Geometry Problem

Click the figure below to see the complete problem 528 about Triangle, Medians, Perpendicular, Measurement.

See also:

Complete Problem 528

Level: High School, SAT Prep, College geometry

## Wednesday, October 13, 2010

### Problem 528: Triangle, Medians, Perpendicular, Measurement

Labels:
measurement,
median,
perpendicular,
triangle

Subscribe to:
Post Comments (Atom)

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