Let the radius of the inscribed circle be $r\;\text{cm}$, the radius of the circumscribed circle be $R\;\text{cm},$ and the side of an equilateral triangle be $a \;\text{cm}.$
Area of the circumscribed circle $ = \dfrac{abc}{4R} = \dfrac{a^{3}}{4R}\;\text{cm}^{2}.\;[\because \text{here}, a = b = c]$
Area of the inscribed circle $A= rS\;\text{cm}^{2}$, where $S = \frac{a+b+c}{2} = \frac{3a}{2}.$
Now, the required ratio $ = \dfrac{A_{IC}}{A_{CC}} = \dfrac{\pi r^{2}}{\pi R^{2}} = \left(\dfrac{r}{R}\right)^{2} \rightarrow (1)$
Now, we have area of the equilateral triangle is equal to the area of inscribed circle $:\dfrac{a^{2}\sqrt{3}}{4} = \dfrac{3ar}{2}$
$\implies r = \dfrac{\sqrt{3}\;a}{6}$
And, area of the equilateral triangle is equal to the area of circumscribed circle $:\dfrac{a^{2}\sqrt{3}}{4} = \dfrac{a^{3}}{4R}$
$\implies R = \dfrac{a}{\sqrt{3}}$
From the equation $(1),$ we get.
$\dfrac{A_{IC}}{A_{CC}} = \left(\dfrac{r}{R}\right)^{2}$
$\implies \dfrac{A_{IC}}{A_{CC}} = \left( \dfrac{\frac{\sqrt{3}\;a}{6}}{\frac{a}{\sqrt{3}}}\right)^{2} = \left[\left(\frac{\sqrt{3}\;a}{6}\right)\left(\frac{\sqrt{3}}{a}\right)\right]^{2} = \left(\dfrac{1}{2}\right)^{2} = \dfrac{1}{4}.$
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