Question

Please solve q 92?

Answer 1

option (1) `6sqrt2`

Explanation:

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`