Difference between revisions of "2014 AMC 12B Problems/Problem 5"
(→Problem) |
m (minor edit) |
||
| Line 1: | Line 1: | ||
| − | ==Problem== | + | == Problem == |
| + | Doug constructs a square window using <math> 8 </math> equal-size panes of glass, as shown. The ratio of the height to width for each pane is <math> 5 : 2 </math>, and the borders around and between the panes are <math> 2 </math> inches wide. In inches, what is the side length of the square window? | ||
| − | |||
<asy> | <asy> | ||
fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray); | fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray); | ||
| Line 12: | Line 12: | ||
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)}\ 32\qquad\textbf{(E)}\ 34 </math> | <math> \textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34 </math> | ||
| − | ==Solution== | + | == Solution == |
| − | |||
Let the height of the panes equal <math>5x</math>, and let the width of the panes equal <math>2x</math>. Now notice that the total width of the borders equals <math>10</math>, and the total height of the borders is <math>6</math>. We have | Let the height of the panes equal <math>5x</math>, and let the width of the panes equal <math>2x</math>. Now notice that the total width of the borders equals <math>10</math>, and the total height of the borders is <math>6</math>. We have | ||
<cmath>10 + 4(2x) = 6 + 2(5x)</cmath> | <cmath>10 + 4(2x) = 6 + 2(5x)</cmath> | ||
| Line 22: | Line 22: | ||
<cmath>10+ 4(2x) = 10 + 16 = \boxed{\textbf{(A)}\ 26}</cmath> | <cmath>10+ 4(2x) = 10 + 16 = \boxed{\textbf{(A)}\ 26}</cmath> | ||
| − | == See | + | == See Also == |
{{AMC12 box|year=2014|ab=B|num-b=4|num-a=6}} | {{AMC12 box|year=2014|ab=B|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 00:10, 19 October 2020
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. 
