As shown in the figure, #AD, BE, and CF# are the medians of #DeltaABC#,
let #AD=m_a, BE=m_b, and CF=m_c#
In #DeltaADC, b^2=m_a^2+(a/2)^2-2*m_a*a/2*cosx ---Eq(1)#
In #DeltaADB, c^2=m_a^2+(a/2)^2-2*m_a*a/2*cosy--- Eq(2)#
Adding #Eq(1) and Eq(2)# together gives,
#b^2+c^2=2m_a^2+a^2/2-2*m_a*a(cosx+cosy)#
as #x+y=180^@, => cosx+cosy=0#
#=> b^2+c^2=2m_a^2+a^2/2#
#=> 4m_a^2=(2b^2+2c^2-a^2) --- Eq(3)#
similarly, #4m_b^2=(2a^2+2c^2-b^2) --- Eq(4)#
and, #4m_c^2=(2a^2+2b^2-c^2) --- Eq(5)#
Adding #Eq(3), Eq(4) and Eq(5)# together, we get:
#4(m_a^2+m_b^2+m_c^2)=3a^2+3b^2+3c^2#
#=> color(red)(m_a^2+m_b^2+m_c^2=3/4(a^2+b^2+c^2))#
Now, given that #a^2+b^2+c^2=120#, and two of the medians are #4 and 5#,
let #m# be the third median,
#=> m^2+4^2+5^2=3/4*120#
#=> m^2=90-16-25=49#
#=> m=sqrt49=7# units
