Drop altitudes from the endpoints of the shorter base to the longer base. In an isosceles trapezoid, this gives two congruent right triangles.

The difference of the bases is \(10-4=6\), so the projection of each leg onto the longer base is \(\dfrac{6}{2}=3\).

Then the height of the trapezoid is:

\(h=\sqrt{5^2-3^2}=\sqrt{25-9}=4\).

The area of the trapezoid is:

\(S=\dfrac{10+4}{2}\cdot 4=7\cdot 4=28\).

Answer: \(28\text{ cm}^2\).

The area of a rhombus is given by

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

Substitute the data:

\(30=\dfrac{10\cdot d_2}{2}=5d_2\), hence \(d_2=6\).

The diagonals of a rhombus intersect at right angles and bisect each other, so the half-diagonals are \(5\) and \(3\).

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

\(a=\sqrt{5^2+3^2}=\sqrt{25+9}=\sqrt{34}\).

Answer: \(\sqrt{34}\text{ cm}\).

The difference of the bases is \(12-4=8\). In an isosceles trapezoid, after dropping the altitudes, this difference is split equally, so in each right triangle the horizontal leg is \(4\).

Since one of the acute angles is \(45^\circ\), in the right triangle the height equals the other leg:

\(h=4\).

Now the area is:

\(S=\dfrac{4+12}{2}\cdot 4=8\cdot 4=32\text{ cm}^2\).

Answer: \(32\text{ cm}^2\).

The diagonals of a parallelogram divide it into four triangles of equal area.

So the area of the whole parallelogram is:

\(S=4\cdot 24=96\text{ cm}^2\).

The altitude to the side \(12\) is \(\dfrac{96}{12}=8\text{ cm}\).

The altitude to the side \(8\) is \(\dfrac{96}{8}=12\text{ cm}\).

The smaller altitude is \(8\text{ cm}\).

Answer: \(8\text{ cm}\).

In a right trapezoid, two angles are right angles, and the other two sum to \(180^\circ\). If one of them is \(135^\circ\), then the other acute angle is \(45^\circ\).

The difference of the bases is \(12-8=4\text{ cm}\).

This difference is the horizontal leg of the right triangle near the slanted side. Since the acute angle is \(45^\circ\), the height of the trapezoid equals the other leg:

\(h=4\text{ cm}\).

Therefore,

\(S=\dfrac{12+8}{2}\cdot 4=10\cdot 4=40\text{ cm}^2\).

Answer: \(40\text{ cm}^2\).

The side \(AD\) consists of the segments \(AH\) and \(HD\), so

\(AD=6+4=10\text{ cm}\).

In a rhombus, all sides are equal, hence \(AB=10\text{ cm}\).

Triangle \(ABH\) is right-angled because \(BH\) is an altitude. By the Pythagorean theorem:

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

The area of the rhombus equals the product of a side and the corresponding height:

\(S=AD\cdot BH=10\cdot 8=80\text{ cm}^2\).

Answer: \(80\text{ cm}^2\).

In a right trapezoid, the longer leg is the slanted one. The difference of the bases is

\(20-10=10\text{ cm}\).

This gives a right triangle whose hypotenuse is \(26\), one leg is \(10\), and the other leg is the height of the trapezoid.

By the Pythagorean theorem:

\(h=\sqrt{26^2-10^2}=\sqrt{676-100}=\sqrt{576}=24\text{ cm}\).

Then the area is:

\(S=\dfrac{20+10}{2}\cdot 24=15\cdot 24=360\text{ cm}^2\).

Answer: \(360\text{ cm}^2\).

The diagonals of a rhombus intersect at right angles and bisect each other. Hence half of the given diagonal is \(8\text{ cm}\).

Let half of the other diagonal be \(x\). Then in a right triangle with hypotenuse \(17\), we have:

\(x^2+8^2=17^2\).

\(x^2+64=289\Rightarrow x^2=225\Rightarrow x=15\).

Therefore, the second diagonal is \(30\text{ cm}\).

