Wednesday, March 6, 2019

Geometry Problem 1422: Triangle, Transversal, Cevian, Area, Quadrilateral, Measurement

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 1422: Triangle, Transversal, Cevian, Area, Quadrilateral, Measurement, Tutoring.

Posted by Antonio Gutierrez at 11:51 AM
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2 comments:

  1. c.t.e.oMarch 6, 2019 at 3:26 PM

    Draw altitudes: h1 from G to BD, h2 from AD, h3 from G to BE, h4 from F to EC
    From Given S : e/c = h2/(3h1), and d/a = (8.h4)/(x.h3)
    substitute h4/h3 with h2/h1 and e/c with d/a (from the given data)
    => S = 10

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  2. Sumith PeirisMarch 6, 2019 at 5:30 PM

    Let h1 = altitude from G in Tr. BDG
    h2 = altitude from F in Tr. ABF
    Similarly for h3 & h4 for Tr.s BEG & BCF

    Let x be the required area

    h1e = 14 & h2c = 42,
    h3d = 16 & h4a = 2x + 16

    h1/h2 = BG/BF = h3/h4

    So (14/e)/(42/c) = (16/d)/((2x + 16)/a)) which gives
    c/3e = 8a/(d(x+8))
    Hence x + 8 = 24.ae/cd = 18
    Therefore x = 10

    Sumith Peiris
    Moratuwa
    Sri Lanka

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