Difference between revisions of "2018 AMC 10A Problems/Problem 9"
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<math>\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24 </math> | <math>\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24 </math> | ||
| − | + | ==Solutions== | |
| − | ==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>. | ||
| − | =Solution 2= | + | ===Solution 2=== |
Let the base length of the small triangle be <math>x</math>. Then, there is a triangle <math>ADE</math> encompassing the 7 small triangles and sharing the top angle with a base length of <math>4x</math>. Because the area is proportional to the square of the side, let the base <math>BC</math> be <math>\sqrt{40}x</math>. Then triangle <math>ADE</math> has an area of 16. So the area is <math>40 - 16 = \boxed{24}</math>. | Let the base length of the small triangle be <math>x</math>. Then, there is a triangle <math>ADE</math> encompassing the 7 small triangles and sharing the top angle with a base length of <math>4x</math>. Because the area is proportional to the square of the side, let the base <math>BC</math> be <math>\sqrt{40}x</math>. Then triangle <math>ADE</math> has an area of 16. So the area is <math>40 - 16 = \boxed{24}</math>. | ||
| + | == See Also == | ||
| + | |||
| + | {{AMC10 box|year=2018|ab=A|num-b=7|num-a=9}} | ||
| + | {{MAA Notice}} | ||
Revision as of 19:56, 9 February 2018
All of the triangles in the diagram below are similar to iscoceles triangle 
, in which 
. Each of the 7 smallest triangles has area 1, and 
has area 40. What is the area of trapezoid 
?
![[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]](http://latex.artofproblemsolving.com/d/1/3/d13adfd6d388ea5d30660f97f3b425db198e8093.png)
Solutions
Solution 1
Let 
be the area of 
. Note that is comprised of the 
small isosceles triangles and a triangle similar to with side length ratio 
(so an area ratio of 
). Thus, we have ![\[x=7+\dfrac{9}{16}x\]](http://latex.artofproblemsolving.com/6/f/5/6f545ed79d97b7ce3867600ae61833b88f6e5194.png)
This gives 
, so the area of 
.
Solution 2
Let the base length of the small triangle be . Then, there is a triangle encompassing the 7 small triangles and sharing the top angle with a base length of 
. Because the area is proportional to the square of the side, let the base 
be 
. Then triangle has an area of 16. So the area is 
.
See Also
| 2018 AMC 10A (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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
