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

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==Solution 1==
 
==Solution 1==
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 <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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Notice <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]</math>.
 
Notice <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]</math>.
 
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>\big(\frac{DE}{BC}\big)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}</math>, then <math>4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\big(\frac{4}{\sqrt{40}}\big)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]</math>
 
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>\big(\frac{DE}{BC}\big)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}</math>, then <math>4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\big(\frac{4}{\sqrt{40}}\big)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]</math>
Thus, <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\big(1-\frac{2}{5}\big)=\frac{3}{5}\cdot 40=\boxed{24}</math>
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Thus, <math>\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\big(1-\frac{2}{5}\big)=\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 by ktong
 
Solution by ktong
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==Solution 5==
 
==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. ∎ --anna0kear
 
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. ∎ --anna0kear
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==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>. --lepetitmoulin
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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>. --OGBooger
  
 
==See Also==
 
==See Also==
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{{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}}
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[[Category:Introductory Geometry Problems]]

Revision as of 01:50, 14 January 2020

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 $\sqrt{40}x$. Then triangle $ADE$ has an area of 16. So the area 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, $\big(\frac{DE}{BC}\big)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}$, then $4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\big(\frac{4}{\sqrt{40}}\big)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]$ Thus, $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\big(1-\frac{2}{5}\big)=\frac{3}{5}\cdot 40=\boxed{24}.$

[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 by ktong

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. ∎ --anna0kear

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 $ABC-16 =\boxed{24}$. --lepetitmoulin

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 $40-16=\boxed{24}$. --OGBooger

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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