Answer 1

It is given that ABC is an equilateral triangle of side 12 cm.
Construction:
Join OA, OB and OC.

Draw:
OP⊥BC
OQ⊥AC
OR⊥AB

Let the radius of the circle be r cm.
Area of ∆AOB + Area of ∆BOC + Area of ∆AOC = Area of ∆ABC

`1/2xxABxxOR+1/2xxBCxxOP+1/2xxACxxOQ=sqrt3/4xx(side)^2`

`1/2xx12xxr+1/2xx12xxr+1/2xx12xxr=sqrt3/4xx(12)^2`

`3xx1/2xx12xxr=sqrt3/4xx12xx12`

`r=2sqrt3=2xx1.73=3.46`

Therefore, the radius of the inscribed circle is 3.46 cm.

Now, area of the shaded region = Area of ∆ABC − Area of the inscribed circle

`=[sqrt3/4xx(12)^2-pi(2sqrt3)^2] cm^2`

`=[36sqrt3-12pi]cm^2`

`=[36xx1.73-12xx3.14] cm^2`

`=[62.28-37.68]cm^2`

=24.6 cm²

Therefore, the area of the shaded region is 24.6 cm²