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Difference between revisions of "2018 AMC 10A Problems/Problem 9"

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All of the triangles in the diagram below are similar to iscoceles triangle <math>ABC</math>, in which <math>AB=AC</math>. Each of the 7 smallest triangles has area 1, and <math>\triangle ABC</math> has area 40. What is the area of trapezoid <math>DBCE</math>?
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{{duplicate|[[2018 AMC 10A Problems/Problem 9|2018 AMC 10A #9]] and [[2018 AMC 12A Problems/Problem 8|2018 AMC 12A #8]]}}
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==Problem==
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All of the triangles in the diagram below are similar to isosceles triangle <math>ABC</math>, in which <math>AB=AC</math>. Each of the <math>7</math> smallest triangles has area <math>1,</math> and <math>\triangle ABC</math> has area <math>40</math>. What is the area of trapezoid <math>DBCE</math>?
  
 
<asy>
 
<asy>
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<math>\textbf{(A) }  16  \qquad        \textbf{(B) }  18  \qquad    \textbf{(C) }  20  \qquad  \textbf{(D) }  22 \qquad  \textbf{(E) }  24 </math>
 
<math>\textbf{(A) }  16  \qquad        \textbf{(B) }  18  \qquad    \textbf{(C) }  20  \qquad  \textbf{(D) }  22 \qquad  \textbf{(E) }  24 </math>
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==Solution 1==
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Let <math>x</math> be the area of <math>ADE</math>. Note that <math>x</math> is comprised of the <math>7</math> small isosceles triangles and a triangle similar to <math>ADE</math> with side length ratio <math>3:4</math> (so an area ratio of <math>9:16</math>). Thus, we have <cmath>x=7+\dfrac{9}{16}x.</cmath> This gives <math>x=16</math>, so the area of <math>DBCE=40-x=\boxed{(E) 24}</math>.
  
 
==Solution 2==
 
==Solution 2==
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Let <math>x</math> be the area of <math>ADE</math>. Note that <math>x</math> is comprised of the <math>7</math> small isosceles triangles and a triangle similar to <math>ADE</math> with side length ratio <math>3:4</math> (so an area ratio of <math>9:16</math>). Thus, we have <cmath>x=7+\dfrac{9}{16}x</cmath> This gives <math>x=16</math>, so the area of <math>DBCE=40-x=\boxed{24}</math>.
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Let the base length of the small triangle be <math>x</math>. Then, there is a triangle <math>ADE</math> encompassing the 7 small triangles and sharing the top angle with a base length of <math>4x</math>. Because the area is proportional to the square of the side, let the base <math>BC</math> be <math>\sqrt{40}x</math>. The ratio of the area of triangle <math>ADE</math> to triangle <math>ABC</math> is <math>\left(\frac{4x}{\sqrt{40}x}\right)^2 = \frac{16}{40}</math>. The problem says the area of triangle <math>ABC</math> is <math>40</math>, so the area of triangle <math>ADE</math> is <math>16</math>. So the area of trapezoid <math>DBCE</math> is <math>40 - 16 = \boxed{24}</math>.
  
 
==Solution 3==
 
==Solution 3==
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Notice <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]</math>.
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Let the base of the small triangles of area 1 be <math>x</math>, then the base length of <math>\Delta ADE=4x</math>. Notice, <math>\left(\frac{DE}{BC}\right)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}</math>, then <math>4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\left(\frac{4}{\sqrt{40}}\right)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]</math>
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Thus, <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\left(1-\frac{2}{5}\right)=\frac{3}{5}\cdot 40=\boxed{24}.</math>
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<asy>
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unitsize(5);
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dot((0,0));
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dot((60,0));
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dot((50,10));
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dot((10,10));
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dot((30,30));
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draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0));
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draw((10,10)--(50,10));
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label("$B$",(0,0),SW);
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label("$C$",(60,0),SE);
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label("$E$",(50,10),E);
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label("$D$",(10,10),W);
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label("$A$",(30,30),N);
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draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10));
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draw((15,15)--(45,15));
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</asy>
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==Solution 4==
 
The area of <math>ADE</math> is 16 times the area of the small triangle, as they are similar and their side ratio is <math>4:1</math>. Therefore the area of the trapezoid is <math>40-16=\boxed{24}</math>.
 
The area of <math>ADE</math> is 16 times the area of the small triangle, as they are similar and their side ratio is <math>4:1</math>. Therefore the area of the trapezoid is <math>40-16=\boxed{24}</math>.
  
