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Difference between revisions of "1997 PMWC Problems/Problem I11"
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==Problem==
 
==Problem==
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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 <math>6750 cm^2</math>.  
+
A rectangle <math>ABCD</math> is made up of five small congruent rectangles as shown in the given figure. Find the perimeter, in cm, of <math>ABCD</math> if its area is <math>6750\text{ cm}^2</math>.  
−
[[Image:ABCD.gif]]
+
 
 +
<asy>
 +
import cse5;
 +
import olympiad;
 +
size(4cm);
 +
pathpen=black;
 +
pair A=(0,0),B=(0,-2.5),C=(3,-2.5),D=(3,0);
 +
D(MP("A",A,W)--MP("B",B,W)--MP("C",C,E)--MP("D",D,E)--cycle);
 +
D((0,-1.5)--(3,-1.5));
 +
D((1,0)--foot((1,0),(0,-1.5),(3,-1.5)));
 +
D((2,0)--foot((2,0),(0,-1.5),(3,-1.5)));
 +
D((1.5,-1.5)--(1.5,-2.5));</asy>
  
 
==Solution==
 
==Solution==
  
−
Let l and w be the length, and width, respectively, of one of the little rectangles.
+
Let <math>l</math> and <math>w</math> be the length, and width, respectively, of one of the small rectangles.
  
 
<math>3w=2l</math>
 
<math>3w=2l</math>
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<math>l=\dfrac{3}{2}w</math>
 
<math>l=\dfrac{3}{2}w</math>
  
−
<math>6750=\dfrac{15}{2}w^2</math>
+
<math>6750= 5lw = \dfrac{15}{2}w^2</math>
  
 
<math>w=30</math>
 
<math>w=30</math>
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<math>2(w+l)+6w=330</math>
 
<math>2(w+l)+6w=330</math>
 +
 +
==See Also==
 +
{{PMWC box|year=1997|num-b=I10|num-a=I12}}
 +
 +
[[Category:Introductory Geometry Problems]]

Latest revision as of 12:44, 13 August 2014

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] import cse5; import olympiad; size(4cm); pathpen=black; pair A=(0,0),B=(0,-2.5),C=(3,-2.5),D=(3,0); D(MP("A",A,W)--MP("B",B,W)--MP("C",C,E)--MP("D",D,E)--cycle); D((0,-1.5)--(3,-1.5)); D((1,0)--foot((1,0),(0,-1.5),(3,-1.5))); D((2,0)--foot((2,0),(0,-1.5),(3,-1.5))); D((1.5,-1.5)--(1.5,-2.5));[/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

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

See Also

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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