Difference between revisions of "2018 AMC 10A Problems/Problem 9"
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==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== | ||
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==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 
, 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)
Contents
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 
, so to find the area of such trapezoid 
, we just take 
, like so. ∎ --anna0kear
Solution 2
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 3
The area of is 16 times the area of the small triangle, as they are similar and their side ratio is 
. Therefore the area of the trapezoid is .
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
