Saturday, February 14, 2015

Geometry Problem 1083: Semicircle, Diameter, Perpendicular, 90 Degrees, Tangent Line, Metric Relations

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

Click the diagram below to enlarge it.

Online Math: Geometry Problem 1083: Semicircle, Diameter, Perpendicular, 90 Degrees, Tangent Line, Metric Relations.

8 comments:

  1. through D draw line perpendicular to AB, end on F;
    Connect GC,
    triangle GDF similar to triangle DGE, ( angle AGD = angle GCD)
    a is radius of circle
    so X^2 = 2a(a+b)

    ReplyDelete
  2. http://s29.postimg.org/m7r8fv3hz/pro_1083.png
    Draw lines and points per sketch
    Lot O is the center of the circle, we have FO⊥AE and GO⊥AB => AFOG is a square
    So CD= 2.AF= 2a
    We also have ∠ (HDG) =∠ (BGC) and ∠ (CDG)= ∠ (BGC)
    so GD is an angle bisector of angle CDH
    DH=GK= a + b
    Relation in the right triangle DGC give DG^2= DK.DC
    or x^2=2a.(a+b)

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  3. May I point out a typo ?
    Instead of DH=GK= a + b, should it not be DH=DK= a + b ?

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  4. Thanks for the correction.
    the corrected statement is " DH=DK=a+b"

    Peter Tran

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  5. https://www.dropbox.com/s/uggeh3qzew0fdld/1083.fig?dl=0

    Complete the rectangle BAEJ and semicircle to full circle.
    Let EJ and diameter GOK intersect at H.
    Then x^2 = GD^2 = GK.GH = 2a(a + b).

    ReplyDelete
  6. Let H be on AG such that HD//BC//GO where O is the centre of the semicircle CGD

    AGOF is a square of side a and the radius of the semicircle is a

    Tr. a CDG and GDH are similar hence

    2a/x = x/(a+b) from which the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Further if DE = c and DF = y then it can be shown that

    y^2 = 2ac

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  8. Moreover if J is the food of the perpendicular from G to CD,

    CJ = a-b and GJ^2 = a^2 - b^2

    ReplyDelete