imath> is even bigger than <imath>\frac{\sqrt{13}}{8}</imath>, so most likely the answer is <imath>\boxed{\textbf{(A)} \frac{3}{8}}</imath>.
 
 
~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 <imath>\sqrt{7^2+1^2}</imath> 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 6 (another cheese) ==
 
Inspired by [https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_16 2022 AMC 10B Problem 16], another right-triangle and rectangle mashup, we can pretend that the problem uses <imath>3-4-5</imath> similarity, because AMC problems do that a lot. In this case, we can verify it works by setting <imath>BC = \frac{3}{4}, AC = \frac{5}{4}</imath> by <imath>3-4-5</imath> similarity, so <imath>CD = \frac{15}{4}</imath> and <imath>CE = \frac{25}{4}</imath>. Then, we see that <imath>BC + CE = \frac{25+3}{4} = 7 = BE = AF</imath>, which is our only other condition, so this setup works and we can bubble in <imath>\frac{1}{2} \cdot \frac{3}{4} \cdot 1 = \boxed{\textbf{(A)} \frac{3}{8}}</imath>. of course, if it weren't <imath>3-4-5</imath>, we can use any other technique.
 
~tiguhbabehwo
 
== Solution 7 (area) ==
 
Connect <imath>AE</imath>, which equals <imath>\sqrt{50}</imath>. Then calculate <imath>ED</imath>, which equals <imath>5</imath>. 
<imath>\triangle ABC \sim \triangle EDC</imath> (AA). 
<imath>\dfrac{AB}{ED} = \dfrac{1}{5}</imath>. 
 
Therefore, 
<imath>\dfrac{\text{Area of } \triangle ABC}{\text{Area of } \triangle EDC} = \dfrac{1}{25}</imath>. 
 
Also, let the area of <imath>\triangle ACE = x</imath>. 
The area of <imath>\triangle ABC = \dfrac{7}{2} - x</imath>. 
The area of <imath>\triangle EDC = \dfrac{25}{2} - x</imath>. 
 
Solve: 
<imath>x = \dfrac{25}{8}</imath> 
 
So, 
<imath>\text{Area of } \triangle ABC = \dfrac{7}{2} - \dfrac{25}{8} = \boxed{\textbf{(A)} \dfrac{3}{8}}</imath>.
 
~sandpiper357
 
== Solution 8 (analytic geometry) ==
<asy>
unitsize(1cm);
pair _A, _B, _C, _D, _E, _F;
_A = (0, 0);
_B = (0, 1);
_E = (7, 1);
_F = (7, 0);
_C = (3/4, 1);
_D = extension(_A, _C, _E, _E + rotate(90)*(_A - _C));
fill(_A -- _B -- _C -- cycle, gray);
draw(_A -- _B -- _E -- _D -- _A -- _F -- _E);
draw((-1, 0) -- (8, 0), Arrows);
draw((0, -1) -- (0, 5), Arrows);
label("$A$", _A, SW);
label("$B$", _B, W);
label("$C$", _C, N);
label("$D$", _D, N);
label("$E$", _E, NE);
label("$F$", _F, SE);
label("$x$", (8, 0), S);
label("$y$", (0, 5), E);
label("$1$", _B, NE);
label("$7$", _F, SW);
label("$(a, 1)$", _C, SE);
label(scale(0.8) * rotate(aTan(4/3)) * "$y =\frac{1}{a}x$", (_A + _D)/2, NW);
label(scale(0.8) * rotate(-aTan(3/4)) * "$y - 1 = -a(x - 7)$", (_D + _E)/2, NE);
</asy>
As shown in the figure, establish a coordinate system with point A as the origin.
 
Suppose <imath>C = (a, 1)</imath>, then the equation of <imath>AC</imath> is <imath>y = \dfrac{1}{a}x</imath>. Since <imath>DE \perp AC</imath>, the slope of <imath>DE</imath> is <imath>-a</imath>, the equation of <imath>DE</imath> is <imath>y - 1 = -a(x - 7) \to ax + y - 7a - 1 = 0</imath>.
 
Notice that the length of <imath>AD</imath> is the distance between point <imath>A</imath> and line <imath>DE</imath>, due to the distance formula, we get:
<cmath>\frac{|a \cdot 0 + 0 - 7a - 1|}{\sqrt{a^2 + 1}} = 5</cmath>
<cmath>(7a + 1)^2 = 25(a^2 + 1)</cmath>
<cmath>(4a - 3)(3a + 4) = 0</cmath>
<cmath>a = \frac{3}{4}</cmath>
 
The area of <imath>\triangle BCA</imath> is therefore <imath>\frac{a \cdot 1}{2}=\frac{3}{8}=\boxed{A}</imath>.
 
