In the above figure, circle with centre I is inscribed in an equilateral triangle ABC.
BC and BA are two tangents from B to circle, therefore must be the angle bisector of `angleB`.
But `angleB =60^(@) ( :. DeltaABC" is an equilateral "Delta)`
`:. angle IBD=30^(@)`
`:. tan30^(@)=r/(a//2)`
`:. r=a/(2sqrt3)`
Now, inside the circle, square PQRS is inscribad.
In the above figure, IPQ is isosceles-right angled triangle.
`:. PQ^(2=IP^2+IQ^2=r^2+r^2=2r^2`
`rArr PQ=sqrt2xxa/(2sqrt3=a/sqrt6`
`:." Area of square"=(a/sqrt6)^(2)=a^2/6" sq. unit"`
