Geometry Problem
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Level: High School, SAT Prep, College geometry
Monday, December 27, 2010
Problem 557: Quadrilateral, Midpoints, Sides, Varignon Parallelogram, Area
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draw AC => EF = 1/2 AC, EF//AC
ReplyDelete=> HG = 1/2 AC, HG//AC
=> EF = HG, EF//EG => EFGH paral
draw BPQ perpendicular to AC, P on EF, Q on AC
=> BP = PQ => SBEF = 1/2 SBAC
pb 146
ReplyDeletejoin the diagonal AC. now in triangle ABC we have E and F are the midpoints of AB and BC respectively. so by the converse of midpoint theorem we get that EF is parallel to AC. similarly GH is parallel to AC. these imply that EF and GH are parallel. similarly by joining the diagonal BD we can find that FG and HE are parallel. so we can conclude that the figure EFGH is a parallelogram as both pairs of opposite sides are parallel. also by joining the diagonals BD and AC we can find that the areas of triangle BEF = 1/4*(area of triangle ABC)
ReplyDeletearea of triangle HDG = 1/4*(area of triangle DAC)
area of triangle AEH = 1/4*(area of triangle ABD)
area of triangle GCF = 1/4*(area of triangle BCD). adding the results we get that the sum of the areas of the triangles BEF, HDG, AEH and GCF equals half of the area of the quadrilateral ABCD. therefore we get that the area of parallelogram EFGH is half of the area of quadrilateral ABCD.
Q. E. D.
since E , F ,G , H are mid points of the sides by mid point theorem pair of opposite sides are parallelso EFGH is a parallelogram and ABFGH , DCFEH can be treated as carpets by carpets theorem area of parallelogram is half the area of quadrilateral
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