1997 PMWC Problems/Problem I11

Problem

A rectangle $ABCD$ is made up of five small congruent rectangles as shown in the given figure. Find the perimeter, in cm, of $ABCD$ if its area is $6750\text{cm}^2$ .

$[asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(3.45cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -19.75, xmax = 39.09, ymin = -10.43, ymax = 20.84; /* image dimensions */ /* draw figures */ draw((0,3.45)--(0,0)); draw((0,0)--(4.29,0)); draw((0,3.45)--(4.29,3.45)); draw((4.29,3.45)--(4.29,0)); draw((0,1.32)--(4.29,1.32)); draw((2.14,0)--(2.14,1.32)); draw((1.43,1.32)--(1.43,3.45)); draw((2.86,1.32)--(2.86,3.45)); /* dots and labels */ label("$B$", (-0.2,0), SW * labelscalefactor); label("$A$", (-0.2,3.45), NW * labelscalefactor); label("$C$", (4.29,0), SE * labelscalefactor); label("$D$", (4.29,3.45), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ //Credit to dasobson for the diagram[/asy]$

Solution

Let $l$ and $w$ be the length, and width, respectively, of one of the small rectangles.

$3w=2l$

$l=\dfrac{3}{2}w$

$6750= 5lw = \dfrac{15}{2}w^2$

$w=30$

$l=45$

The perimeter of the big rectangle is

$2(w+l)+6w=330$

1997 PMWC (Problems)
Preceded by Problem I10	Followed by Problem I12
I: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 T: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10

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1997 PMWC Problems/Problem I11

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