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

Latest revision as of 14:11, 20 March 2024

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, as it is given in the problem. 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 (CREATIVE THINKING)

https://youtu.be/jLLTtK0maQQ

~Education, the Study of Everything

Video Solution

https://youtu.be/q3aWZI-AkHs

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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.

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 ==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 <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>.
+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>.
 ==Video Solution (CREATIVE THINKING)==

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