document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1032429>Question 1032429</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The perimeter of a square and a rectangle are equal. One side of the rectangle is 11 cm and the area of the square is 4 cm^2 more than the area of the rectangle.  Find the side of square </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #647113 by </B><A HREF=/tutors/aboutme.mpl?userid=jorel555><B>jorel555(1290)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jorel555><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=647113\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x=one side of the square; y=one side of the rectangle:<BR>\n" );
document.write( "The perimeters are equal, so 4x=2y+22, or y=2x-11. The areas are given by x^2, and 11(2x-11). Since the square is 4 cm^2 larger than the rectangle, we get:\r<BR>\n" );
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document.write( "x^2-4=11(2x-11)<BR>\n" );
document.write( "x^2-4=22x-121<BR>\n" );
document.write( "x^2-22x+117=0<BR>\n" );
document.write( "(x-13)(x-9)=0<BR>\n" );
document.write( "x=13, or 9\r<BR>\n" );
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document.write( "Plugging 13 into our original equation, we get a square 13X13 or 169cm^2, and a rectangle (26-11)X 11, or 165cm^2, which satisfies our requirements!!!!!!!\r<BR>\n" );
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