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In a !ABC , medians AD and BE are drawn....

In a `!ABC` , medians AD and BE are drawn. If AD = 4, `angleDAB=pi//6` and `angleABE=pi//3` then the area of `!ABC` is

A

`64/(3sqrt3)`

B

`8/(3sqrt3)`

C

`16/(3sqrt3)`

D

`32/(3sqrt3)`

Text Solution

Verified by Experts

In `DeltaABG`, we have
`cospi/6=(AG)/(AB)rArr(sqrt3)/2=8/(3AB)rArrAB=16/(3sqrt3)`
`therefore` Area of `DeltaABD=1/2ABxxADsinpi/6`
`rArr` Area of `DeltaABD=1/2xx16/(3sqrt3)xx4xx1/2=16/(3sqrt3)` sq. units
`rArr` Area of `DeltaABC=2` Area of `DeltaABD` `=32/(3sqrt3)` sq. units
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