Q92
Given that G is the centroid of #Delta ABC#
#GA=2sqrt3,GB=2sqrt2and GC=2#
AP median is produced such that #GP=PO#
#B,O and C,O# are joined. The quadrilateral CBCO is a parallelogram,
As #AG:GP=2:1# we get
#DeltaBGP=1/3DeltaABP=1/3*1/2DeltaABC#
Again #DeltaBGO=2DeltaBGP#
So #DeltaABC=3DeltaBGO#
Now #GO=GA=2sqrt3#
#BO=GC=2#
and #GB=2sqrt2#
So #GO^2-(2sqrt3)^2=12#
#BO^2+BG^2=2^2+(2sqrt2)^2=12#
So #Delta BGO# is rigtangled at B
So #DeltaBGO=1/2BO×BG#
Area of
#Delta ABC=3DeltaBGO=3/2xxBOxxBG=3/2xx2xx2sqrt2=6sqrt2#