Saturday, May 7, 2011

Geometry Problem 600: Circle, Area, Tangent, Semicircle, Tangency Point

Geometry Problem
Click the figure below to see the complete problem 600.

 Geometry Problem 600: Circle, Area, Tangent, Semicircle, Tangency Point.

6 comments:

  1. Let r1= radius of circle dia. AB, r2= radius of circle dia. BC and r3=radius of circle dia. CD
    Apply Pythagoras theorem to circles with radius r1 and r2 we get
    (r1+r2)^2-(r1-r2)^2= 81 or 4.r1.r2=81
    Similarly with circles r2 and r3
    (r2+r3)^2-(r2-r3)^2=1 or 4.r2.r3=1
    With circles r1 and r3
    (r1+2.r2+r3)^2-(r1-r3)^2=100 or 4.r2^2+4.r1.r3=18

    Solve system equations above we get r1=3/2 , r2= 27/2 , r3=1/6
    Replace values of r1, r2 and r3 in following formula we will get the result
    Area S= ½( pi.(r1+r2+r3)^2 – pi.r1^2-pi.r2^2-pi.r3^2))

    Peter Tran

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  2. Minor typo correction
    solution of system of equations give:
    r1=27/2 ; r2=3/2 and r3=1/6

    Peter Tran

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  3. Comment:
    2S+π(r₁²+r₂²+r₃²)=π(r₁+r₂+r₃)²
    S=π(r₁r₂+r₂r₃+r₃r₁)=π(81+1+100)/4=91π/2

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  4. Pravin,
    Should the final answer not be 91π/4 ?
    Ajit

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  5. Applying Problem 812.

    Take a=9, b=1.

    Then
    S = π/4 (a^2 + b^2 + ab)
    = π/4 (81 + 1 + 9)
    = 91π/4

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  6. r1 --> Big circle
    r2 --> medium circle
    r3 --> small circle
    r4 --> Biggest circle

    Using formula of tangent of two circles we will have 3 equation:
    1. (r1 + r2)^2 = (r1 - r2)^2 + 81
    2. (r1 + r2 + r3)^2 = (r1 - r3)^2 + 100
    3. (r2 + r3)^2 = (r2 - r3)^2 + 1

    from the 1st equation: r1 . r2 = 81/4
    from the 3rd equation: r2 . r3 = 1/4

    r2 = 1/4r3

    With the similarity of right angle triangle:
    4. (r3 + r2)/(r1 + 2r2 + r3) = 1/10
    5. (r2 - r3)/(r1 - r2) = 1/10

    from the 4th equation and 5th equation:
    9r3 + 8r2 = 10r2 - 9r3
    9r3 = r2
    9r3 = 1/4r3
    r3 = 1/6
    r2 = 3/2
    r1 = 27/2
    r4 = 27

    Area of S = 91π/4

    Thank u

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