Difference between revisions of "2014 AMC 12B Problems/Problem 5"
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fill((0,24)--(26,24)--(26,26)--(0,26)--cycle,gray); | fill((0,24)--(26,24)--(26,26)--(0,26)--cycle,gray); | ||
</asy> | </asy> | ||
| − | <math> \textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D) | + | <math> \textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34 </math> |
==Solution== | ==Solution== | ||
Revision as of 11:06, 5 April 2015
Problem
Doug constructs a square window using 
equal-size panes of glass, as shown. The ratio of the height to width for each pane is 
, and the borders around and between the panes are 
inches wide. In inches, what is the side length of the square window?
![[asy] fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray); fill((6,0)--(8,0)--(8,26)--(6,26)--cycle,gray); fill((12,0)--(14,0)--(14,26)--(12,26)--cycle,gray); fill((18,0)--(20,0)--(20,26)--(18,26)--cycle,gray); fill((24,0)--(26,0)--(26,26)--(24,26)--cycle,gray); fill((0,0)--(26,0)--(26,2)--(0,2)--cycle,gray); fill((0,12)--(26,12)--(26,14)--(0,14)--cycle,gray); fill((0,24)--(26,24)--(26,26)--(0,26)--cycle,gray); [/asy]](http://latex.artofproblemsolving.com/4/a/4/4a4bd527ea5f7e41a6d5541d0fc5b06615b22b69.png)
Solution
Let the height of the panes equal 
, and let the width of the panes equal 
. Now notice that the total width of the borders equals 
, and the total height of the borders is 
. We have
![\[10 + 4(2x) = 6 + 2(5x)\]](http://latex.artofproblemsolving.com/c/3/1/c31292295a131eb993a821ed21726cc9ea13ebc4.png)
![\[x = 2\]](http://latex.artofproblemsolving.com/9/6/9/969f2a62fcd94063dcee2eff234579ad20ad8c10.png)
Now, the total side length of the window equals
![\[10+ 4(2x) = 10 + 16 = \boxed{\textbf{(A)}\ 26}\]](http://latex.artofproblemsolving.com/5/d/d/5dd5e50ec1c8b28f264628aeddca055d921744a5.png)
See also
| 2014 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 4 |
Followed by Problem 6 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
