document.write( "Question 175285: The original dimensions of a rectangle were 15cm by 20cm. When both dimensions were decreased by the same amount, the area of the new rectangle is half the area of the original rectangle. Find the dimensions of the new rectangle. \n" ); document.write( "
Algebra.Com's Answer #130374 by nerdybill(7384)\"\" \"About 
You can put this solution on YOUR website!
Let x = amount decreased
\n" ); document.write( ".
\n" ); document.write( "\"area of new\" = 1/2(\"area of original\")
\n" ); document.write( ".
\n" ); document.write( "(15-x)(20-x) = 1/2(15)(20)
\n" ); document.write( "Applying FOIL to the left:
\n" ); document.write( "300-15x-20x+x^2 = (15)(10)
\n" ); document.write( "x^2-35x+300 = 150
\n" ); document.write( "x^2-35x+150 = 0
\n" ); document.write( "Factoring:
\n" ); document.write( "(x-30)(x-5) = 0
\n" ); document.write( ".
\n" ); document.write( "Two possible solutions:
\n" ); document.write( "x = {5, 30}
\n" ); document.write( ".
\n" ); document.write( "We can toss out the 30 because we'll get negative measurements.
\n" ); document.write( "Therefore,
\n" ); document.write( "x = 5 cm
\n" ); document.write( ".
\n" ); document.write( "Dimensions:
\n" ); document.write( "(15-5)cm by (20-5)cm
\n" ); document.write( "or
\n" ); document.write( "10cm by 15cm \n" ); document.write( "
\n" );