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

## 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

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