# Area Problem 6 Nov

**Question**

Area of triangle if radius of the circle is !!1cm!!.

!!a)" "4(sqrt3)!!

!!b)" "7 + 4(sqrt3)!!

!!c)" "4(sqrt3) - 6!!

!!c)" "6 + 4(sqrt3)!!

**Solution**

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.

Now,

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!!