In a `!XYZ` ,let x , y , z be the length of the side opposite to angles. X , Y, Z respectively and 2s = x + y + z. If `(s-x)/4=(s-y)/3=(s-z)/2` and area of the incircle of the triangle XYZ is `(8pi)/3`, the area of `DeltaXYZ` is

Options

A. `6sqrt6` sq. units
B. `3sqrt6` sq. units
C. `12sqrt6` sq. units
D. `6sqrt3` sq. units

Explanation

We have,
`(s-x)/4=(s-y)/3=(s-z)/2`
`rArr(s-x)/4=(s-y)/3=(s-z)/2=((s-x)+(s-y)+(s-z))/(4+3+2)`
`rArr(s-x)/4=(s-y)/3=(s-z)/2=s/9`
`rArrs-x=(4s)/9,s-y=(3s)/9,s-z=(2s)/9`
It is given that the area of the in-circle is `(8pi)/3`
`thereforepir^(2)=(8pi)/3`
`rArr(Delta^(2))/(s^(2))=8/3`
`rArrs(s-x)(s-y)(s-z)=8/3s^(2)`
`rArrsxx(4s)/9xx(3s)/9xx(2s)/9=8/3s^(2)`
`rArrs=9`
Putting s = 9 in (i), we get `6sqrt6 `