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Difference between revisions of "2018 AMC 10A Problems/Problem 9"
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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>
 
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>
 +
 +
<asy>
 +
unitsize(5);
 +
dot((0,0));
 +
dot((60,0));
 +
dot((50,10));
 +
dot((10,10));
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dot((30,30));
 +
draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0));
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draw((10,10)--(50,10));
 +
label("$B$",(0,0),SW);
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label("$C$",(60,0),SE);
 +
label("$E$",(50,10),E);
 +
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));
 +
draw((15,15)--(45,15));
 +
</asy>
 +
 +
  
 
Solution by ktong
 
Solution by ktong

Revision as of 12:30, 17 February 2018

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

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

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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