In the given figure, ABC is a right angled triangle, right -angled at A in which `AB=6 cm, BC=10 cm ` and I is the incentre of `DeltaABC`.
Find the area of the shaded region. [Take `pi=3.14`]

Explanation

From right `DeltaBAC`, we have
`AC^(2)=BC^(2)-AB^(2)={(10)^(2)-6^(2)}=(100-36)=64`
`impliesAC=sqrt(64)=8cm`
`ar(DeltaABC)=(1/2xx"base"xx"height")=(1/2xxABxxAC)`
`=(1/2xx6xx8)cm^(2)=24cm^(2)`
Let the radius of the incricle be `r`. Then
`ar(DeltaABC)=ar(DeltaIAb)+ar(DeltaIBC)+(DeltaICA)`
`implies24=(1/2xxABxxr)+(1/2xxBCxxr)+(1/2xxCAxxr)`
`=1/2r(AB+BC+CA)=1/2r(6+10+8)=12r`
`impliesr=2cm`

`:.` area of the shaded region ltbgt `=ar(Delta ABC)-("area of incircle with" r=2cm)`
`=(24-pir^(2))cm^(2)=(24-3.14xx2xx2)cm^(2)`
`=(24-12.56)cm^(2)=11.44cm^(2)`
Hence the required area is `11.44cm^(2)`