Difference between revisions of "1997 PMWC Problems/Problem I11"

m (Solution)
Line 21: Line 21:
 
<math>2(w+l)+6w=330</math>
 
<math>2(w+l)+6w=330</math>
  
==See also==
+
==See Also==
 
{{PMWC box|year=1997|num-b=I10|num-a=I12}}
 
{{PMWC box|year=1997|num-b=I10|num-a=I12}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 15:04, 15 May 2012

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 $6750 cm^2$. ABCD.gif

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