First find the radius of the inscribed circle of the equilateral triangle.

For an equilateral triangle with side \(a\): \(r=\dfrac{a\sqrt{3}}{6}\).

Substitute \(a=8\sqrt{6}\):

\(r=\dfrac{8\sqrt{6}\cdot\sqrt{3}}{6}=\dfrac{8\sqrt{18}}{6}=\dfrac{8\cdot3\sqrt{2}}{6}=4\sqrt{2}\text{ cm}\).

The diameter of the circle equals the diagonal of the inscribed square: \(d=2r=8\sqrt{2}\text{ cm}\).

If the side of the square is \(x\), then \(x\sqrt{2}=8\sqrt{2}\).

Therefore, \(x=8\text{ cm}\).

The radius of the circle inscribed in the square equals half the side of the square: \(R=\dfrac{6\sqrt{3}}{2}=3\sqrt{3}\text{ cm}\).

This circle is the circumcircle of the equilateral triangle.

For an equilateral triangle, \(R=\dfrac{a}{\sqrt{3}}\), so \(a=R\sqrt{3}\).

Therefore, \(a=3\sqrt{3}\cdot\sqrt{3}=9\text{ cm}\).

For a regular hexagon inscribed in a circle, the side equals the radius of the circle.

Hence, \(r=3\sqrt{3}\text{ cm}\).

This circle is inscribed in the equilateral triangle.

For an equilateral triangle, \(r=\dfrac{a\sqrt{3}}{6}\).

Thus, \(a=\dfrac{6r}{\sqrt{3}}=\dfrac{6\cdot3\sqrt{3}}{\sqrt{3}}=18\text{ cm}\).

The radius of the circle circumscribed about a square equals half the diagonal of the square.

The diagonal of the square is: \(d=12\sqrt{6}\cdot\sqrt{2}=12\sqrt{12}=24\sqrt{3}\text{ cm}\).

So the radius of the circle is: \(r=\dfrac{d}{2}=12\sqrt{3}\text{ cm}\).

This circle is inscribed in the regular hexagon, so its radius is the apothem of the hexagon.

For a regular hexagon, \(r=\dfrac{a\sqrt{3}}{2}\).

Therefore, \(a=\dfrac{2r}{\sqrt{3}}=\dfrac{2\cdot12\sqrt{3}}{\sqrt{3}}=24\text{ cm}\).

Using the circumference formula: \(2\pi R=12\pi\sqrt{3}\).

Hence, \(R=6\sqrt{3}\text{ cm}\).

For an equilateral triangle, \(R=\dfrac{a}{\sqrt{3}}\).

Therefore, \(a=R\sqrt{3}=6\sqrt{3}\cdot\sqrt{3}=18\text{ cm}\).

From the condition: \(\pi r^2=8\pi\).

Therefore, \(r^2=8\), so \(r=2\sqrt{2}\text{ cm}\).

The side of the square equals the diameter of the circle: \(a=2r=4\sqrt{2}\text{ cm}\).

The distance from the center of the square to a vertex is half the diagonal: \(\dfrac{a\sqrt{2}}{2}=\dfrac{4\sqrt{2}\cdot\sqrt{2}}{2}=4\text{ cm}\).

Using the circumference formula: \(2\pi r=6\pi\sqrt{3}\).

Hence, \(r=3\sqrt{3}\text{ cm}\).

For a regular hexagon, the apothem is \(r=\dfrac{a\sqrt{3}}{2}\).

Therefore, \(a=\dfrac{2r}{\sqrt{3}}=\dfrac{2\cdot3\sqrt{3}}{\sqrt{3}}=6\text{ cm}\).

From the area formula for a circle: \(\pi R^2=27\pi\).

So \(R^2=27\), hence \(R=3\sqrt{3}\text{ cm}\).

For an equilateral triangle, \(R=\dfrac{a}{\sqrt{3}}\).

Therefore, \(a=R\sqrt{3}=3\sqrt{3}\cdot\sqrt{3}=9\text{ cm}\).

Using the circumference formula: \(2\pi R=2\pi\sqrt{2}\).

Therefore, \(R=\sqrt{2}\text{ cm}\).

For a square, the distance from the center to a side equals the inradius, which is related to the circumradius by \(r=\dfrac{R}{\sqrt{2}}\).

Then \(r=\dfrac{\sqrt{2}}{\sqrt{2}}=1\text{ cm}\).

From the area formula for a circle: \(\pi r^2=36\pi\).

Therefore, \(r=6\text{ cm}\).

In an equilateral triangle, the incenter and circumcenter coincide, and \(R=2r\).

Therefore, the distance from the center of the circle to a vertex is \(R=2\cdot6=12\text{ cm}\).

The distance from the center of a regular hexagon to a side is the apothem, i.e. the inradius.

For a regular hexagon: \(r=\dfrac{a\sqrt{3}}{2}\).

Substitute \(a=4\sqrt{3}\):

\(r=\dfrac{4\sqrt{3}\cdot\sqrt{3}}{2}=\dfrac{4\cdot3}{2}=6\text{ cm}\).

From the area formula of a circle: \(\pi R^2=64\pi\).

Therefore, \(R=8\text{ cm}\).

For a regular hexagon inscribed in a circle, the side equals the radius of the circle.

So the side of the hexagon is \(8\text{ cm}\).

The distance from the incenter of an equilateral triangle to a vertex equals the circumradius.

For an equilateral triangle: \(R=\dfrac{a}{\sqrt{3}}\).

Substitute \(a=12\sqrt{3}\): \(R=\dfrac{12\sqrt{3}}{\sqrt{3}}=12\text{ cm}\).

The square is inscribed in the circle, so the diameter of the circle equals the diagonal of the square.

The diagonal of the square is: \(d=8\sqrt{2}\cdot\sqrt{2}=16\text{ cm}\).

Therefore, the radius of the circle is \(8\text{ cm}\).

For a regular hexagon inscribed in the same circle, the side equals the radius.

So the side of the hexagon is \(8\text{ cm}\).

For an equilateral triangle, the circumradius is \(R=\dfrac{a}{\sqrt{3}}\).

Substitute \(a=4\sqrt{6}\): \(R=\dfrac{4\sqrt{6}}{\sqrt{3}}=4\sqrt{2}\text{ cm}\).

This same circle is the circumcircle of the square.

For a square, the side is related to the circumradius by \(a=R\sqrt{2}\).

Therefore, \(a=4\sqrt{2}\cdot\sqrt{2}=8\text{ cm}\).

For an equilateral triangle, the circumradius is \(R=\dfrac{a}{\sqrt{3}}=\dfrac{10\sqrt{3}}{\sqrt{3}}=10\text{ cm}\).

The regular hexagon is inscribed in the same circle, and its side equals the radius of that circle.

Therefore, the side of the hexagon is \(10\text{ cm}\).