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.

Posted by Antonio Gutierrez at 2:48 PM
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4 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

    ReplyDelete

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

    Peter Tran

    ReplyDelete

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

    ReplyDelete

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

    ReplyDelete

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