Wednesday, August 4, 2010

Problem 491: Square, Right Triangle, Incircle, Inscribed Circle, Radius

Geometry Problem
Click the figure below to see the complete problem 491 about Square, Right Triangle, Incircle, Inscribed Circle, Radius.

Problem 491. Square, Right Triangle, Incircle, Inscribed Circle, Radius
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Complete Problem 491

Level: High School, SAT Prep, College geometry

5 comments:

  1. In ▲AHD
    1) x = (a+b-c)/2 => AE = a/2 - x
    2) from pythagore theorem
    x = (-a + a√3)/4

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  2. Let the tangent from A to the nearest circle=z. We can now write the following equations immediately: 2z + 2r = a -----(1) and a^2=(z+3r)^2+(z+r)^2 -----(2) from which z can be eliminated to obtain: r=a(√3-1)/4
    Ajit

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  3. I'm missing something.... how do you get the 2z + 2r = a equation from?

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  4. In any right angled triangle such as ABE, we've: AE+BE-AB=2*(In-radius) and hence in our notation: (z+r)+(z+3r)-a=2r or a=2z+2r.
    Ajit

    ReplyDelete
    Replies
    1. How was BE found to be z+3r?

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