Area of triangle if radius of the circle is !!1cm!!.
!!b)" "7 + 4(sqrt3)!!
!!c)" "4(sqrt3) - 6!!
!!c)" "6 + 4(sqrt3)!!
In the below triangle we can see that there are three similar circles arranged.
Since radius of all circles are equal to !!1!!. So, !!OP = PQ = QO = 2cm!!.
Hence triangle QOP are equilateral trianle.
Now angle !!QOP = 60!! degree, so angle !!TOR = 120!! degree.
In the quadilateral !!BTOR!!, !!OT and OR!! are perpendicular to side !!AB and BC!!.
Using summation of quadilateral property.
Angle !!TBR + 90 + 90 + 120 = 360!!
!!=>!! Angle !!TBR = 60!! degree.
Similarly we can prove for other angle as well.
Hence triangle ABC is equilateral triangle.
Radius of the circle !!= 1cm = OR!!
Also !!tan30 = (OP)/(BR)!!
!!=> BR = (OP)/tan30 = 1/(1/sqrt3) = sqrt3!!
Similarly, !!SC = sqrt3!!
Also we can see that !!OP = 1 + 1 = 2!!
and !!OP = RS!!
Hence side of triangle !!= BC = BR + RS + SC =sqrt3 + 2 + sqrt3 = 2 + 2sqrt3 !!
We know that area of equlateral triangle !!= ((sqrt3)/4)xx(Side)^2 = ((sqrt3)/4)xx(2 + 2sqrt3)^2 = 6 + 4sqrt3!!