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

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==Solution 7==
 
==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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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>. 
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-ConcaveTriangle
  
 
==Video Solution==
 
==Video Solution==

Revision as of 01:08, 22 January 2021

The following problem is from both the 2018 AMC 12A #8 and 2018 AMC 10A #9, 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 $\sqrt{40}x$. 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, $\left(\frac{DE}{BC}\right)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}$, 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

Video Solution

https://youtu.be/ZiZVIMmo260

https://youtu.be/tq4rWskQDh8

~savannahsolver

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

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