document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=605862>Question 605862</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If i wanted to frame a 5\" by 7\" picture with a 1\" frame, how much material would i need for my frame? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #381836 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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=381836\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The area of the picture (without the frame) is 5*7 = 35 square inches\r<BR>\n" );
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document.write( "Now add 1 inch to each side (this will mean you're effectively adding 1+1 = 2 inches to each dimension): 5+2 = 7 and 7+2 = 9\r<BR>\n" );
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document.write( "So the overall dimensions of the picture and the frame combined are 7\" by 9\"\r<BR>\n" );
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document.write( "The area of this is 7*9 = 63 square inches.\r<BR>\n" );
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document.write( "The area of the frame alone is then the difference between the area of the picture+frame and the picture alone (a drawing may help see this)\r<BR>\n" );
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document.write( "So the area of the frame is then: \r<BR>\n" );
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document.write( "(Area of picture+frame) - (Area of picture only) = 63 - 35 = 28\r<BR>\n" );
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document.write( "So the area of the frame only is 28 square inches.\r<BR>\n" );
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document.write( "So if you have a thin strip of material that is 1 inch in width and 28 inches in length, then this would be a starting point for the frame. </SPAN>\n" );
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