anonymous
  • anonymous
Let the incircle of ΔABC have radius 2 and let it be tangent to BC at X. Suppose |BX| = 3 and |XC| = 4. What is the exact length of ΔABC? No trig functions should be used.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
According to my calculations, the longest side should be 5. As long as ABC is a right triangle.
anonymous
  • anonymous
It would be an arbitrary triangle, though, no guarantees whatsoever that it would be a right triangle.
anonymous
  • anonymous
Alright. Give me a minute u have this down in my Geometry math journal I will just have to find it.

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mathmate
  • mathmate
|dw:1437781734957:dw|
anonymous
  • anonymous
Right, I agree with the picture. Just wasn't able to come up with something using that. Tried for a while and either nothing worked or something that I thought might work had a bad assumption.
mathmate
  • mathmate
The in-circle is tangent to all three sides, so the centre joined to the vertices bisect the vertex angles. Call them \(\alpha, \beta, \gamma\) at vertices A, B and C respectively. Name the centre as O, then \(\angle OBX =tan^{-1}2/3\) = 33.69 \(\angle OCX =tan^{-1}2/4\) =15.86 We can then calculate \(\angle A =2(90-33.69-15.86)\) =80.90 Then you can use the fact that the longest side is opposite the greatest angle of a triangle to find the longest side.
anonymous
  • anonymous
Oh, I apologize, I forgot to add in that the answer must be exact.
mathmate
  • mathmate
|dw:1437782481651:dw|
mathmate
  • mathmate
The answer that you get will be pretty much exact! lol
anonymous
  • anonymous
Yeah, but not allowed to do that. Basically we have to avoid trig functions.
anonymous
  • anonymous
Completely spaced on that requirement when typing up the question, my bad.
anonymous
  • anonymous
I can't find it at the moment. So, since all three sides are all different lengths, we can assume it is a Scalene triangle or a right triangle. The first thing we should do is draw and label a Scalene triangle. (I will post pictures of my work in just a second)
anonymous
  • anonymous
@ganeshie8 Any idea?
Loser66
  • Loser66
|dw:1437789668137:dw|
Mertsj
  • Mertsj
|dw:1437791155114:dw|
Loser66
  • Loser66
@Mertsj from where you have BX/CX = AB/AC??
Mertsj
  • Mertsj
Angle bisector theorem
Loser66
  • Loser66
It is apply if AOX is a line, that is A, O, X on the same line. This case is not that.
Mertsj
  • Mertsj
Good point.
Loser66
  • Loser66
|dw:1437791652962:dw|
anonymous
  • anonymous
I dont suppose there's an easy way to get |AO| then, huh?
ganeshie8
  • ganeshie8
is the perimeter 21 ?
ganeshie8
  • ganeshie8
It seems calculating the area two different ways gives two independent equations which can be solved easily |dw:1437844198135:dw|
ganeshie8
  • ganeshie8
That is first way, we may use \(\href{https:///en.wikipedia.org/wiki/Heron%27s_formula}{\text{Heron's formula}}\) to work the area again : |dw:1437844485456:dw|
anonymous
  • anonymous
Isn't 14 + 2d just the perimeter?
ganeshie8
  • ganeshie8
\[14+2d = \sqrt{12d(7+d)} \implies d = 3.5\]
ganeshie8
  • ganeshie8
good observation, it just happens that perimeter = area in our case :)
anonymous
  • anonymous
What 2nd calculation for the area gives 14 + 2d?
ganeshie8
  • ganeshie8
just addup the areas of individual triangles keeping in mind that \(2\) is the height in each triangle
ganeshie8
  • ganeshie8
|dw:1437844853560:dw|
anonymous
  • anonymous
Ah, gotcha. And then yeah, Heron's formula for the other calculation. Alright, that works perfectly :D
ganeshie8
  • ganeshie8
Ofcourse the price for avoiding trig is that we need to remember Heron's formula
anonymous
  • anonymous
Im perfectly fine with Heron's formula and a lot of the other formulas. Been doing too much math to forget it, just geometry is so much of an issue of whether or not you see the trick or where to start, lol. Sometimes the solution is easy, but your running through all the connections and things you know in your head and you just never come up with the one that leads to an easy solution.
anonymous
  • anonymous
Thanks once again :)
ganeshie8
  • ganeshie8
Haha ikr! sometimes u just have to wait, think in shower, sleep over it till the right method/idea strikes ;)
anonymous
  • anonymous
I don't have that option during exams, though. Bummer! D: Haha.

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