anonymous
  • anonymous
Circles, tangent everywhere, all the way up the triangle. Sum of perimeters of all the triangles (Bit difficult to draw this, wait a minute)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1350233869190:dw|
anonymous
  • anonymous
Base 20, and sides 26
anonymous
  • anonymous
Sum of perimeters of all the circles, I mean...

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anonymous
  • anonymous
wait r u trying to teach or is this something you really need help on? cuz ur at level 99
anonymous
  • anonymous
Think of it as a tutorial (or not):-)
anonymous
  • anonymous
so circles dont matter so i would add 20 to 26?
anonymous
  • anonymous
I already corrected that above, it is the perimeter of all the circles
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
Hint This is just Pythagorus and a splash of imagination....
experimentX
  • experimentX
|dw:1350234945917:dw| is something missing ... like side of triangle or ... radius of circle of first and second??
anonymous
  • anonymous
Maybe you didn't see it, sorry.. I posted above "Base 20, and sides 26"
anonymous
  • anonymous
Maybe I should close it and repost it with out the error and including the figures :-(
klimenkov
  • klimenkov
No. I will solve it.
klimenkov
  • klimenkov
Do you have an answer?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
Wow, this a lot of typing....
anonymous
  • anonymous
Anyone just want to write out the infinite sum...?
klimenkov
  • klimenkov
Try to check my solution. Lets sign the radiuses of the circles \(r_1, r_2,...\). Then their length will be \(2\pi r_1+2\pi r_2+...=2\pi(r_1+r_2+...)\). Now watch the picture.|dw:1350235644893:dw|From the picture you can see that radiuses take a half of the height of the triangle. So I conclude that the whole perimeter of the circumferences is \(\frac 12\cdot\sqrt{26^2-10^2}=12\)
klimenkov
  • klimenkov
Sorry, forgot to multiply by \(2\pi\). Answer is \(24\pi\).
anonymous
  • anonymous
That's right, pi*(d1 + d2 +.....) = pi * altitude of triangle
anonymous
  • anonymous
I think I've seen this done before using a geometric progression.
anonymous
  • anonymous
@CliffSedge Yup, u could do that, lot more work though....
anonymous
  • anonymous
Right, I will close this one....

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