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

(added a new solution.)
m
Line 22: Line 22:
  
 
==Solution 1==
 
==Solution 1==
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=24, like so. ∎ --anna0kear
+
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 <math>7+5+3+1=16</math>, so to find the area of such trapezoid <math>BCED</math>, we just take <math>40-16=\boxed{24}</math>, like so. ∎ --anna0kear
  
 
==Solution 2==
 
==Solution 2==
Line 28: Line 28:
  
 
==Solution 3==
 
==Solution 3==
The area of <math>ADE</math> is 16 times the area of the small triangle, as they are similar and their side ratio is <math>4:1</math>. Therefore the area of the trapezoid is <math>40-16= \boxed{24}</math>.
+
The area of <math>ADE</math> is 16 times the area of the small triangle, as they are similar and their side ratio is <math>4:1</math>. Therefore the area of the trapezoid is <math>40-16=\boxed{24}</math>.
  
 
== See Also ==
 
== See Also ==
 
 
{{AMC10 box|year=2018|ab=A|num-b=8|num-a=10}}
 
{{AMC10 box|year=2018|ab=A|num-b=8|num-a=10}}
 
{{AMC12 box|year=2018|ab=A|num-b=7|num-a=9}}
 
{{AMC12 box|year=2018|ab=A|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:47, 8 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

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 2

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 3

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 $40-16=\boxed{24}$.

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS