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As shown in the figure, AD, BE, and CFAD,BE,andCF 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
