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 March 3rd, 2014, 02:23 AM #1 Newbie   Joined: Mar 2014 Posts: 2 Thanks: 0 Help with an incircle problem? Hi there, I am new so please forgive my longwinded question. I know general geometry & trigonometry but my son in highschool has given me a tough question (for me). The hypotenuse of a right-angled triangle is 15cm. A circle of radius 2cm is drawn inside the triangle so that it touches each of the 3 sides. What is the perimeter of the triangle? Could someone please tell me the formula or names of the formulas to work out this one?
 March 3rd, 2014, 08:50 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Help with an incircle problem? [color=#000000]The radius of the inscribed circle is given by the formula $r=\frac{2\cdot \text{Area}}{Perimeter}\Leftrightarrow \cancel{2}=\frac{\cancel{2}\cdot \text{Area}}{Perimeter}\Leftrightarrow \text{Area}=\text{Perimeter}$. If we denote the two vertical sides of the right triange by x and y then the above formula gives $x+y+15=\frac{x\cdot y}{2}(1)$. Combining (1) with the formula which is given by applying the Pythagorean theorem $x^2+y^2=15^2(2)$ we get four solutions $\left\{\begin{array}{c}\{x\to -15,y\to 0\}\\\{x\to 0,y\to -15\}\\\left\{x\to \frac{1}{2} \left(19-\sqrt{89}\right),\;y\to \frac{1}{2} \left(19+\sqrt{89}\right)\right\}\\\left\{x\to \frac{1}{2} \left(19+\sqrt{89}\right),\;y\to \frac{1}{2} \left(19-\sqrt{89}\right)\right\}\end{array}\right\}$ (there is also an easy way to find x+y). Since a side of a triangle cannot be equal to zero, we get only the last two solutions and so choosing the third pair (we will end up with the same solution with the fourth one) $\text{Perimeter}=x+y+15=19+15=34$.[/color]
 March 3rd, 2014, 11:40 AM #3 Newbie   Joined: Mar 2014 Posts: 2 Thanks: 0 Re: Help with an incircle problem? Thank you so much. I had found the formula for the radius of the incircle, but I had been stumped on how to combine that with the area and the perimeter of the triangle! Cheers.
 March 3rd, 2014, 02:18 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 Perimeter = 2(hypotenuse + inradius)  (easily seen from a diagram)                = 34cm.

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