Difference between revisions of "1995 AJHSME Problems/Problem 6"
Talkinaway (talk | contribs) (Created page with "==Problem== Figures <math>I</math>, <math>II</math>, and <math>III</math> are squares. The perimeter of <math>I</math> is <math>12</math> and the perimeter of <math>II</math> i...") |
|||
| Line 29: | Line 29: | ||
* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
| + | {{MAA Notice}} | ||
Revision as of 23:15, 4 July 2013
Problem
Figures 
, 
, and 
are squares. The perimeter of is 
and the perimeter of is 
. The perimeter of is
![[asy] draw((0,0)--(15,0)--(15,6)--(12,6)--(12,9)--(0,9)--cycle); draw((9,0)--(9,9)); draw((9,6)--(12,6)); label("$III$",(4.5,4),N); label("$II$",(12,2.5),N); label("$I$",(10.5,6.75),N); [/asy]](http://latex.artofproblemsolving.com/5/8/c/58cc78e77fd3ec27c31db8cb87221ab494e6797e.png)
Solution
Since the perimeter of , each side is 
.
Since the perimeter of is , each side is 
.
The side of is equal to the sum of the sides of and . Therefore, the side of is 
.
Since is also a square, it has an perimeter of 
, and the answer is 
.
See Also
| 1995 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
