2018 AMC 10A Problems/Problem 9
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 
.
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/b/d/3/bd3f50d2af0033703b18ceb0ed872071e292a382.png)
Solution by ktong
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
