In the given figure, ∆ ABC is an equilateral triangle and radius of each smaller circle is 5 cm. Find the perimeter of ∆ABC.

30(\(\sqrt {3}\) + 2)

In ∆ABC,

\(\angle\) ABC = 60°                   ( because ∆ABC is equilateral)

In ∆OBP,

\(\angle\) OBP = \(\frac{1}{2}\) x \(\angle\) ABC = 30°

Now,

In ∆OBP,

tan θ = \(\frac{Perpendicular}{Base}\)

⇒ tan 30° = \(\frac{1}{\sqrt {3}}\) = \(\frac{5}{BP}\)

BP = 5\(\sqrt {3}\)

CQ =  BP = 5\(\sqrt {3}\)

BC = BP + PQ + QC = 5\(\sqrt {3}\) + 20 + 5\(\sqrt {3}\) = 10\(\sqrt {3}\) + 20

∆ABC is an equilateral triangle, therefore

AB =BC = CA

Perimeter = 3 x side = 3 x BC = 3 x (10\(\sqrt {3}\) + 20) = 30(\(\sqrt {3}\) + 2)