Difference between revisions of "2014 AMC 10B Problems/Problem 5"

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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?
 
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>
 
<asy>
 
fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray);
 
fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray);
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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)}}\ 32\qquad\textbf{(E)}\ 34 </math>
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<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]]
 
[[Category: Introductory Geometry Problems]]
  
 
==Solution==
 
==Solution==
  
We note that the total length must be the same as the total height because the window a square. 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 4 widths and 5 bars of length 2. This is equal to two heights of the small rectangles as well as 3 bars of 2. 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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We note that the total length must be the same as the total height because the window is a square. 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==
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https://youtu.be/q3aWZI-AkHs
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~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 16:37, 13 January 2021

Problem

Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5 : 2$, and the borders around and between the panes are $2$ 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]

$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$

Solution

We note that the total length must be the same as the total height because the window is a square. Calling the width of each small rectangle $2x$, and the height $5x$, we can see that the length is composed of $4$ widths and $5$ bars of length $2$. This is equal to two heights of the small rectangles as well as $3$ bars of $2$. Thus, $4(2x) + 5(2) = 2(5x) + 3(2)$. We quickly find that $x = 2$. The total side length is $4(4) + 5(2) = 2(10) + 3(2) = 26$, or $\boxed{\textbf{(A)}}$.

Video Solution

https://youtu.be/q3aWZI-AkHs

~savannahsolver

See Also

2014 AMC 10B (ProblemsAnswer KeyResources)
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

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