document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1207590>Question 1207590</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In triangle $PQR,$ let $X$ be the intersection of the angle bisector of $\angle P$ with side $QR$, and let $Y$ be the foot of the perpendicular from $X$ to side $PR$.  If $PQ = 10,$ $QR = 10,$ and $PR = 12,$ then compute the length of $XY$. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #845568 by </B><A HREF=/tutors/aboutme.mpl?userid=Timnewman><B>Timnewman(323)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Timnewman><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=Timnewman><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=845568\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Hello, here is the solution to your mathematics problem. I hope you find this helpful.<BR>\n" );
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document.write( "Let's break down the problem step by step:\r<BR>\n" );
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document.write( "1. Draw the triangle $PQR$ with the given side lengths: $PQ = 10$, $QR = 10$, and $PR = 12$.\r<BR>\n" );
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document.write( "2. Draw the angle bisector of $\angle P$ and label the intersection point with $QR$ as $X$.\r<BR>\n" );
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document.write( "3. Draw the perpendicular from $X$ to $PR$ and label the foot of the perpendicular as $Y$.\r<BR>\n" );
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document.write( "4. Let's use the angle bisector theorem, which states that the ratio of the lengths of the sides is equal to the ratio of the distances from the angle bisector to the sides. In this case:\r<BR>\n" );
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document.write( "$$\frac{PQ}{QR} = \frac{PX}{XQ}$$\r<BR>\n" );
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document.write( "Since $PQ = QR = 10$, we get:\r<BR>\n" );
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document.write( "$$\frac{10}{10} = \frac{PX}{XQ}$$\r<BR>\n" );
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document.write( "$$1 = \frac{PX}{XQ}$$\r<BR>\n" );
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document.write( "$$PX = XQ$$\r<BR>\n" );
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document.write( "This means that $X$ is the midpoint of $QR$.\r<BR>\n" );
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document.write( "1. Since $QR = 10$, we have:\r<BR>\n" );
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document.write( "$$XQ = \frac{1}{2}QR = \frac{1}{2}(10) = 5$$\r<BR>\n" );
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document.write( "1. Now, let's use the Pythagorean theorem in triangle $PXQ$:\r<BR>\n" );
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document.write( "$$PX^2 + XQ^2 = PQ^2$$\r<BR>\n" );
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document.write( "$$PX^2 + 5^2 = 10^2$$\r<BR>\n" );
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document.write( "$$PX^2 + 25 = 100$$\r<BR>\n" );
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document.write( "$$PX^2 = 75$$\r<BR>\n" );
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document.write( "$$PX = \sqrt{75} = 5\sqrt{3}$$\r<BR>\n" );
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document.write( "1. Now, let's use the Pythagorean theorem in triangle $XYR$:\r<BR>\n" );
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document.write( "$$XY^2 + YR^2 = XR^2$$\r<BR>\n" );
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document.write( "Since $XR = PX = 5\sqrt{3}$, we get:\r<BR>\n" );
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document.write( "$$XY^2 + YR^2 = (5\sqrt{3})^2$$\r<BR>\n" );
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document.write( "$$XY^2 + YR^2 = 75$$\r<BR>\n" );
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document.write( "1. We can also use the Pythagorean theorem in triangle $PYR$:\r<BR>\n" );
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document.write( "$$PY^2 + YR^2 = PR^2$$\r<BR>\n" );
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document.write( "$$PY^2 + YR^2 = 12^2$$\r<BR>\n" );
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document.write( "$$PY^2 + YR^2 = 144$$\r<BR>\n" );
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document.write( "1. Subtract the two equations:\r<BR>\n" );
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document.write( "$$XY^2 - PY^2 = -69$$\r<BR>\n" );
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document.write( "$$XY^2 = PY^2 - 69$$\r<BR>\n" );
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document.write( "1. We know that $PY = PX = 5\sqrt{3}$, so:\r<BR>\n" );
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document.write( "$$XY^2 = (5\sqrt{3})^2 - 69$$\r<BR>\n" );
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document.write( "$$XY^2 = 75 - 69$$\r<BR>\n" );
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document.write( "$$XY^2 = 6$$\r<BR>\n" );
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document.write( "$$XY = \sqrt{6}$$\r<BR>\n" );
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document.write( "Therefore, the length of $XY$ is $\sqrt{6}$.\r<BR>\n" );
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document.write( "Note: This problem is a classic example of using the angle bisector theorem and the Pythagorean theorem to solve a triangle problem.\r<BR>\n" );
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document.write( "If you're satisfied with my solution and interested in further learning, I'm available for a one-on-one online session. I'd be happy to share my knowledge with you and provide personalized guidance. </SPAN>\n" );
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