Saturday, March 30, 2019

Geometry Problem 1430: Regular Octagon, Area, Equivalent Figures

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 1430: Regular Octagon, Area, Equivalent Figures, Tutoring.

Posted by Antonio Gutierrez at 1:28 PM
Labels: , , , ,

4 comments:

  1. c.t.e.oMarch 30, 2019 at 2:39 PM

    OE meet DF at P, extend DF, draw from G // OE, S intersection
    => OPSG is a rectangle

    ReplyDelete
  2. Peter TranMarch 30, 2019 at 4:37 PM

    https://photos.app.goo.gl/xk498LjP5akrivYf8

    Draw points L, K per sketch
    Note that triangle OKD is a right isosceles triangle
    And Triangle CLD congruent to DKE ( case ASA)
    S(CKD)+ S(DKE)=1/2.DK(CL+KE)= ½.OK.(CO)= S( COK)
    So red areas= yellow area

    ReplyDelete
  3. Sailendra TApril 10, 2019 at 10:38 PM

    Let AE=x and AP=x1

    Area of Yellow region = (x-2x1)*x/2 ----------(1)
    Area of the Octagon = 4*S(ABH)+S(BDFH)= (x-2x1)x --------(2)

    From (1) and (2) It can be observed that Area of yellow region is half of Area of Octagon
    Q.E.D

    ReplyDelete

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