document.write( "Question 23801: The combined area of a square and a rectangle is 64 square centimeters. The width of the rectangle is 2 cm more than the length of a side of the square and the length of the rectangle is 2 cm more than its width. Find the dimensions of the square and the rectangle. \r
\n" ); document.write( "\n" ); document.write( "thanks a lot.. i'll really appreciate it if someone answered this one..
\n" ); document.write( "thanks again..
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Algebra.Com's Answer #12567 by venugopalramana(3286)\"\" \"About 
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The combined area of a square and a rectangle is 64 square centimeters. The width of the rectangle is 2 cm more than the length of a side of the square and the length of the rectangle is 2 cm more than its width. Find the dimensions of the square and the rectangle.
\n" ); document.write( "thanks a lot.. i'll really appreciate it if someone answered this one..
\n" ); document.write( "thanks again..\r
\n" ); document.write( "\n" ); document.write( "LET SIDE OF SQUARE =X CM.
\n" ); document.write( "2 CM MORE = X+2=WIDTH OF RECTANGLE
\n" ); document.write( "FURTHER 2 CM MORE =X+2+2=X+4=LENGTH OF RECTANGLE
\n" ); document.write( "AREA OF SQUARE = SIDE *SIDE =X*X=X^2
\n" ); document.write( "AREA OF RECTANGLE = LENGTH * WIDTH = (X+2)(X+4)=X^2+2X+4X+8=X^2+6X+8
\n" ); document.write( "TOTAL ARTEA OF SQUARE AND RECTANGLE = X^2+X^2+6X+8=2X^2+6X+8
\n" ); document.write( "=2(X^2+3X+4)=64...GIVEN
\n" ); document.write( "X^2+3X+4=64/2=32
\n" ); document.write( "X^2+3X+4-32=0
\n" ); document.write( "X^2+3X-28=0
\n" ); document.write( "X^2+7X-4X-28=0
\n" ); document.write( "X(X+7)-4(X+7)=0
\n" ); document.write( "(X+7)(X-4)=0
\n" ); document.write( "X-4=0...OR...X=4
\n" ); document.write( "SO SIDE OF SQUARE =4 CM.
\n" ); document.write( "WIDTH OF RECTANGLE =6 CM AND LENGTH OF RECTANGLE = 8 CM.
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