document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=116771>Question 116771</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A rectangle having area 120 square meters has sides of length X+3 meters and x+10 meters.  What is the length of a diagonal of the rectangle? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #84901 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><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=84901\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The area of a triangle is equal to its length times its width. For this problem the width<BR>\n" );
document.write( "is (x+3) meters and the length is (x + 10) meters. The product of these two is the area and, therefore,<BR>\n" );
document.write( "is equal to 120 square meters. So we can write:<BR>\n" );
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document.write( "(x + 3)(x + 10) = 120<BR>\n" );
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document.write( "Multiplying out the left side results in:<BR>\n" );
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document.write( "x^2 + 13x + 30 = 120<BR>\n" );
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document.write( "To solve this equation, get it into the standard form of:<BR>\n" );
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document.write( "ax^2 + bx + c = 0<BR>\n" );
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document.write( "by subtracting 120 from both sides to get rid of the 120 on the right side. When you do this<BR>\n" );
document.write( "subtraction to both sides you get:<BR>\n" );
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document.write( "x^2 + 13x - 90 = 0<BR>\n" );
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document.write( "Factor the left side to get:<BR>\n" );
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document.write( "(x + 18)(x - 5) = 0<BR>\n" );
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document.write( "This equation will be true if either of the two factors is equal to zero because a multiplication<BR>\n" );
document.write( "by zero on the left side will make the left side equal the zero on the right side.<BR>\n" );
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document.write( "Set the two factors equal to zero (one at a time) and solve for x. First:<BR>\n" );
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document.write( "x + 18 = 0 ... subtract 18 from both sides and get that <BR>\n" );
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document.write( "x = -18<BR>\n" );
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document.write( "That won't work because if x = -18, then each of the two sides would have a negative<BR>\n" );
document.write( "size. Negative width and negative length doesn't make sense.<BR>\n" );
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document.write( "Next set the second factor equal to zero:<BR>\n" );
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document.write( "x - 5 = 0<BR>\n" );
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document.write( "solve for x by adding 5 to both sides to get rid of the -5 on the left side. This makes<BR>\n" );
document.write( "the equation become:<BR>\n" );
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document.write( "x = 5<BR>\n" );
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document.write( "That looks good. When x = 5, then the width (which is x + 3) is 8. And the length (which<BR>\n" );
document.write( "is x + 10 = 15.<BR>\n" );
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document.write( "So the sides of the rectangle are 8 and 15. When you draw the diagonal of the rectangle, then<BR>\n" );
document.write( "it forms the hypotenuse of a right triangle which has legs of 8 and 15. By the Pythagorean theorem<BR>\n" );
document.write( "you know that the sum of the squares of these two legs equals the square of the hypotenuse or<BR>\n" );
document.write( "diagonal. In equation form this is:<BR>\n" );
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document.write( "8^2 + 15^2 = H^2<BR>\n" );
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document.write( "Squaring out the terms on the left side results in:<BR>\n" );
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document.write( "64 + 225 = H^2<BR>\n" );
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document.write( "Add the terms on the left side and get:<BR>\n" );
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document.write( "289 = H^2<BR>\n" );
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document.write( "Solve for H by taking the square root of both sides to get:<BR>\n" );
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document.write( "17 = H<BR>\n" );
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document.write( "Since all the dimensions in this problem are in meters (or meters^2) the answer is that <BR>\n" );
document.write( "the length of the Diagonal of the square is 17 meters.<BR>\n" );
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document.write( "Hope this helps you to understand the problem and how to go about getting the answer to it.<BR>\n" );
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document.write( " </SPAN>\n" );
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