document.write( "Question 383851: Originally, the dimensions of a rectangle were 18cm by 24cm. When both dimensions were decreased by the same amount, the area of the rectangle decreased by 117 sq. cm. Find the dimensions of the new rectangle. \n" ); document.write( "
Algebra.Com's Answer #271746 by mananth(16946)\"\" \"About 
You can put this solution on YOUR website!
rectangle dimensions
\n" ); document.write( "18 by 24 cm
\n" ); document.write( "Area = 18*24=432 cm^2
\n" ); document.write( "..
\n" ); document.write( "let them be decreased by x
\n" ); document.write( "..
\n" ); document.write( "new dimensions =
\n" ); document.write( "18-x
\n" ); document.write( "24-x
\n" ); document.write( "..
\n" ); document.write( "area = (18-x)(24-x)
\n" ); document.write( "area = 432-18x-24x+x^2
\n" ); document.write( "area=432-42x+x^2
\n" ); document.write( "..
\n" ); document.write( "Original area - reduced rectangle area = 117
\n" ); document.write( "432-(432-42x+x^2)=117
\n" ); document.write( "432-432+42x-x^2=117
\n" ); document.write( "42x-x^2=117
\n" ); document.write( "x^2-42x+117=0
\n" ); document.write( "x^2-39x-3x+117=0
\n" ); document.write( "x(x-39)-3(x-39)=0
\n" ); document.write( "(x-3)(x-39)=0
\n" ); document.write( "So x= 3. x= 39 is not practical.
\n" ); document.write( "dimensions of new rectangle = 15cm by 21cm
\n" ); document.write( "..
\n" ); document.write( "m.ananth@hotmail.ca\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );