Sunday, September 28, 2014

Geometry Problem 1046: Rectangle, Circle, Circumradius, Metric Relations

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

Click the figure below to view the complete problem 1046.

Online Math: Geometry Problem 1046: Rectangle, Circle, Circumradius, Metric Relations, Polya's Mind Map.

Posted by Antonio Gutierrez at 10:58 PM
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6 comments:

  1. Jacob HA (EMK2000)September 29, 2014 at 9:54 AM

    EA²+EC² = AC² = 4R²
    EB²+ED² = BD² = 4R²

    EA²+EB²+EC²+ED² = 8R²

    ReplyDelete
  2. AjitSeptember 29, 2014 at 10:08 PM

    By the symmetry of the figure, B, O & D are collinear making BD & AC diameters of the circumcircle each = 2R. And EB²+ED²=BD²=4R². Likewise, EA²+EC²=4R². Hence the result.

    ReplyDelete
  3. AnonymousSeptember 30, 2014 at 1:11 AM

    the solution is obtained by applying the Pythagorean theorem to the diagonals AC and BD
    AExAE + ECxEC = ACxAC = 4 (rxr)
    BExBE + EDxED = BDxBD = 4 (rxr) then
    AExAE +ECxEC + BExBE +EDxED = 8 (rxr)

    ReplyDelete
  4. ion raduOctober 3, 2014 at 11:22 AM

    EA^2+ EC^2 =AC^2=4R^2;EB^2+ ED^2 =BD^2=4R^2=>EA^2+ EC^2+EB^2+ ED^2=2.4R^2=8R^2

    ReplyDelete
  5. AnonymousOctober 5, 2014 at 9:35 AM

    we have two times the application of the Pythagorean theorem applied to the two diagonals of the rectangle Square and add them give exactly 8 (rxr)

    ReplyDelete
  6. Sumith PeirisOctober 3, 2015 at 7:14 AM

    The result follows easily from applying Pythagoras to right Tr.s AEC and BED

    ReplyDelete

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