document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=830795>Question 830795</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a lot with 45ft of street frontage , containing 1280 sq. yds. , would have a depth of? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #500893 by </B><A HREF=/tutors/aboutme.mpl?userid=JulietG><B>JulietG(1812)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=JulietG><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/cgi-bin/show-face.mpl?user=JulietG><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=500893\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You have to make the assumption first that the lot is a rectangle.<BR>\n" );
document.write( "In that case, you'll use the formula for Area.  Area = Width * Length.<BR>\n" );
document.write( "We know the area, and we know the width.<BR>\n" );
document.write( "Where most people will get this wrong is not catching that the measurements are different.  First, we need to convert ft to yards. (3 feet in a yard.)  The *new* measurement of street frontage is 15 (45/3)<BR>\n" );
document.write( "Now we can plug in the values and find the other measurement.<BR>\n" );
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document.write( "Area (1280) = Length (x) * Width (15)<BR>\n" );
document.write( "1280 = 15x<BR>\n" );
document.write( "Divide each side by 15 <BR>\n" );
document.write( "length = 85 1/3 yards<BR>\n" );
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document.write( "Since it's \"1/3\", I'm suspecting your teacher wants the answer in feet.  So we'll convert back by multiplying 85 1/3 yards by 3 feet in a yard.<BR>\n" );
document.write( "3 * 85.33 = 256 feet<BR>\n" );
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