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February 22nd, 2013, 06:04 PM   #1
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Area of triangle ABC

[attachment=0:3g7xy306]Area of triangle ABC.JPG[/attachment:3g7xy306]
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 Area of triangle ABC.JPG (15.6 KB, 265 views)

 February 22nd, 2013, 06:53 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond Re: Area of triangle ABC $\text{Let }\bar{AH}\text{ be }h.\text{ Let }B\text{ and }C\text{ be the measure of the angles at }B\text{ and }C. \\ 45^{\circ}\,+\,B\,+\,C\,=\,180^{\circ} \\ C\,=\,135^{\circ}\,-\,B \\ \tan(B)\,=\,\frac{h}{2} \\ \tan(135^{\circ}\,-\,B)\,=\,\frac{h}{3} \\ \frac{\tan(135^{\circ}\,-\,B)}{\tan(B)}\,=\,\frac23 \text{From }\tan(a\,\pm\,b)\,=\,\frac{\tan(a)\,\pm\,\tan(b)}{ 1\,\mp\,\tan(a)\tan(b)} \frac{\tan(135^{\circ}\,-\,B)}{\tan(B)}\,=\,\frac{1\,+\,\tan(B)}{\tan^2(B)\ ,-\,\tan(B)}\,=\,\frac23\,\Rightarrow\,\tan(B)\,=\,3 \tan(B)\,=\,\frac{h}{2}\,=\,3\,\Rightarrow\,h\,=\, 6\text{ so the area of }ABC\,=\,15$
 February 22nd, 2013, 09:15 PM #3 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: Area of triangle ABC greg1313: great ,you got it
 February 22nd, 2013, 10:47 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Re: Area of triangle ABC Letting the lowers case letters correspond to the sides opposite the given vertices, the law of cosines gives: $25=b^2+c^2-\sqrt{2}bc$ By Pythagoras, we find: $h^2=b^2-9=c^2-4\,\therefore\,c^2=b^2-5$ and so substituting into the first equation, we arrive at the factored quartic: $$$b^2-45$$$$b^2-10$$=0$ Since $h>3$ as $A=45^{\circ}$ then: $b^2=45\,\therefore\,h=6$ and the area is: $A=\frac{5\cdot6}{2}=15$
 February 23rd, 2013, 12:05 AM #5 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: Area of triangle ABC now try to prove it with no trigonometry using geometry only
 February 23rd, 2013, 08:30 AM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond Re: Area of triangle ABC Let a, b and c be the sides opposite A, B and C. Let h be AH. Construct a circumcircle around ABC. By the Inscribed Angle theorem and some calculations, the radius of the circumcircle is 5/?2, so its diameter is 10/?2. Given that the diameter of the circumcircle is abc/(2T), where T is the area of the triangle, we have 10/?2 = 5bc/(2T). Since T = 5h/2, we have 10/?2 = bc/h. 50h² = b²c². b² - 9 = c² - 4, 50h² = 50(b² - 9) = b²(b² - 5) ? b = ?10, ?45. Only ?45 fits ABC, so T = 15.
February 24th, 2013, 09:08 PM   #7
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Re: Area of triangle ABC

The following graph may give you a hint :
h=2r=6
the area is :
$\dfrac {5\times 6}{2}=15$

[attachment=0:2nsrutdq]Area of ABC.JPG[/attachment:2nsrutdq]
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 Area of ABC.JPG (15.5 KB, 94 views)

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