2016 AMC 10B Problems/Problem 11
Contents
Problem
Carl decided to in his rectangular garden. He bought 
fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly 
yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?
Solution
If the dimensions are 
, then one side will have 
posts (including corners) and the other 
.
The total number of posts is 
.
![[asy]size(7cm);fill((0,0)--(5,0)--(5,7)--(0,7)--cycle,lightgreen); for(int i=0;i<5;++i)dot((i,0),red);for(int i=0;i<7;++i)dot((5,i),blue);dot((5,7)); draw(arc((0,0),.5,-90,-270)--arc((4,0),.5,90,-90)--cycle,gray+dotted); draw(arc((5,0),.5,-180,0)--arc((5,6),.5,0,180)--cycle,gray+dotted); draw((0,-1)--(5,-1),Arrows);draw((6,0)--(6,7),Arrows); label("$a$",(0,-1)--(5,-1),S);label("$b$",(6,0)--(6,7),E); label("$a$",(1,1));label("$b$",(4,5)); [/asy]](http://latex.artofproblemsolving.com/c/6/4/c642ad2571631bd6b03e98e56e8c7d16aeb92a72.png)
Solve the system 
to get 
. Then the area is 
which is 
.
Solution 2
To do this problem, we have to draw a rectangle. We also have to keep track of the fence posts. Putting a post on each corner leaves us with only 
posts. Now there are twice as many posts on the longer side then the shorter side. From this we can see that we can put 
posts on the longer side and posts on the shorter side. On the shorter side, we have 
spaces between the posts. On the longer side, we have 7 spaces between the 8 fence posts. There are yards between each post. Therefore, the answer is 
.
See Also
| 2016 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 | ||
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