document.write( "Question 287100: the length of a rectangle is 5cm. greater than twice its width. if the perimeter is 52 cm., what is the dimensions of the rectangle? \n" ); document.write( "
Algebra.Com's Answer #208228 by sharon1948(6)\"\" \"About 
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Problem #287100
\n" ); document.write( "1. the length of a rectangle is 5cm. greater than twice its width. if the perimeter is 52 cm., what is the dimensions of the rectangle
\n" ); document.write( "2. Let w=width
\n" ); document.write( " 2w+5= length
\n" ); document.write( "I am a visual person so I draw the rectangle out and label it.
\n" ); document.write( "Next write the formula of the Area of a rectangle.
\n" ); document.write( " A=lw
\n" ); document.write( "With the information you have,replace it into the formula:
\n" ); document.write( " 52=(2w+5)w (I rewrite the formula for my understanding)
\n" ); document.write( " w(2w+5)=52
\n" ); document.write( "Next we need to factor the equation and set it to equal 0:
\n" ); document.write( " 2w^2 +5w=52
\n" ); document.write( "subtract 52 from the right and rewrite the equation:
\n" ); document.write( " 2w^2+5w-52=0\r
\n" ); document.write( "\n" ); document.write( "Solve by grouping
\n" ); document.write( "2w^2+13w ] [ -8w -52 =0
\n" ); document.write( "w(2w+13) -4(2w+13) =0
\n" ); document.write( "(w-4) (2w+13)\r
\n" ); document.write( "\n" ); document.write( "Width = 4\r
\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "52= ((2)4+5)4
\n" ); document.write( "52= (13)4
\n" ); document.write( "52=52\r
\n" ); document.write( "\n" ); document.write( "Length = 13\r
\n" ); document.write( "\n" ); document.write( "The dimensions of the rectangle are width equals 4cm and the length equals 13cm.
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