Geometry Problem
Click the figure below to see the complete problem 528 about Triangle, Medians, Perpendicular, Measurement.
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Complete Problem 528
Level: High School, SAT Prep, College geometry
Wednesday, October 13, 2010
Problem 528: Triangle, Medians, Perpendicular, Measurement
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Solution sent by Peter Tran:
ReplyDeleteLet G is the intersection of AD and CE
We have AD^2= ½*(AB^2+AC^2-BC^2/2) …….. Formula to calculate the median
AG= 2/3*AD .
Replace AC=b , AB=c and BC=a we get AG^2=2/9*(c^2+b^2-a^2/2)
Similarly we also have CG^2=2/9*(a^2b^2-C^2/2)
But AC^2=AG^2+CG^2 .
Replace values of AG^2 and CG^2 from above formulas we will get a^2+c^2=5.b^2
Peter Tran