Thursday, December 6, 2018

Geometry Problem 1405: Four Congruent Circles, Congruent Chords

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

Details: Click on the figure below.

Geometry Problem 1405: Four Congruent Circles, Congruent Chords, Tutoring.

Posted by Antonio Gutierrez at 7:40 PM
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3 comments:

  1. c.t.e.oDecember 7, 2018 at 1:51 AM

    From ang CFH and CFP => CH = CP
    From congr of circles AC = BC
    => Tr ACH congr Tr BCP (SAS)

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  2. Sumith PeirisDecember 7, 2018 at 3:39 AM

    ABC, BCD, BDF are all equilateral triangles
    ABF is a straight line and a diameter of circle B and so AH is perpendicular to FP

    S(BFP) = 1/2.BP/2.FP......(1) (since < BPF = 30 since D is the centre of TR BFP)

    S(AFP) = 1/2.AH. FP ......(2)

    Since S(AFP) = 2.S(BFP) from (1) and (2) we have

    AH = BP

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. AnonymousDecember 9, 2018 at 1:15 PM

    CBF equal 120*. Angle CPF equal 60*. Angle CHP=FAC=CPE=60*.Triangle CHP equilateral. PC=PH=HC. R(C,60*) B-->A P-->H BP=AH. In triangle APH AP in second power=AH second power+ HC in second power=BP in second power + PC in second power. This means AP in second power=Bp in second power + PC in second power.

    Erina,NJ

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