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October 3rd, 2012, 04:15 PM   #1
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Right triangle area

Right triangle ABC, the 2 legs being a and b.

Let radius of incircle = r

Show that area of ABC = (a - r)(b - r)

Kinda fun!
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October 3rd, 2012, 06:08 PM   #2
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Re: Right triangle area







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October 3rd, 2012, 07:07 PM   #3
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Re: Right triangle area

[attachment=0:14dz3n3u]Denistriangle.jpg[/attachment:14dz3n3u]

By Pythagoras:





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October 3rd, 2012, 07:09 PM   #4
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Re: Right triangle area

Nice one Greg. I went this way:

(a - r) (b - r) = area?

Since r = (a + b - c) / 2 :
[a - (a + b - c) / 2] * [b - (a + b - c) / 2] = area

Expand and simplify:
(c^2 - a^2 - b^2 + 2ab) / 4 = area
Since c^2 - a^2 - b^2 = 0, then:
2ab / 4 = area
ab / 2 = area

Anything "wrong"? Doesn't "feel" 100% ok !
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October 4th, 2012, 02:19 AM   #5
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Nothing is wrong, but there is no need to repeat "= area" until the end.

(a - r)(b - r) = (a - (a + b - c)/2)(b - (a + b - c)/2)
= (c + a - b)(c - a + b)/4
= (c - (a - b))/4
= (c - (a + b) + 2ab)/4
= ab/2
= area of triangle ABC.
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October 4th, 2012, 03:40 AM   #6
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Re: Right triangle area

OOOOOOOOOOOOOOOOOOOKKKKKKKKKKKKKKKKKKKKKKK! Thanks, Skip.
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