document.write( "Question 1067641: The length of a rectangle is
\n" );
document.write( "44
\n" );
document.write( "inches
\n" );
document.write( "greater
\n" );
document.write( "than twice the width. If the diagonal is 2 inches more than the length, find the dimensions
\n" );
document.write( "Length 2w+44
\n" );
document.write( "Diagonal 2w+46 \n" );
document.write( "
Algebra.Com's Answer #682752 by LinnW(1048)![]() ![]() You can put this solution on YOUR website! The length and with are the legs of a triangle, \n" ); document.write( "and the diagonal is the hypotenuse. \n" ); document.write( "So \n" ); document.write( "subtract \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "factoring, \n" ); document.write( " \n" ); document.write( "So w = -10 or w = 18 \n" ); document.write( "Since we need a positive width, w = 18 \n" ); document.write( "The length is 2w+44 = 2(18) + 44 = 36 + 44 = 80 \n" ); document.write( "The diagonal is 2w + 46 = 2(18) + 46 = 36 + 46 = 82 \n" ); document.write( "Notice that 18^2 + 80^2 = 82^2 \n" ); document.write( " |