Difference between revisions of "2014 AMC 10B Problems/Problem 5"
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==Problem== | ==Problem== | ||
− | Doug constructs a square window using 8 equal-size panes of glass, as shown. The ratio of the height to width | + | 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? |
− | <math> \textbf {(A) } 26 \qquad \textbf {(B) } 28 \qquad \textbf {(C) } 30 \qquad \textbf {(D) } 32 \qquad \textbf {(E) } 34</math> | + | <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== | ||
− | 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{(A)}</math>. | + | 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>. |
+ | |||
+ | ==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 15:37, 13 January 2021
Contents
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?
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 , 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
~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 |
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