See complete Problem 90
Quadrilateral and Triangle areas, Midpoints. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Geometry Problem 90
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See complete Problem 90
Quadrilateral and Triangle areas, Midpoints. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Let X be the mid-pt of BC,
ReplyDeleteXM//BA and XN//CD,
[EBM]+[ENC]=[EBX]+[EXC]=[EBC]
S_1=[BMNC]=1/2[BANC]=1/2*S/2=S/4
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-
ReplyDelete(ABD)/2+(ABM)/2+(AMD)/2=(ABD)/2-(ABD)/2+(ABM)/2+(AMD)/2=(ABC)/4+(ACD)/4=
(ABCD)/4.