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May 25th, 2015, 03:15 PM   #1
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inscribed circle of a triangle

Can anyone help me with this problem? I have no idea how to solve it.
Attached Images inscribed circle.jpg (48.9 KB, 14 views) May 26th, 2015, 09:09 AM #2 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond As triangle $\displaystyle ADF$ is equlateral, $\displaystyle \triangle{ADF}=\frac{x^2\sqrt3}{4}$. By the secant-tangent theorem $\displaystyle x^2=\overline{AG}\cdot\overline{AE}$ and $\displaystyle \frac{\triangle{ADF}}{\overline{AG}\cdot\overline{ AE}}=\frac{\sqrt3}{4}$. If $\displaystyle \overline{BD}=4$ and $\displaystyle \overline{CF}=2,\overline{BC}=6$, as the distance from a point outside a circle to the points of tangency is equal. Using the fact that the area of a triangle is equivalent to the radius of the incircle ($\displaystyle r$) times the semiperimeter ($\displaystyle s$) we may write $\displaystyle \frac{\sqrt3}{2}(x+2)\left(\frac{x+4}{2}\right)=rs = \frac{x}{\sqrt3}(6+x)$ which can be simplified and rearranged to $\displaystyle x^2+6x-24=0$. I used the law of cosines to determine the radius of the incircle in terms of $\displaystyle x$. Thanks from matisolla Tags circle, inscribed, triangle Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post HawkI Math 0 March 30th, 2015 03:08 PM sf7 Geometry 11 April 19th, 2014 07:05 AM Daltohn Calculus 11 October 27th, 2013 08:19 AM graviton120 Algebra 6 July 25th, 2009 03:16 AM Daltohn Algebra 0 December 31st, 1969 04:00 PM

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