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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>.
 
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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==Solution 2==
 
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Dissect the diagram:
 

Revision as of 12:30, 9 February 2018

All of the triangles in the diagram below are similar to iscoceles 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 $DBCE=40-x=\boxed{24}$.

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