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:

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

Half of the given diagonal is \(5\text{ cm}\).

Let half of the other diagonal be \(x\).

Then in the right triangle:

\(13^2=5^2+x^2\).

\(169=25+x^2\), so \(x^2=144\), hence \(x=12\).

Therefore, the second diagonal is \(24\text{ cm}\).

The area of the rhombus is:

\(S=\dfrac{10\cdot 24}{2}=120\text{ cm}^2\).

The altitude to the base bisects the base.

Half of the base is \(5\text{ cm}\).

The height is:

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

The area of the triangle is:

\(S=\dfrac{1}{2}\cdot 10\cdot 12=60\text{ cm}^2\).

Let the second leg be \(x\).

Using the area formula for a right triangle:

\(\dfrac{1}{2}\cdot 10\cdot x=30\).

Hence \(5x=30\), so \(x=6\text{ cm}\).

Then the hypotenuse is:

\(c=\sqrt{10^2+6^2}=\sqrt{136}=2\sqrt{34}\text{ cm}\).

Find the other side of the rectangle:

\(b=\sqrt{10^2-8^2}=\sqrt{100-64}=6\text{ cm}\).

The area of the rectangle is:

\(S=8\cdot 6=48\text{ cm}^2\).

The diagonals of a rectangle divide it into 4 triangles of equal area.

Therefore, the area of each triangle is:

\(\dfrac{48}{4}=12\text{ cm}^2\).

In a right trapezoid, the shorter leg is the height, so \(h=6\text{ cm}\).

The longer leg is \(10\text{ cm}\).

Find the difference of the bases from the right triangle:

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

So the longer base is \(8+8=16\text{ cm}\).

The area of the trapezoid is:

\(S=\dfrac{8+16}{2}\cdot 6=12\cdot 6=72\text{ cm}^2\).

The area of a trapezoid can be found by the formula \(S=mh\), where \(m\) is the midline.

Find the height. The leg is \(8\text{ cm}\) and makes an angle of \(30^\circ\) with the base, so

\(h=8\sin 30^\circ=8\cdot \dfrac12=4\text{ cm}\).

Then

\(S=12\cdot 4=48\text{ cm}^2\).

Find the height to the base:

\(30=\dfrac{1}{2}\cdot 10\cdot h\), so \(h=6\text{ cm}\).

The height bisects the base, so half of the base is \(5\text{ cm}\).

The equal side is the hypotenuse of a right triangle:

\(a=\sqrt{5^2+6^2}=\sqrt{61}\text{ cm}\).

Find the other leg:

\(b=\sqrt{20^2-16^2}=\sqrt{400-256}=12\text{ cm}\).

The area of the original right triangle is:

\(S=\dfrac{1}{2}\cdot 16\cdot 12=96\text{ cm}^2\).

The triangle formed by the midsegments is similar to the original one with scale factor \(\dfrac12\).

Therefore, its area is 4 times smaller:

\(S_{\text{midsegment}}=\dfrac{96}{4}=24\text{ cm}^2\).

In a right trapezoid, the shorter leg is the height, so \(h=4\text{ cm}\).

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

With a \(45^\circ\) angle, the difference of the bases equals the height, so it is \(4\text{ cm}\).

Since the shorter base is \(4\text{ cm}\), the longer base is \(8\text{ cm}\).

Therefore,

\(S=\dfrac{4+8}{2}\cdot 4=6\cdot 4=24\text{ cm}^2\).

First find the height of the trapezoid:

\(180=\dfrac{6+24}{2}\cdot h=15h\).

Therefore, \(h=12\text{ cm}\).

The difference of the bases is \(24-6=18\text{ cm}\), so each side segment is \(9\text{ cm}\).

The leg is the hypotenuse of a right triangle with sides \(12\) and \(9\):

\(c=\sqrt{12^2+9^2}=\sqrt{144+81}=15\text{ cm}\).

A right angle measures \(90^\circ\). If the diagonal divides it in the ratio \(1:2\), the two angles are \(30^\circ\) and \(60^\circ\).

Consider the right triangle where the diagonal is the hypotenuse of length \(8\text{ cm}\).

In a \(30^\circ-60^\circ-90^\circ\) triangle, the legs are \(4\text{ cm}\) and \(4\sqrt{3}\text{ cm}\).

Therefore, the area of the rectangle is:

\(S=4\cdot 4\sqrt{3}=16\sqrt{3}\text{ cm}^2\).