~[https://artofproblemsolving.com/wiki/index.php/User:Reda_mandymath reda_mandymath]
 
== Solution 9 ==
From the Pythagorean Theorem, <imath>AE=5\sqrt{2}</imath> and we also discover <imath>\triangle ADE</imath> is a <imath>45°-45°-90°</imath> right triangle. This means that <imath>DE=5</imath>. As <imath>\triangle ABC \sim \triangle EDC</imath> by AA similarity and <imath>AB=1</imath>, we can set up a system of linear equations in terms of <imath>a</imath> and <imath>b</imath> where they represent <imath>BC</imath> and <imath>AC</imath> respectively. Since <imath>CD=5CB</imath> and <imath>CE=5AC</imath>, we have <cmath>5a+b=5</cmath> and <cmath>a+5b=7</cmath> and solving the equations results in <imath>a=\frac{3}{4}</imath>. The area of <imath>\triangle ABC</imath> is <imath>\boxed{\textbf{(A)} \frac{3}{8}}</imath>. <imath>b</imath> is unnecessary to find.
 
Note that <imath>\triangle ABC</imath> is a <imath>3-4-5</imath> triangle.
 
~ruihl123
 
==Chinese Video Solution==
 
https://www.bilibili.com/video/BV1nhkUByE3V/
 
~metrixgo
 
== Video Solution (Fast and Easy to Understand) ==
https://youtu.be/RvU1P9qRu84?si=Ynf6wWPNB1EuF_mq ~ Pi Academy
https://youtu.be/RvU1P9qRu84?si=Ynf6wWPNB1EuF_mq ~ Pi Academy
==Video Solution by SpreadTheMathLove==
https://www.youtube.com/watch?v=dAeyV60Hu5c


==Video Solution==
==Video Solution==
Line 105: Line 226:


~MK
~MK
==Video Solution by Daily Dose of Math==
https://youtu.be/5Fjos1vBt0A
~Thesmartgreekmathdude


==See Also==
==See Also==

Latest revision as of 20:55, 11 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 a 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 ${1}\cdot \frac{\frac{3}{4}}{2}$ 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 shi, as aldzandrtc would have said)

Minor edit by pyraminx

Minor edit by wisewigglyjaguar

Minor edit by jeffyang2025

Solution 2 (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 (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 4 (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 5 (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 6 (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 7 (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, $\text{Area of } \triangle ABC = \dfrac{7}{2} - \dfrac{25}{8} = \boxed{\textbf{(A)} \dfrac{3}{8}}$.

~sandpiper357

Solution 8 (analytic geometry)

[asy] unitsize(1cm); pair _A, _B, _C, _D, _E, _F; _A = (0, 0); _B = (0, 1); _E = (7, 1); _F = (7, 0); _C = (3/4, 1); _D = extension(_A, _C, _E, _E + rotate(90)*(_A - _C)); fill(_A -- _B -- _C -- cycle, gray); draw(_A -- _B -- _E -- _D -- _A -- _F -- _E); draw((-1, 0) -- (8, 0), Arrows); draw((0, -1) -- (0, 5), Arrows); label("$A$", _A, SW); label("$B$", _B, W); label("$C$", _C, N); label("$D$", _D, N); label("$E$", _E, NE); label("$F$", _F, SE); label("$x$", (8, 0), S); label("$y$", (0, 5), E); label("$1$", _B, NE); label("$7$", _F, SW); label("$(a, 1)$", _C, SE); label(scale(0.8) * rotate(aTan(4/3)) * "$y =\frac{1}{a}x$", (_A + _D)/2, NW); label(scale(0.8) * rotate(-aTan(3/4)) * "$y - 1 = -a(x - 7)$", (_D + _E)/2, NE); [/asy] As shown in the figure, establish a coordinate system with point A as the origin.

Suppose $C = (a, 1)$, then the equation of $AC$ is $y = \dfrac{1}{a}x$. Since $DE \perp AC$, the slope of $DE$ is $-a$, the equation of $DE$ is $y - 1 = -a(x - 7) \to ax + y - 7a - 1 = 0$.

Notice that the length of $AD$ is the distance between point $A$ and line $DE$, due to the distance formula, we get: \[\frac{|a \cdot 0 + 0 - 7a - 1|}{\sqrt{a^2 + 1}} = 5\] \[(7a + 1)^2 = 25(a^2 + 1)\] \[(4a - 3)(3a + 4) = 0\] \[a = \frac{3}{4}\]

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

~reda_mandymath

Solution 9

From the Pythagorean Theorem, $AE=5\sqrt{2}$ and we also discover $\triangle ADE$ is a $45°-45°-90°$ right triangle. This means that $DE=5$. As $\triangle ABC \sim \triangle EDC$ by AA similarity and $AB=1$, we can set up a system of linear equations in terms of $a$ and $b$ where they represent $BC$ and $AC$ respectively. Since $CD=5CB$ and $CE=5AC$, we have \[5a+b=5\] and \[a+5b=7\] and solving the equations results in $a=\frac{3}{4}$. The area of $\triangle ABC$ is $\boxed{\textbf{(A)} \frac{3}{8}}$. $b$ is unnecessary to find.

Note that $\triangle ABC$ is a $3-4-5$ triangle.

~ruihl123

Chinese Video Solution

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

~metrixgo

Video Solution (Fast and Easy to Understand)

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

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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