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So we also know that side BC must be less than 3 but more than 1. That's proven by the triangle leg addition postulate.
Answer is D.1^2 + 2^2 = 2m^2 + m^2/210 = 5m^2m = sqrt(2).
There is at least one geometric solution, using similarity
AD=BC=2xBD=DC=xalso AB=1,AC=2using Stewart's TheoremAB^2 . CD + AC^2 . BD = BC . (AD^2 + BD.CD)1.x + 4.x = 2x(4x^2 + x.x)x=1/sqrt2AD=2x=2/sqrt2=sqrt2so option D is answer.
So we also know that side BC must be less than 3 but more than 1. That's proven by the triangle leg addition postulate.
ReplyDeleteAnswer is D.
ReplyDelete1^2 + 2^2 = 2m^2 + m^2/2
10 = 5m^2
m = sqrt(2).
There is at least one geometric solution, using similarity
ReplyDeleteAD=BC=2x
ReplyDeleteBD=DC=x
also AB=1,AC=2
using Stewart's Theorem
AB^2 . CD + AC^2 . BD = BC . (AD^2 + BD.CD)
1.x + 4.x = 2x(4x^2 + x.x)
x=1/sqrt2
AD=2x=2/sqrt2=sqrt2
so option D is answer.