Monday, May 19, 2008

Geometry Problem 90



See complete Problem 90
Quadrilateral and Triangle areas, Midpoints. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.
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2 comments:

  1. AnonymousSeptember 5, 2008 at 10:54 AM

    Let X be the mid-pt of BC,
    XM//BA and XN//CD,
    [EBM]+[ENC]=[EBX]+[EXC]=[EBC]
    S_1=[BMNC]=1/2[BANC]=1/2*S/2=S/4

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  2. APOSTOLIS MANOLOUDISApril 27, 2016 at 2:34 PM

    Area triangle ABC is (ABC) then (BMD)=(ABD)-(ABM)-(AMD).But (MEN)=(MED)-(NED)-(NMD)=(MEC)+(MCD)-(BED)/2-(BMD)/2=(AEC)/2+(ACD)/2-(BED)/2-(ABD)/2+(ABM)/2+(AMD)/2=(AED)/2-(BED)/2-
    (ABD)/2+(ABM)/2+(AMD)/2=(ABD)/2-(ABD)/2+(ABM)/2+(AMD)/2=(ABC)/4+(ACD)/4=
    (ABCD)/4.

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