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March 2nd, 2010, 07:12 PM  #1 
Member Joined: Dec 2009 Posts: 66 Thanks: 0  Inscribed Circles
Alright, I'm stuck on this problem on the ICTM Regional 2007 exam. The question reads: "The inscribed circle of a right triangle divides the hypotenuse of this right triangle into two segments of lengths 6 and 14. Find the area of the triangle." Under every attempt, I've gotten the area to be 20 root(21) whereas the real answer is 84. After quickly looking over my calculations I realized my flaw in assuming angles, and now understand why the answer I got was wrong. How do you solve this problem? Thanks in advance! 
March 2nd, 2010, 08:57 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond  Re: Inscribed Circles
From the center of the circle, draw lines to the tangent points and to the two vertices where the legs meet the hypotenuse. You now have two pairs of congruent right triangles and it can be shown that the length of the legs of the triangle we wish to find the area of are 14 + r and 6 + r, where r is the radius of the circle. Using the Pythagorean Theorem, 400 = (6 + r)^2 + (14 + r)^2. Solve for r and then use that to determine the lengths of the sides and finally, the area of the triangle. 
March 2nd, 2010, 09:54 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: Inscribed Circles Hello, Recipe! Quote:
Code:  A o :  * 6 :  * :  * * * D :  * o : 6  * * * : * * * a  * : * O * * 14 : F o     * * * : * : * * :  : * : a6 * : * * :  * : * * :  * : * *  C o*o*o B b14 E 14 :             b                 : [color=beige]. . [/color] [color=beige]. . [/color] [color=beige]. . . . . . . . . . . . [/color] [color=beige]. . . . . . . . . . .[/color]  
March 3rd, 2010, 09:47 AM  #4 
Senior Member Joined: Mar 2007 Posts: 428 Thanks: 0  Re: Inscribed Circles
Since it seems yet again to be the policy to simply give full solution rather than point the way, let's look at it slightly differently. Let the radius of the incircle be "r". Then (r + 14)^2 + (r + 6)^2 = 20^2, or r^2 + 20r  84 = 0 The area of the triangle will be half the product of the legs: (r + 14)(r + 6) = r^2 + 20r + 84 = (r^2 + 20r  84) + 168 = 0 + 168 = 168. So the area is 84. 
March 3rd, 2010, 11:05 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,927 Thanks: 2205 
As above, r² + 20r = 84. Area of triangle = inradius × perimeter/2 = r(r + 14 + r + 6 + 20)/2 = r² + 20r = 84. I corrected two typing errors in the previous solutions. 
March 3rd, 2010, 02:58 PM  #6  
Senior Member Joined: Mar 2007 Posts: 428 Thanks: 0  Re: Quote:
 
March 3rd, 2010, 05:28 PM  #7 
Member Joined: Dec 2009 Posts: 66 Thanks: 0  Re: Inscribed Circles
Thanks a lot everyone! I understand it now, my problem was recognizing (or remembering) that two segments drawn from the same point to tangent points of a circle are congruent. 

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