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**Hint:**Here first we will let the radius of the circle be $r$ and the altitude of the triangle to be $h$ such that it is equal to $x+r$ illustrated in the figure. Then we will apply the Pythagoras theorem to get the value of $r$.

The Pythagoras theorem states that the square of the hypotenuse is equal to the sum of squares of base and height of a right-angled triangle.

\[{\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {altitude} \right)^2}\]

**Complete step by step solution:**

Let the radius of the circle

\[OA = OC = r\]

Let the height/altitude of the isosceles \[\Delta ABC\] be $h$ such that:

\[h = r + x\]……………………………..(1)

Now since it is given that the base of the triangle is 6cm

Hence,

\[AB = 6cm\]

which implies that:

\[AM = MB = 3cm\]

Also, it is given that one of the equal sides of the triangle is 12 cm

Therefore,

\[AC = 12cm\]

Now since \[\Delta AOM\] is another right angled triangle and r is the hypotenuse of the triangle

Therefore, applying Pythagoras theorem in \[\Delta AOM\]

According to Pythagoras theorem:

\[{\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {altitude} \right)^2}\]

Here,

$AC$ is the hypotenuse,

$AM$is the base,

$MC$ is the altitude.

Putting in the values we get:-

\[{\left( {AC} \right)^2} = {\left( {AM} \right)^2} + {\left( {MC} \right)^2} \]

${\left( {12} \right)^2} = {\left( 3 \right)^2} + {\left( h \right)^2}$

$ 144 = 9 + {h^2} $

${h^2} = 144 - 9$

$ {h^2} = 135 $

$ h = \sqrt {135} $

Now, applying Pythagoras theorem in \[\Delta AOM\]

$OA$ is hypotenuse,

$AM$ is base,

$OM$ is the altitude.

Therefore putting in the values in the formula of Pythagoras theorem we get:-

\[{r^2} = {3^2} + {x^2}\]

From equation 1 we get:-

\[x = h - r\]

Putting in this value in above equation we get:-

\[{r^2} = {3^2} + {\left( {h - r} \right)^2}\]

Now applying the following identity:

\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]

We get:-

\[{r^2} = 9 + {h^2} + {r^2} - 2hr\]

\[2hr = 9 + {h^2} \]

Solving for $r$,

$ r = \dfrac{{9 + {h^2}}}{{2h}} $

Solving it further and putting the known values we get:-

\[\Rightarrow r = \dfrac{{9 + {{\left( {\sqrt {135} } \right)}^2}}}{{2\left( {\sqrt {135} } \right)}} \]

On simplifying the above values,

$\Rightarrow r = \dfrac{{9 + 135}}{{2\sqrt {135} }}$

On further simplifications, we get

$\Rightarrow r = \dfrac{{144}}{{2\sqrt {135} }}$

Now rationalizing the value of r by multiplying both the numerator and denominator by \[\sqrt {135} \] in order to make the denominator a whole number we get:-

\[\Rightarrow r = \dfrac{{144}}{{2\sqrt {135} }} \times \dfrac{{\sqrt {135} }}{{\sqrt {135} }}\]

Simplifying it further we get:-

\[\Rightarrow r = \dfrac{{144\sqrt {135} }}{{2 \times 135}} \]

$\Rightarrow r = \dfrac{{144\sqrt {135} }}{{270}} $

On further simplification,

$\Rightarrow r = \dfrac{{8 \times 3\sqrt {15} }}{{15}}$

$ r = \dfrac{{8\sqrt {15} }}{5} $

**$\therefore$ The radius of the circle is \[\dfrac{{8\sqrt {15} }}{5}cm\]. Hence, option E is the correct option.**

**Note:**

In an isosceles triangle, two sides are equal and the corresponding angles opposite to them are also equal.

Also, the altitude from a vertex to the opposite side bisects it into two halves.

The student can also consider $\Delta BOM$ instead of $\Delta AOM$ and then apply the Pythagoras theorem to get the value of $r$.

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