In a `!ABC` the length of the median AD to the side BC is 4 units. If `angleA=60^(@)` and the area of the triangle is `2sqrt3` sq. units. The length of side BC, is

Options

A. `2sqrt3`
B. `4sqrt3`
C. 6
D. 8

Explanation

We have,
`angleA=60^(@)`
`rArrcosA=1/2rArr(b^(2)+c^(2)-a^(2))/(2bc)=1/2rArrb^(2)+c^(2)=bc`…(i)
It is given that
`Delta=2sqrt3` sq. units
`rArr1/2bcsinA=2sqrt3rArr1/2bcxxsin60^(@)=2sqrt3`
`rArrbc=8`

From (i) and (ii), we get
`b^(2)+c^(2)-a^(2)=8rArrb^(2)+c^(2)=a^(2)+8`
In `DeltaABD`, we have
`AB^(2)+AC^(2)=2(AD^(2)+BD^(2))`
`rArrc^(2)+b^(2)=2(16+(a^(2))/4)rArr2(b^(2)+c^(2))=64+a^(2)`
From (iii) and (iv), we get
`2(a^(2)+8)=64+aa^(2)rArra^(2)=48rArra=4sqrt3`