The area of the rhombus is:

\(S=\dfrac{16\cdot 30}{2}=240\text{ cm}^2\).

Answer: \(240\text{ cm}^2\).

First find the other side of the rectangle using the Pythagorean theorem:

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

The area of the rectangle is:

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

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

Therefore, the area of each triangle is:

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

Answer: each of the four triangles has area \(12\text{ cm}^2\).

Since the trapezoid is isosceles, the angles at the same base are equal. If one angle is \(120^\circ\), then the acute angle is \(60^\circ\).

The leg is \(8\text{ cm}\), and the smaller base is also \(8\text{ cm}\).

The height of the trapezoid is:

\(h=8\sin 60^\circ=8\cdot \dfrac{\sqrt3}{2}=4\sqrt3\text{ cm}\).

The projection of each leg onto the larger base is:

\(8\cos 60^\circ=4\text{ cm}\).

So the larger base is:

\(8+4+4=16\text{ cm}\).

Hence the area is:

\(S=\dfrac{8+16}{2}\cdot 4\sqrt3=12\cdot 4\sqrt3=48\sqrt3\text{ cm}^2\).

Answer: \(48\sqrt3\text{ cm}^2\).

In a right trapezoid, the smaller leg is the height. Therefore, the height is \(6\text{ cm}\), and the slanted leg is \(10\text{ cm}\).

Find the difference of the bases from the right triangle:

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

Then the larger base is:

\(8+8=16\text{ cm}\).

The area of the trapezoid is:

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

Answer: \(72\text{ cm}^2\).

The side \(BC\) consists of the segments \(BH\) and \(HC\), so

\(BC=7+5=12\text{ cm}\).

In a parallelogram, opposite sides are equal, hence \(CD=AB=13\text{ cm}\).

Triangle \(DHC\) is right-angled because \(DH\) is an altitude. Therefore, by the Pythagorean theorem:

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

The area of the parallelogram is:

\(S=BC\cdot DH=12\cdot 12=144\text{ cm}^2\).

Answer: \(144\text{ cm}^2\).

The area of a trapezoid equals the product of its midline and height:

\(S=m\cdot h\).

Find the height. A leg of length \(8\) forms an angle of \(150^\circ\) with the base, so the height is

\(h=8\sin 150^\circ=8\cdot \dfrac12=4\text{ cm}\).

Therefore,

\(S=10\cdot 4=40\text{ cm}^2\).

Answer: \(40\text{ cm}^2\).

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

If one angle is \(135^\circ\), then the acute angle at the slanted side is \(45^\circ\).

In the right triangle near the slanted side, with an angle of \(45^\circ\), the legs are equal, so the difference of the bases equals the height:

\(d=6\text{ cm}\).

Hence the larger base is:

\(6+6=12\text{ cm}\).

The area of the trapezoid is:

\(S=\dfrac{6+12}{2}\cdot 6=9\cdot 6=54\text{ cm}^2\).

Answer: \(54\text{ cm}^2\).

Using the trapezoid area formula:

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

Hence \(h=12\text{ cm}\).

The difference of the bases is \(24-6=18\text{ cm}\). After dropping the altitudes, it is split equally, so the horizontal leg is \(9\text{ cm}\).

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

\(c=\sqrt{12^2+9^2}=\sqrt{144+81}=\sqrt{225}=15\text{ cm}\).

Answer: \(15\text{ cm}\).

The area of a rhombus equals the product of a side and the corresponding height:

\(S=a\cdot h\).

Substitute the data:

\(18=a\cdot 3\Rightarrow a=6\text{ cm}\).

The height of a rhombus is expressed through the side and an angle:

\(h=a\sin \alpha\).

Thus,

\(3=6\sin \alpha\Rightarrow \sin \alpha=\dfrac12\).

Hence one angle of the rhombus is \(30^\circ\). The other angle is supplementary to it, so it is \(150^\circ\).

Answer: \(30^\circ\) and \(150^\circ\).