Consider one diagonal of the isosceles trapezoid.

Its horizontal projection is \(\dfrac{12+6}{2}=9\text{ cm}\), and the height is \(3\sqrt{7}\text{ cm}\).

So the diagonal length is:

\(d=\sqrt{9^2+(3\sqrt{7})^2}=\sqrt{81+63}=12\text{ cm}\).

The intersection point of the diagonals divides them in the ratio of the bases:

\(12:6=2:1\).

Therefore, a diagonal of length \(12\text{ cm}\) is divided into segments \(8\text{ cm}\) and \(4\text{ cm}\).

The hypotenuse is \(\sqrt{9^2+12^2}=15\text{ cm}\).

The smaller acute angle is opposite the shorter leg, so point \(M\) divides the hypotenuse in the ratio \(5:10=1:2\).

Place the triangle so that the right-angle vertex is \((0,0)\), and the ends of the legs are \((12,0)\) and \((0,9)\).

Then the point \(M\), which is \(5\text{ cm}\) from the smaller-angle vertex, has coordinates \((8,3)\).

Hence the distance to the right-angle vertex is:

\(OM=\sqrt{8^2+3^2}=\sqrt{64+9}=\sqrt{73}\text{ cm}\).

The centers of circles inscribed in the same angle lie on the angle bisector.

The distances from the angle vertex to the centers are proportional to the radii.

Let the distance to the center of the smaller circle be \(x\). Then the distance to the center of the larger circle is \(\dfrac{6}{4}x=\dfrac{3}{2}x\).

Since the circles do not touch each other, the distance between their centers equals the difference of these distances:

\(\dfrac{3}{2}x-x=13\).

\(\dfrac{x}{2}=13\), so \(x=26\text{ cm}\).

Place point \(B\) at the origin and side \(BC\) on the \(x\)-axis.

Then \(B=(0,0)\), \(C=(7,0)\), \(K=(2,0)\).

Find the coordinates of \(A=(x,y)\) from:

\(AB=6\Rightarrow x^2+y^2=36\), \(AC=8\Rightarrow (x-7)^2+y^2=64\).

Subtracting, we get \(x=\dfrac{3}{2}\), and then \(y=\dfrac{3\sqrt{15}}{2}\).

Point \(M\) is the midpoint of \(AB\), so \(M=\left(\dfrac{3}{4},\dfrac{3\sqrt{15}}{4}\right)\).

Then

\(MK^2=\left(2-\dfrac34\right)^2+\left(\dfrac{3\sqrt{15}}{4}\right)^2=\dfrac{25}{16}+\dfrac{135}{16}=10\).

Therefore, \(MK=\sqrt{10}\text{ cm}\).

The centers of the circles lie on the angle bisector, and the distances from the vertex to the centers are proportional to the radii.

Let the distance to the center of the smaller circle be \(x\). Then the distance to the center of the larger circle is \(\dfrac{8}{6}x=\dfrac{4}{3}x\).

Since the circles are externally tangent, the distance between the centers equals the sum of the radii:

\(6+8=14\text{ cm}\).

On the other hand, this distance equals the difference of the distances from the vertex:

\(\dfrac{4}{3}x-x=14\).

\(\dfrac{x}{3}=14\), hence \(x=42\text{ cm}\).

Since the bases of the trapezoid are parallel, \(\angle BCA=\angle CAD\).

Given that \(\angle BAC=\angle CDA\).

Therefore, triangles \(BAC\) and \(CDA\) are similar.

From the similarity:

\(\dfrac{BC}{CA}=\dfrac{CA}{AD}\).

Substitute \(BC=5\), \(AD=20\):

\(\dfrac{5}{CA}=\dfrac{CA}{20}\).

Hence \(CA^2=100\), so \(CA=10\text{ cm}\).

In a trapezoid, the intersection point of the diagonals divides them in the ratio of the bases:

\(\dfrac{AO}{OC}=\dfrac{AD}{BC}=\dfrac{20}{5}=4\).

Thus, \(AO:OC=4:1\), and the whole diagonal \(AC=10\text{ cm}\).

So \(AO=\dfrac{4}{5}\cdot10=8\text{ cm}\).

Let the base be \(a\), and the equal sides are \(30\text{ cm}\).

Using the side formula with the circumradius:

\(30=2R\sin\alpha=50\sin\alpha\), where \(\alpha\) is a base angle.

Therefore, \(\sin\alpha=\dfrac{3}{5}\), and \(\cos\alpha=\dfrac{4}{5}\).

The vertex angle is \(180^\circ-2\alpha\), so

\(\sin(180^\circ-2\alpha)=\sin2\alpha=2\sin\alpha\cos\alpha=2\cdot\dfrac35\cdot\dfrac45=\dfrac{24}{25}\).

Hence the base is:

\(a=2R\sin(180^\circ-2\alpha)=50\cdot\dfrac{24}{25}=48\text{ cm}\).

The diagonal of the rectangle is:

\(AC=\sqrt{9^2+12^2}=15\text{ cm}\).

The point \(N\) lies on the diagonal with \(AN=5\text{ cm}\), so it divides the diagonal in the ratio \(1:2\).

Place the rectangle in coordinates: \(A=(0,0)\), \(C=(12,9)\), \(D=(0,9)\).

Then point \(N\) has coordinates \((4,3)\).

Therefore,

\(DN=\sqrt{(4-0)^2+(3-9)^2}=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}\text{ cm}\).

Let the altitude be dropped to side \(BC\), dividing it into segments of \(16\text{ cm}\) and \(9\text{ cm}\).

Then \(BC=25\text{ cm}\).

By the Pythagorean theorem for the two right triangles:

\(AB=\sqrt{12^2+16^2}=20\text{ cm}\), \(AC=\sqrt{12^2+9^2}=15\text{ cm}\).

So the triangle has sides \(15\), \(20\), \(25\), hence it is right-angled.

For a right triangle, the circumradius equals half the hypotenuse:

\(R=\dfrac{25}{2}\text{ cm}\).

Find the second side of the rectangle:

\(BC=\sqrt{13^2-5^2}=\sqrt{169-25}=12\text{ cm}\).

Place the rectangle as follows: \(A=(0,0)\), \(B=(5,0)\), \(C=(5,12)\), \(D=(0,12)\).

Then point \(N\) has coordinates \((5,4)\).

The equation of line \(AN\) is \(y=\dfrac{4}{5}x\).

The extension of side \(CD\) has equation \(y=12\).

Find the intersection point:

\(12=\dfrac{4}{5}x\Rightarrow x=15\).

Hence \(M=(15,12)\), and \(D=(0,12)\).

Therefore, \(MD=15\text{ cm}\).

The diagonals of a rhombus are perpendicular and bisect each other.

The half-diagonals are \(12\text{ cm}\) and \(9\text{ cm}\).

Hence the side of the rhombus is:

\(BC=\sqrt{12^2+9^2}=15\text{ cm}\).

Choose coordinates: \(A=(-12,0)\), \(C=(12,0)\), \(B=(0,9)\), \(D=(0,-9)\).

Point \(E\) divides side \(BC\) in the ratio \(BE:EC=5:10=1:2\).

So \(E=\left(4,6\right)\).

Then

\(AE=\sqrt{(4+12)^2+(6-0)^2}=\sqrt{16^2+6^2}=\sqrt{292}=2\sqrt{73}\text{ cm}\).

Place the trapezoid as follows: \(A=(0,0)\), \(D=(8,0)\), \(B=(0,h)\), \(C=(6,h)\).

Then since \(CD=10\text{ cm}\), we have

\((8-6)^2+h^2=10^2\).

\(4+h^2=100\), so \(h^2=96\), hence \(h=4\sqrt{6}\text{ cm}\).

