Difference between revisions of "2016 AMC 8 Problems/Problem 22"
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fill((0,0)--(1,4)--(1.5,3)--cycle, black); | fill((0,0)--(1,4)--(1.5,3)--cycle, black); | ||
fill((3,0)--(2,4)--(1.5,3)--cycle, black); | fill((3,0)--(2,4)--(1.5,3)--cycle, black); | ||
| − | label("A",(3.05,4.2)); | + | label("$A$",(3.05,4.2)); |
| − | label("B",(2,4.2)); | + | label("$B$",(2,4.2)); |
| − | label("C",(1,4.2)); | + | label("$C$",(1,4.2)); |
| − | label("D",(0,4.2)); | + | label("$D$",(0,4.2)); |
| − | label("E", (0,-0.2)); | + | label("$E$", (0,-0.2)); |
| − | label("F", (3,-0.2)); | + | label("$F$", (3,-0.2)); |
</asy> | </asy> | ||
Revision as of 11:39, 23 November 2016
Rectangle 
below is a 
rectangle with 
. What is the area of the "bat wings" (shaded area)?
![[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); [/asy]](http://latex.artofproblemsolving.com/7/f/d/7fd8455ce082d7e31a8c1ead67467b4bcf747e8d.png)
Solution
The area of the trapezoid containing the shaded region and the two isosceles triangles is 
. Next we find the height of each triangle to calculate their area. The triangles are similar, and are in a 
ratio, so the height of the bigger one is 3, while the height of the smaller one is 1. Thus, their areas are 
and 
. Subtracting these areas from the trapezoid, we get 
. Therefore, the answer is 
.
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
