document.write( "Question 74859: The length of a rectangle is 5 cm more than 2 times its width. If the area of the rectangle is 75 cm2, find the dimensions of the rectangle to the nearest thousandth. \n" ); document.write( "
Algebra.Com's Answer #53783 by psbhowmick(878)![]() ![]() You can put this solution on YOUR website! Let the width = x cm. \n" ); document.write( "Two times width = 2x cm. \n" ); document.write( "5cm more than this value = (2x + 5) cm \n" ); document.write( "So, the length = (2x + 5) cm.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Hence area = Length X Width = \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "But, according to the problem this area is 75 \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So, \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So either (x-5) = 0 or (2x+15) = 0 \n" ); document.write( "i.e. either x = 5 or x = -7.5\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "But width of a rectangle (x) cannot be negative. \n" ); document.write( "So negative answer is discarded. \n" ); document.write( "Hence x = 5.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So the width = 5 cm and length = 2x5 + 5 = 15 cm. \n" ); document.write( " |