The midpoint of side \(CD\) has coordinates

\(M\left(\dfrac{8+6}{2},\dfrac{0+4\sqrt6}{2}\right)=(7,2\sqrt6)\).

Then

\(AM=\sqrt{7^2+(2\sqrt6)^2}=\sqrt{49+24}=\sqrt{73}\text{ cm}\).

The altitude to the base bisects the base, so each half is \(6\text{ cm}\).

Find the height:

\(h=\sqrt{10^2-6^2}=\sqrt{100-36}=8\text{ cm}\).

The area of the triangle is:

\(S=\dfrac{1}{2}\cdot 12\cdot 8=48\text{ cm}^2\).

The circumradius is given by:

\(R=\dfrac{abc}{4S}\).

Substitute \(a=10\), \(b=10\), \(c=12\):

\(R=\dfrac{10\cdot10\cdot12}{4\cdot48}=\dfrac{1200}{192}=\dfrac{25}{4}\text{ cm}\).

Consider triangle \(DAB\).

Since \(EF\parallel AB\), triangles \(DEF\) and \(DAB\) are similar.

The similarity ratio is:

\(\dfrac{DE}{DA}=\dfrac{2}{6}=\dfrac{1}{3}\).

Therefore, \(\dfrac{DF}{DB}=\dfrac{1}{3}\), i.e. \(DF=\dfrac{DB}{3}\).

Now find diagonal \(DB\) in triangle \(DAB\).

Since \(\angle ADC=30^\circ\), the adjacent angle of the parallelogram is \(\angle DAB=150^\circ\).

By the cosine rule:

\(DB^2=AD^2+AB^2-2\cdot AD\cdot AB\cos150^\circ\).

\(DB^2=6^2+(3\sqrt3)^2-2\cdot6\cdot3\sqrt3\cdot\left(-\dfrac{\sqrt3}{2}\right)=36+27+54=117\).

So \(DB=3\sqrt{13}\text{ cm}\).

Hence \(DF=\dfrac{DB}{3}=\sqrt{13}\text{ cm}\).

The altitude to the base of an isosceles triangle bisects the base.

Hence half the base is \(3\text{ cm}\), and the height is:

\(h=\sqrt{15^2-3^2}=\sqrt{225-9}=6\sqrt6\text{ cm}\).

Place the points as follows: \(A=(-3,0)\), \(C=(3,0)\), \(B=(0,6\sqrt6)\).

Since \(BM=5\text{ cm}\) and \(BC=15\text{ cm}\), point \(M\) divides side \(BC\) in the ratio \(1:2\).

Therefore, \(M\) is one-third of the way from \(B\) to \(C\), so \(M=(1,4\sqrt6)\).

Then

\(AM=\sqrt{(1+3)^2+(4\sqrt6)^2}=\sqrt{16+96}=\sqrt{112}=4\sqrt7\text{ cm}\).

Place the rectangle in coordinates: \(A=(0,0)\), \(B=(6,0)\), \(D=(0,8)\), \(C=(6,8)\).

Since \(AE=2\text{ cm}\), we have \(E=(0,2)\).

The equation of diagonal \(AC\) is:

\(y=\dfrac{8}{6}x=\dfrac{4}{3}x\).

The equation of line \(BE\), passing through \((6,0)\) and \((0,2)\), is:

\(y=-\dfrac{1}{3}x+2\).

Find the intersection point:

\(\dfrac{4}{3}x=-\dfrac{1}{3}x+2\Rightarrow \dfrac{5}{3}x=2\Rightarrow x=\dfrac{6}{5}\).

Then \(y=\dfrac{8}{5}\).

Point \(K\) lies on diagonal \(AC\), whose length is \(\sqrt{6^2+8^2}=10\text{ cm}\).

The division ratio is \(\dfrac{x_K}{x_C}=\dfrac{6/5}{6}=\dfrac{1}{5}\).

Therefore, \(AK=\dfrac{1}{5}\cdot 10=2\text{ cm}\).