document.write( "Question 149785: 60% of the area of Rectangle A is the same as 80% of the area of Rectangle B. Rectangle B is 25cm2 smaller than Rectangle A. Given that they both have a length of 25 cm, find the difference in their perimeter.\r
\n" ); document.write( "\n" ); document.write( "Rmks: Pls try not to use algebra.
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Algebra.Com's Answer #109898 by kmcruz09(38)\"\" \"About 
You can put this solution on YOUR website!
I don't think you can solve this one without using algebra. Anways, here's the solution.\r
\n" ); document.write( "\n" ); document.write( "60% of the area of Rectangle A is the same as 80% of the area of Rectangle B. Rectangle B is 25cm2 smaller than Rectangle A. Given that they both have a length of 25 cm, find the difference in their perimeter.\r
\n" ); document.write( "\n" ); document.write( "let x = width of rectangle A
\n" ); document.write( "let y = width of rectangle B\r
\n" ); document.write( "\n" ); document.write( "then Area of rectangle A = 25x
\n" ); document.write( "and Area of rectangle B = 25y\r
\n" ); document.write( "\n" ); document.write( "\"0.6%2825x%29+=+0.8%2825y%29\"
\n" ); document.write( "\"150x+=+200y\"
\n" ); document.write( "\"x+=+4y%2F3\"
\n" ); document.write( "\"25x+=+25y+%2B+25\"
\n" ); document.write( "\"x+=+y%2B1\"
\n" ); document.write( "\"y%2B1=4y%2F3\"
\n" ); document.write( "\"3y%2B3=4y\"
\n" ); document.write( "\"y=3\"
\n" ); document.write( "\"x=4\"\r
\n" ); document.write( "\n" ); document.write( "PERIMETERa = \"2%28x%2B25%29\"
\n" ); document.write( "= \"58\"
\n" ); document.write( "PERIMETERb = \"2%28y%2B25%29\"
\n" ); document.write( "= \"56\"\r
\n" ); document.write( "\n" ); document.write( "PERIMETERa - PERIMETERb = 2cm
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