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(A) $3cm$ (B) $4cm$ (C) $2\sqrt 3 cm$ (D) $2cm$

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Hint: If $a$ is the side of the equilateral triangle, then the circumradius of the equilateral triangle is $\dfrac{a}{{\sqrt 3 }}$.

If $a$ is the side of the equilateral triangle then according to the question, the side of the equilateral triangle is given as $3\sqrt 3 $.

$ \Rightarrow a = 3\sqrt 3 $.

$\therefore $ And we know that the circumradius of an equilateral triangle

is $\dfrac{a}{{\sqrt 3 }}$. If $R$ is the circumradius, then we’ll get:

$

\Rightarrow R = \dfrac{a}{{\sqrt 3 }}, \\

\Rightarrow R = \dfrac{{3\sqrt 3 }}{{\sqrt 3 }}, \\

\Rightarrow R = 3 \\

$

Therefore, the radius of the circumcircle of the given equilateral triangle is $3cm$.

(A) is the correct option.

Note: In an equilateral triangle circumcenter, the incenter and centroid lies at the same point. If $a$ is the side of the triangle, then $\dfrac{{a\sqrt 3 }}{2}$ is its altitude. And centroid divides the altitude in the ratio $2$: $1$. Larger divided part is circumradius and the smaller part is inradius. Thus:

Circumradius $ = \dfrac{a}{{\sqrt 3 }}$ and inradius $ = \dfrac{a}{{2\sqrt 3 }}$.

If $a$ is the side of the equilateral triangle then according to the question, the side of the equilateral triangle is given as $3\sqrt 3 $.

$ \Rightarrow a = 3\sqrt 3 $.

$\therefore $ And we know that the circumradius of an equilateral triangle

is $\dfrac{a}{{\sqrt 3 }}$. If $R$ is the circumradius, then we’ll get:

$

\Rightarrow R = \dfrac{a}{{\sqrt 3 }}, \\

\Rightarrow R = \dfrac{{3\sqrt 3 }}{{\sqrt 3 }}, \\

\Rightarrow R = 3 \\

$

Therefore, the radius of the circumcircle of the given equilateral triangle is $3cm$.

(A) is the correct option.

Note: In an equilateral triangle circumcenter, the incenter and centroid lies at the same point. If $a$ is the side of the triangle, then $\dfrac{{a\sqrt 3 }}{2}$ is its altitude. And centroid divides the altitude in the ratio $2$: $1$. Larger divided part is circumradius and the smaller part is inradius. Thus:

Circumradius $ = \dfrac{a}{{\sqrt 3 }}$ and inradius $ = \dfrac{a}{{2\sqrt 3 }}$.