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

Latest revision as of 13:12, 5 July 2013

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 $\boxed{C}$

1997 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == See also ==
 {{AHSME box|year=1997|num-b=4|num-a=6}}
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Difference between revisions of "1997 AHSME Problems/Problem 5"

Latest revision as of 13:12, 5 July 2013

Problem

Solution

See also