document.write( "Question 261632: The perimeter of a rectangle is 100 meters. Its length is 7 meters less than twice its width. Find its area. \n" ); document.write( "
Algebra.Com's Answer #192766 by MathTherapy(10552)![]() ![]() You can put this solution on YOUR website! The perimeter of a rectangle is 100 meters. Its length is 7 meters less than twice its width. Find its area.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let width be W\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Then length = 2W - 7, since its length is 7 meters less than twice its width\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Since its perimeter is 100 meters, and since perimeter = 2W + 2L, we will then have:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "2W + 2(2W - 7) = 100\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "2W + 4W - 14 = 100\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "6W = 114\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "W, or width = \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Since its width, or W = 19, then L, or its length = 31 (2*19 - 7)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Since its width = 19, and length = 31, then its area = 19 * 31 = |