In the given figure, 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 `sqrt(3)=1.73` and `pi=3.14`]

Answer

`r=2sqrt(3),` ar (shaded region) `=24.6cm^(2)`

Explanation

Draw `AOD_|_BC`. Then D is the midpoint of BC.
`:.AC=12cm` and `DC=6cm`
`AD^(2)=AC^(2)-DC^(2)=(12)^(2)-6^(2)=108impliesAD=sqrt(108)=6sqrt(3)cm`.
`:.h=6sqrt(3)m`. Now `3r=h=6sqrt(3)impliesr=2sqrt(3)cm`.
So, the radius of inscribed circle is `2sqrt(3)` cm
ar (shaded region)`=[((sqrt(3))/4xx12xx12)-{3.14xx(2sqrt(3))^(2)}]cm^(2)`
`=[(36xx1.73)-(12xx3.14)]cm^(2)`
`=(62.28-37.68)cm^(2)`
`=24.6cm^(2)`
Note `h=3r` in case of incircle.