My Math Forum Inscribed Circles

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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: 7,934 Thanks: 1128 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:
 The inscribed circle of a right triangle divides the hypotenuse into two segments of lengths 6 and 14.[color=beige] .[/color]Find the area of the triangle.
Code:
-   A o
:     |   *     6
:     |       *
:     |       * * *    D
:     |   *           o
:   6 | *               * *
:     |*                 *    *
a     |                           *
:     *         O         *           *   14
:   F o - - - - *         *               *
:     *         :         *                   *
:     |         :                                 *
: a-6 |*      :          *                            *
:     | *       :       *                                 *
:     |   *     :     *                                       *
-   C o-------*-o-*-----------------------------------------------o B
b-14   E                    14
: - - - - - - - - - - - - b - - - - - - - - - - - - - - - - :

$\text{We have right triangle }ABC.$

$\text{Inscribed circle with center }O\text{, tangent at }D,E,F.$

$\text{Given: }\:AD\,=\,6,\;DB \,=\,14$
$\text{Note that: }\:AF\,=\,6,\;EB\,=\,14$

$\text{Let: }\:a\,=\,AC,\;b\,=\,BC,\:\text{ where }a^2\,+\,b^2\:=\:20^2\;\;[1]$
$\text{Then: }\:FC\,=\,a-6,\;CE \,=\,b-14$

$\text{Since }FOEC\text{ is a square: }\;a\,-\,6 \:=\:b\,-\,14\;\;\;\Rightarrow\;\;\;a\,-\,b \:=\:-8\;\;[2]$

$\text{Square [2]: }\;(a\,-\,b)^2 \:=\:(-8)^2 \;\;\;\Rightarrow\;\;\;a^2\,-\,2ab \,+\,b^2\;=\;64$

[color=beige]. . [/color]$\text{and we have: }\;a^2\,+\,b^2\,-\,2ab \;=\;64\;\;[3]$

$\text{From [1]: }\:a^2\,+\,b^2\:=\:400$

[color=beige]. . [/color]$\text{Substitute into [3]: }\;400\,-\,2ab \;=\;64 \;\;\;\Rightarrow\;\;\;2ab \:=\:336$

$\text{Divide by 4: }\;\;\underbrace{\frac{1}{2}\,ab} \;=\;84$
[color=beige]. . . . . . . . . . . . [/color]$\uparrow$
[color=beige]. . . . . . . . . . .[/color]$This\text{ is the area of the triangle !}$

 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: 20,617 Thanks: 2072 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:
 Originally Posted by skipjack I corrected two typing errors in the previous solutions.
If mine, thanks. Typing never was a strong point. Yet, I play classical piano with ease!

 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.

 Tags circles, inscribed

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Daro666 Algebra 0 October 24th, 2012 12:11 PM aleschio Algebra 0 September 21st, 2012 12:09 PM RiDo Algebra 2 June 21st, 2012 01:31 AM tiba Algebra 3 June 20th, 2012 05:56 AM croatianboy Algebra 2 June 4th, 2012 08:45 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top