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.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

According to my calculations, the longest side should be 5. As long as ABC is a right triangle.
It would be an arbitrary triangle, though, no guarantees whatsoever that it would be a right triangle.
Alright. Give me a minute u have this down in my Geometry math journal I will just have to find it.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

|dw:1437781734957:dw|
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.
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.
Oh, I apologize, I forgot to add in that the answer must be exact.
|dw:1437782481651:dw|
The answer that you get will be pretty much exact! lol
Yeah, but not allowed to do that. Basically we have to avoid trig functions.
Completely spaced on that requirement when typing up the question, my bad.
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)
@ganeshie8 Any idea?
|dw:1437789668137:dw|
|dw:1437791155114:dw|
@Mertsj from where you have BX/CX = AB/AC??
Angle bisector theorem
It is apply if AOX is a line, that is A, O, X on the same line. This case is not that.
Good point.
|dw:1437791652962:dw|
I dont suppose there's an easy way to get |AO| then, huh?
is the perimeter 21 ?
It seems calculating the area two different ways gives two independent equations which can be solved easily |dw:1437844198135:dw|
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|
Isn't 14 + 2d just the perimeter?
\[14+2d = \sqrt{12d(7+d)} \implies d = 3.5\]
good observation, it just happens that perimeter = area in our case :)
What 2nd calculation for the area gives 14 + 2d?
just addup the areas of individual triangles keeping in mind that \(2\) is the height in each triangle
|dw:1437844853560:dw|
Ah, gotcha. And then yeah, Heron's formula for the other calculation. Alright, that works perfectly :D
Ofcourse the price for avoiding trig is that we need to remember Heron's formula
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.
Thanks once again :)
Haha ikr! sometimes u just have to wait, think in shower, sleep over it till the right method/idea strikes ;)
I don't have that option during exams, though. Bummer! D: Haha.

Not the answer you are looking for?

Search for more explanations.

Ask your own question