In the isosceles trapezoid, drop perpendiculars from the ends of the shorter base to the longer base.

The difference of the bases is \(14-6=8\text{ cm}\), so each end segment is \(4\text{ cm}\).

The height is found from the right triangle:

\(h=\sqrt{5^2-4^2}=\sqrt{25-16}=3\text{ cm}\).

The area is:

\(S=\dfrac{14+6}{2}\cdot 3=10\cdot 3=30\text{ cm}^2\).

The difference of the bases is \(12-6=6\text{ cm}\).

Since the trapezoid is isosceles, dropping the heights gives segments of \(3\text{ cm}\) on each side.

With a \(45^\circ\) angle, the height equals this segment:

\(h=3\tan 45^\circ=3\text{ cm}\).

So the area is:

\(S=\dfrac{6+12}{2}\cdot 3=9\cdot 3=27\text{ cm}^2\).

In a right trapezoid, two angles are right angles and the other two are supplementary.

If one angle is \(135^\circ\), then the other non-right angle is \(45^\circ\).

The difference of the bases is \(14-6=8\text{ cm}\).

With a \(45^\circ\) angle, the height equals this horizontal segment, so \(h=8\text{ cm}\).

Therefore,

\(S=\dfrac{14+6}{2}\cdot 8=10\cdot 8=80\text{ cm}^2\).

The area of a rhombus is given by:

\(S=\dfrac{d_1d_2}{2}\).

Substitute the given values:

\(24=\dfrac{12\cdot d_2}{2}\), so \(d_2=4\text{ cm}\).

The diagonals of a rhombus are perpendicular and bisect each other, so the half-diagonals are \(6\text{ cm}\) and \(2\text{ cm}\).

The side of the rhombus is the hypotenuse of a right triangle:

\(a=\sqrt{6^2+2^2}=\sqrt{40}=2\sqrt{10}\text{ cm}\).

The difference of the bases is \(16-6=10\text{ cm}\).

The longer leg is the slanted side of length \(26\text{ cm}\).

The height is found by Pythagoras' theorem:

\(h=\sqrt{26^2-10^2}=\sqrt{676-100}=24\text{ cm}\).

The area of the trapezoid is:

\(S=\dfrac{16+6}{2}\cdot 24=11\cdot 24=264\text{ cm}^2\).