Difference between revisions of "2014 AMC 10B Problems/Problem 5"
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==Problem== | ==Problem== | ||
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| + | 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? | ||
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| + | <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> | ||
| + | |||
| + | <math> \textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34 </math> | ||
| + | [[Category: Introductory Geometry Problems]] | ||
==Solution== | ==Solution== | ||
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| + | We note that the total length must be the same as the total height, as it is given in the problem. Calling the width of each small rectangle <math>2x</math>, and the height <math>5x</math>, we can see that the length is composed of <math>4</math> widths and <math>5</math> bars of length <math>2</math>. This is equal to two heights of the small rectangles as well as <math>3</math> bars of <math>2</math>. Thus, <math>4(2x) + 5(2) = 2(5x) + 3(2)</math>. We quickly find that <math>x = 2</math>. The total side length is <math>4(4) + 5(2) = 2(10) + 3(2) = 26</math>, or <math>\boxed{\textbf{(A)}}</math>. | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/jLLTtK0maQQ | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/q3aWZI-AkHs | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=B|num-b=4|num-a=6}} | {{AMC10 box|year=2014|ab=B|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 14:11, 20 March 2024
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
We note that the total length must be the same as the total height, as it is given in the problem. Calling the width of each small rectangle 
, and the height 
, we can see that the length is composed of 
widths and 
bars of length . This is equal to two heights of the small rectangles as well as 
bars of . Thus, 
. We quickly find that 
. The total side length is 
, or 
.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
See Also
| 2014 AMC 10B (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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
