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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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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 7 smallest triangles has area 1, and <math>\triangle ABC</math> has area 40. What is the area of trapezoid <math>DBCE</math>?
  
 
<asy>
 
<asy>

Revision as of 17:06, 10 February 2018

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]

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

Solutions

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 $DBCE=40-x=\boxed{24}$.

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, $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\big(1-\frac{2}{5}\big)=\frac{3}{5}\cdot 40=\boxed{24}$

Solution by ktong

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

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