As shown in the figure, #A A', B B' and C C'# are the three medians of #DeltaABC, and G# is the centroid,
As centroid divides each median in the ratio #1:2#,
#=> GA=2/3*A A'=2/3*24=16, G A'=1/3*24=8#
similarly, #GB=12, GB'=6, GC=8, GC'=4#
Extend #GB'# to #D# such that #B'D=GB'=6#,
#=> AGCD# is a parallelogram,
#=> CD=GA=16, and AD=GC=8#,
Use Heron's formula to find area of #DeltaGCD#,
#A=sqrt(s(s-a)(s-b)(s-c))#, where a,b,and c are the side length of the triangle, and #s=(a+b+c)/2#
#=> s=(8+12+16)/2=18#
#=> A_(DeltaGCD)=sqrt(18*10*6*2)=12sqrt15#
diagonals divides a parallelogram into 4 equal areas,
#=># area of #DeltaGCB'=(12sqrt15)/2=6sqrt15#
the three medians divides a triangle into 6 equal areas,
#=># area of #DeltaABC=6*6sqrt15=36sqrt15 " unit"^2#
Sol 2: use formula
#A=4/3(sqrt((s-m_1)(s-m_2)(s-m_3))#
where #m_1,m_2 and m_3# are the three medians of the triangle and #s=(m_1+m_2+m_3)/2#
#=> s=(12+18+24)/2=27#
#=> A=4/3*sqrt(27(27-12)(27-18)(27-24))=36sqrt15 " unit"^2#
