document.write( "Question 1049886: A rectangle’s length is about 19 inches greater than its width. If the perimeter is 50 inches, what is the area of the rectangle? \n" ); document.write( "
Algebra.Com's Answer #665447 by Cromlix(4381)![]() ![]() You can put this solution on YOUR website! Hi there, \n" ); document.write( "Make width = x \n" ); document.write( "Length = 19 + x \n" ); document.write( "Perimeter = 2 x Width + 2 x Length \n" ); document.write( "50 = 2(x) + 2(19 + x) \n" ); document.write( "50 = 2x + 38 + 2x \n" ); document.write( "4x = 12 \n" ); document.write( "x = 3. \n" ); document.write( "Width = 3 inches \n" ); document.write( "Length = 22 inches. \n" ); document.write( "Area = Width x Length \n" ); document.write( "Area = 3 x 22 \n" ); document.write( "Width = 66 inches^2 \n" ); document.write( "Hope this helps :-) \n" ); document.write( " |