Friday, November 7, 2014

Geometry Problem 1057: Triangle, Area, Circle with Perpendicular Diameters, Tangent, Secant

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Online Math: Geometry Problem 1057: Triangle, Area, Circle with Perpendicular Diameters, Tangent, Secant.
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2 comments:

  1. AjitNovember 7, 2014 at 8:34 PM

    Tr. BFE = Tr. BEA - Tr. BFA = Tr. BCA - Tr. BHA. Now draw a perpendicular HX to AB.It can be proven easily that HX =(2R²-25)/2R if R be the radius of the circle.
    Hence Tr. BFE = (R*R) - ((2R²-25)/2R)*(2R/2) = 25/2 = 12.5 sq. units.

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  2. Ivan BazarovNovember 23, 2014 at 9:51 AM

    <CBF=<ECF from tangent angle theorem, <FAB=<FCB from inscribed angles, and <FAB=<CEF from parallel property. Therefore <FCB=<FEC, making triangle CFB similar to triangle ECF. FB/CF=CF/FE, which results in CF^2=25=FB*FE=2*[FEB].

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