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anonymous
 one year ago
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.
anonymous
 one year ago
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.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0According to my calculations, the longest side should be 5. As long as ABC is a right triangle.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It would be an arbitrary triangle, though, no guarantees whatsoever that it would be a right triangle.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright. Give me a minute u have this down in my Geometry math journal I will just have to find it.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437781734957:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, 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
 one year ago
Best ResponseYou've already chosen the best response.1The incircle 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 \(\angle A =2(9033.6915.86)\) =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
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I apologize, I forgot to add in that the answer must be exact.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437782481651:dw

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1The answer that you get will be pretty much exact! lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, but not allowed to do that. Basically we have to avoid trig functions.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Completely spaced on that requirement when typing up the question, my bad.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 Any idea?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1@Mertsj from where you have BX/CX = AB/AC??

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1It is apply if AOX is a line, that is A, O, X on the same line. This case is not that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I dont suppose there's an easy way to get AO then, huh?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3is the perimeter 21 ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3It seems calculating the area two different ways gives two independent equations which can be solved easily dw:1437844198135:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3That 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
 one year ago
Best ResponseYou've already chosen the best response.0Isn't 14 + 2d just the perimeter?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[14+2d = \sqrt{12d(7+d)} \implies d = 3.5\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3good observation, it just happens that perimeter = area in our case :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What 2nd calculation for the area gives 14 + 2d?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3just addup the areas of individual triangles keeping in mind that \(2\) is the height in each triangle

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1437844853560:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah, gotcha. And then yeah, Heron's formula for the other calculation. Alright, that works perfectly :D

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Ofcourse the price for avoiding trig is that we need to remember Heron's formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im 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
 one year ago
Best ResponseYou've already chosen the best response.0Thanks once again :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Haha ikr! sometimes u just have to wait, think in shower, sleep over it till the right method/idea strikes ;)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't have that option during exams, though. Bummer! D: Haha.
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