In Fig 4, a circle is inscribed in an equilateral triangle ABC of side 12 cm. Find the radius of inscribed circle and the area of the shaded region.[Use π=3.14 and √3=1.73]
Solution
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