2018 AMC 10A Problems/Problem 9
Contents
Problem
All of the triangles in the diagram below are similar to isosceles 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)
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/3/e/a/3ea700ecd921b39f8a6b989743cedaeb2f8f3291.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 
.
Solution 3
Notice ![$\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]$](http://latex.artofproblemsolving.com/a/0/b/a0b30da36d6f449cf51fcabb18d09c0b8eb73237.png)
.
Let the base of the small triangles of area 1 be , then the base length of 
. Notice, 
, then ![$4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\big(\frac{4}{\sqrt{40}}\big)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]$](http://latex.artofproblemsolving.com/d/a/2/da2895569a61277de2a288fa73163b955ce61646.png)
Thus, ![$\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\big(1-\frac{2}{5}\big)=\frac{3}{5}\cdot 40=\boxed{24}.$](http://latex.artofproblemsolving.com/3/2/e/32e9b39a3dcec005491d32ecc88921a6bf3dc9b4.png)
Solution 4
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 
.
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 
, so to find the area of such trapezoid 
, we just take , like so.
Solution 6
The combined area of the small triangles is , and from the fact that each small triangle has an area of 
, we can deduce that the larger triangle above has an area of 
(as the sides of the triangles are in a proportion of 
, so will their areas have a proportion that is the square of the proportion of their sides, or 
). Thus, the combined area of the top triangle and the trapezoid immediately below is 
. The area of trapezoid is thus the area of triangle 
.
Solution 7
You can assume for the base of one of the smaller triangles to be 
and the height to be 
, giving an area of 1. The larger triangle above the 7 smaller ones then has base 
and height 
, giving it an area of . Then the area of triangle is 
and .
Video Solution
~savannahsolver
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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