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== See Also ==
+
==Solution 5==
 +
You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be <math>7+5+3+1=16</math>, so to find the area of such trapezoid <math>BCED</math>, we just take <math>40-16=\boxed{24}</math>, like so.
 +
 
 +
==Solution 6==
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The combined area of the small triangles is <math>7</math>, and from the fact that each small triangle has an area of <math>1</math>, we can deduce that the larger triangle above has an area of <math>9</math> (as the sides of the triangles are in a proportion of <math>\frac{1}{3}</math>, so will their areas have a proportion that is the square of the proportion of their sides, or <math>\frac {1}{9}</math>). Thus, the combined area of the top triangle and the trapezoid immediately below is <math>7 + 9 = 16</math>. The area of trapezoid <math>BCED</math> is thus the area of triangle <math>ABC-16 =\boxed{24}</math>.
 +
 
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==Solution 7==
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You can assume for the base of one of the smaller triangles to be <math>\frac{1}{a}</math> and the height to be <math>2a</math>, giving an area of 1. The larger triangle above the 7 smaller ones then has base <math>\frac{3}{a}</math> and height <math>6a</math>, giving it an area of <math>9</math>. Then the area of triangle <math>ADE</math> is <math>16</math> and <math>40-16=\boxed{24}</math>.
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==Solution 8==
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You can construct another trapezoid directly above the one shown, with it's bottom length as the top length of the original. Its area would then be 9/16 of the original. Repeating this process infinitely gives us the sequence <math>7\cdot\left(1+\left(\frac{9}{16}\right)+\left(\frac{9}{16}\right)^2+\left(\frac{9}{16}\right)^3\dots\right)</math> . Using the infinite geometric series sum formula gives us <math>7\cdot\left(\frac{1}{1-\frac{9}{16}}\right)=7\cdot\frac{16}{7}=16</math>. The triangle's area would thus be <math>40-16=\boxed{24}</math>. 
 +
 
 +
-ConcaveTriangle
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==Solution 9 (weird)==
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Note that the area of an isosceles triangle is equivalent to the square of its height. Using this information, the height of the smallest isosceles triangle is <math>1</math>, and thus its base is <math>2.</math>
 +
 
 +
Let <math>h</math> be the height of the top triangle. We can set up a height-to-base similarity ratio, using the top triangle and <math>\triangle{ADE}</math>. The top triangle has a base of <math>3\cdot{2}=6</math>, and <math>DE=4\cdot{2}=8.</math> The height of <math>\triangle{ADE}</math> is <math>h+1</math>, therefore our ratio is <math>\frac{h}{6}=\frac{h+1}{8}</math>, which yields <math>h=3</math> as our answer.
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To find the area of the trapezoid, we can take the area of <math>\triangle{ABC}</math> and subtract the area of <math>\triangle{ADE},</math> whose base is <math>8</math> and height <math>3+1=4</math>. It follows that the area of <math>\triangle{ADE}=16</math>, and subtracting this from <math>40</math> gives us <math>40-16=\boxed{\text{D) }24}</math>.
 +
 
 +
-Benedict T (countmath1)
 +
 
 +
==Video Solution (HOW TO THINK CREATIVELY!)==
 +
https://youtu.be/lgkkomUwSs0
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
 
 +
==Video Solutions==
 +
 
 +
https://youtu.be/HJALwsbHZXc
 +
 
 +
~Whiz
 +
 
 +
https://youtu.be/ZiZVIMmo260
 +
 
 +
~IceMatrix
 +
 
 +
https://youtu.be/tq4rWskQDh8
 +
 
 +
~savannahsolver
 +
 
 +
https://youtu.be/4_x1sgcQCp4?t=2959
 +
 
 +
~pi_is_3.14
 +
 
 +
==See Also==
 +
 
 
{{AMC10 box|year=2018|ab=A|num-b=8|num-a=10}}
 
{{AMC10 box|year=2018|ab=A|num-b=8|num-a=10}}
 
{{AMC12 box|year=2018|ab=A|num-b=7|num-a=9}}
 
{{AMC12 box|year=2018|ab=A|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 13:44, 3 July 2023

The following problem is from both the 2018 AMC 10A #9 and 2018 AMC 12A #8, so both problems redirect to this page.

Problem

All of the triangles in the diagram below are similar to isosceles triangle $ABC$, in which $AB=AC$. Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$. What is the area of trapezoid $DBCE$?

[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]

$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$

Solution 1

Let $x$ be the area of $ADE$. Note that $x$ is comprised of the $7$ small isosceles triangles and a triangle similar to $ADE$ with side length ratio $3:4$ (so an area ratio of $9:16$). Thus, we have \[x=7+\dfrac{9}{16}x.\] This gives $x=16$, so the area of $DBCE=40-x=\boxed{(E) 24}$.

