Radius of circle $=\frac{12}{8} \mathrm{~cm}=1.5 \mathrm{~cm} \Rightarrow$ circumference $=2 \times \pi \times 1.5 \mathrm{~cm}=3 \pi \mathrm{cm}$
Side of the equilateral triangle $=\pi \mathrm{cm}$
Area of the triangle $=\frac{1}{2} \pi^{2} \sin 60^{\circ} \mathrm{cm}^{2}=\frac{\sqrt{3}}{4} \pi^{2} \mathrm{~cm}^{2} \Rightarrow R=\frac{\sqrt{3}}{4}$
IS. 3 Given that the perimeter of an equilateral triangle equals to that of a circle with radius $\frac{12}{Q} \mathrm{~cm}$.
If the area of the triangle is $R \pi^{2} \mathrm{~cm}^{2}$, find the value of $R$.