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Difference between revisions of "1997 PMWC Problems/Problem I11"
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<math>\text{Perimeter}=\boxed{330}</math>
 
<math>\text{Perimeter}=\boxed{330}</math>
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[https://aops.com/wiki/index.php/User:Am24 ~AM24]
  
 
==See Also==
 
==See Also==

Latest revision as of 20:23, 5 September 2025

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

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

Solution 2

We can label the short side of the smaller rectangles $a$ and the long side $b$. Additionally, since all the rectangles are congruent we can say that the area of one of the rectangles is $\frac{6750}{5}=1350$. Using our labels we can get the equations for our systems, $ab=1350 \text{and} (a+b)(2b)=6750$.

$2ab+2b^2=6750 \text{ and } 2ab=2700$

Using elimination,

$2b^2=4050$

$b^2=2025 \implies b=45$ which also means $a=30$.

We want to find the area of the big rectangle which is $2(a+b+2b)$

Plugging in our values we get:

$2(75+90)$

$2(165)$

$\text{Perimeter}=\boxed{330}$

~AM24

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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