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Difference between revisions of "1997 AHSME Problems/Problem 5"

(Created page with "== See also == {{AHSME box|year=1997|num-b=4|num-a=6}}")
 
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==Problem==
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A rectangle with perimeter <math>176</math> is divided into five congruent rectangles as shown in the diagram.  What is the perimeter of one of the five congruent rectangles?
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<asy>
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defaultpen(linewidth(.8pt));
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draw(origin--(0,3)--(4,3)--(4,0)--cycle);
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draw((0,1)--(4,1));
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draw((2,0)--midpoint((0,1)--(4,1)));
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real r = 4/3;
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draw((r,3)--foot((r,3),(0,1),(4,1)));
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draw((2r,3)--foot((2r,3),(0,1),(4,1)));</asy>
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<math> \mathrm{(A)\ } 35.2 \qquad \mathrm{(B) \ }76 \qquad \mathrm{(C) \  } 80 \qquad \mathrm{(D) \  } 84 \qquad \mathrm{(E) \  }86  </math>
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==Solution==
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Let <math>l</math> represent the length of one of the smaller rectangles, and let <math>w</math> represent the width of one of the smaller rectangles, with <math>w < l</math>.
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From the large rectangle, we see that the top has length <math>3w</math>, the right has length <math>l + w</math>, the bottom has length <math>2l</math>, and the left has length <math>l + 2</math>.
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Since the perimeter of the large rectangle is <math>176</math>, we know that <math>172 = 3w + l + w + 2l + l + w</math>, or <math>172 = 5w + 4l</math>
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From the top and bottom of the large rectangle, we know that <math>3w = 2l</math>, or <math>l = 1.5w</math>.
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Plugging that into the first equation, we get <math>176 = 5w + 4(1.5)w</math>
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<math>176 = 11w</math>
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<math>w = 16</math>
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<math>l = 1.5w = 24</math>
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<math>P = 2l + 2w = 2(16 + 24) = 80</math>, and the answer is <math>\boxed{C}</math>
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== See also ==
 
== See also ==
 
{{AHSME box|year=1997|num-b=4|num-a=6}}
 
{{AHSME box|year=1997|num-b=4|num-a=6}}

Revision as of 16:49, 8 August 2011

Problem

A rectangle with perimeter $176$ is divided into five congruent rectangles as shown in the diagram. What is the perimeter of one of the five congruent rectangles? [asy] defaultpen(linewidth(.8pt)); draw(origin--(0,3)--(4,3)--(4,0)--cycle); draw((0,1)--(4,1)); draw((2,0)--midpoint((0,1)--(4,1))); real r = 4/3; draw((r,3)--foot((r,3),(0,1),(4,1))); draw((2r,3)--foot((2r,3),(0,1),(4,1)));[/asy]

$\mathrm{(A)\ } 35.2 \qquad \mathrm{(B) \ }76 \qquad \mathrm{(C) \ } 80 \qquad \mathrm{(D) \ } 84 \qquad \mathrm{(E) \ }86$

Solution

Let $l$ represent the length of one of the smaller rectangles, and let $w$ represent the width of one of the smaller rectangles, with $w < l$.

From the large rectangle, we see that the top has length $3w$, the right has length $l + w$, the bottom has length $2l$, and the left has length $l + 2$.

Since the perimeter of the large rectangle is $176$, we know that $172 = 3w + l + w + 2l + l + w$, or $172 = 5w + 4l$

From the top and bottom of the large rectangle, we know that $3w = 2l$, or $l = 1.5w$.

Plugging that into the first equation, we get $176 = 5w + 4(1.5)w$

$176 = 11w$

$w = 16$

$l = 1.5w = 24$

$P = 2l + 2w = 2(16 + 24) = 80$, and the answer is $\boxed{C}$

See also

1997 AHSME (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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions
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