Wednesday, December 16, 2009

Problem 403. Circular sector, 90 degrees, Circle, Semicircle, Area

Proposed Problem
Click the figure below to see the complete problem 403 about Circular sector, 90 degrees, Circle, Semicircle, Area.

 Problem 403. Circular sector, 90 degrees, Circle, Semicircle, Area.
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Complete Problem 403
Level: High School, SAT Prep, College geometry

2 comments:

  1. Let R radius OC and r radius of circle D.
    Let x distance of D to segment OB. Then x^2 = (R+r)^2-(R-r)^2=4Rr.
    Radius OB is equal to 2R, and then:

    2R=r+sqrt(x^2+r^2) -> R=2r -> S = (pi/4)*R^2 (1)

    A(AOB) = (pi/4)*(2R)^2 - (pi/2)*R^2 = (pi/2)*R^2 = S + S1 + S2 + S3 (2)

    By eq (1) and (2), we've: S = A(AOB)/2 = S1 + S2 + S3

    MIGUE

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  2. Let the radius of circle D be r, that of the semicircle be R so that the radius of the quarter circle is 2R

    Writing the altitude of Tr. CDO in 2 ways we have

    (2R-r)^2 - r^2 = (R+r)^2 - (R-r)^2 which yields upon simplification that R = 2r

    So S1 + S2 +S3 = 1/4pi (2R)^2 -1/2piR^2 - pir^2 which when simplified using R = 2r is = to pir^2 = S

    Sumith Peiris
    Moratuwa
    Sri Lanka

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