Triangle drive

Let triangle ABC be right angled at B. AD is the median to BC and measures 5 units. Similarly, CE is the median to AB and measures (40)^1/2. Now consider triangle EBC. (AB/2)^2+ BC^2 = EC^2 (AB/2)^2 + BC^2= 40 - 1 Similarly, in triangle ABD, AB^2+ (BC/2)^2= AD^2 AB^2+(BC/2)^2=25 - 2 On solving 1 and 2, we get AB=4, BC=6. Then hypotenuse AC=(52)^1/2 or s(13)^1/2.

NOTE : I am going to use 's' to symbolize square....so 3s = 9

1) let triangle be ABC right angled at B where AE and CD are the medians. AB = 2BD and BC = 2BE. AE = root40 and CD =5.

2) using appolonius theorm

ABs + ACs = 2(AEs + BEs) => ACs + 4BDs - 2BEs = 80 let this be equation 1 similarly

ACs + BCs = 2(CDs + BDs) => ACs - 2BDs + 4BEs = 50 let this be equation 2

from pythagoras ACs = ABs + BCs => ACs = 4(BDs + BEs) let this be equation 3

3) solving the resultant equation we get BD = 3 , BE = 2 , AC = 2sqrt(13)