Solution 2

Let the base length of the small triangle be $x$. Then, there is a triangle $ADE$ encompassing the 7 small triangles and sharing the top angle with a base length of $4x$. Because the area is proportional to the square of the side, let the base $BC$ be $\sqrt{40}x$. The ratio of the area of triangle $ADE$ to triangle $ABC$ is $\left(\frac{4x}{\sqrt{40}x}\right)^2 = \frac{16}{40}$. The problem says the area of triangle $ABC$ is $40$, so the area of triangle $ADE$ is $16$. So the area of trapezoid $DBCE$ is $40 - 16 = \boxed{24}$.

Solution 3

Notice $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]$. Let the base of the small triangles of area 1 be $x$, then the base length of $\Delta ADE=4x$. Notice, $\left(\frac{DE}{BC}\right)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}$, then $4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\left(\frac{4}{\sqrt{40}}\right)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]$ Thus, $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\left(1-\frac{2}{5}\right)=\frac{3}{5}\cdot 40=\boxed{24}.$

[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]

Solution 4

The area of $ADE$ is 16 times the area of the small triangle, as they are similar and their side ratio is $4:1$. Therefore the area of the trapezoid is $40-16=\boxed{24}$.

Solution 5

You can see that we can create a "stack" of 5 triangles congruent to the 7 small triangles shown here, arranged in a row above those 7, whose total area would be 5. Similarly, we can create another row of 3, and finally 1 more at the top, as follows. We know this cumulative area will be $7+5+3+1=16$, so to find the area of such trapezoid $BCED$, we just take $40-16=\boxed{24}$, like so.

Solution 6

The combined area of the small triangles is $7$, and from the fact that each small triangle has an area of $1$, we can deduce that the larger triangle above has an area of $9$ (as the sides of the triangles are in a proportion of $\frac{1}{3}$, so will their areas have a proportion that is the square of the proportion of their sides, or $\frac {1}{9}$). Thus, the combined area of the top triangle and the trapezoid immediately below is $7 + 9 = 16$. The area of trapezoid $BCED$ is thus the area of triangle $ABC-16 =\boxed{24}$.

Solution 7

You can assume for the base of one of the smaller triangles to be $\frac{1}{a}$ and the height to be $2a$, giving an area of 1. The larger triangle above the 7 smaller ones then has base $\frac{3}{a}$ and height $6a$, giving it an area of $9$. Then the area of triangle $ADE$ is $16$ and $40-16=\boxed{24}$.

Solution 8

You can construct another trapezoid directly above the one shown, with it's bottom length as the top length of the original. Its area would then be 9/16 of the original. Repeating this process infinitely gives us the sequence $7\cdot\left(1+\left(\frac{9}{16}\right)+\left(\frac{9}{16}\right)^2+\left(\frac{9}{16}\right)^3\dots\right)$ . Using the infinite geometric series sum formula gives us $7\cdot\left(\frac{1}{1-\frac{9}{16}}\right)=7\cdot\frac{16}{7}=16$. The triangle's area would thus be $40-16=\boxed{24}$.

-ConcaveTriangle

Solution 9 (weird)

Note that the area of an isosceles triangle is equivalent to the square of its height. Using this information, the height of the smallest isosceles triangle is $1$, and thus its base is $2.$

Let $h$ be the height of the top triangle. We can set up a height-to-base similarity ratio, using the top triangle and $\triangle{ADE}$. The top triangle has a base of $3\cdot{2}=6$, and $DE=4\cdot{2}=8.$ The height of $\triangle{ADE}$ is $h+1$, therefore our ratio is $\frac{h}{6}=\frac{h+1}{8}$, which yields $h=3$ as our answer.

To find the area of the trapezoid, we can take the area of $\triangle{ABC}$ and subtract the area of $\triangle{ADE},$ whose base is $8$ and height $3+1=4$. It follows that the area of $\triangle{ADE}=16$, and subtracting this from $40$ gives us $40-16=\boxed{\text{D) }24}$.

-Benedict T (countmath1)

Video Solution (HOW TO THINK CREATIVELY!)

https://youtu.be/lgkkomUwSs0

~Education, the Study of Everything


Video Solutions

https://youtu.be/HJALwsbHZXc

~Whiz

https://youtu.be/ZiZVIMmo260

~IceMatrix

https://youtu.be/tq4rWskQDh8

~savannahsolver

https://youtu.be/4_x1sgcQCp4?t=2959

~pi_is_3.14

See Also

2018 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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
2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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