A trapez has parallel sides $A B$ and $C D$, and the intersection point of its diagonals is $M$. The area of triangle $A B M$ is 2, and the area of triangle $C D M$ is 8. What is the area of the trapezoid?
I. solution. The triangles $A B M$ and $C D M$ are similar because their angles are equal. We know that in the case of similar triangles, the ratio of corresponding sides is the square root of the ratio of the areas of the triangles, so

$$
\frac{C D}{A B}=\sqrt{\frac{8}{2}}=2, \text { i.e., } C D=2 \cdot A B
$$

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To calculate the area of the trapezoid, we need the distance between the bases $A B$ and $C D$, which is the sum of the heights of the triangles $A B M$ and $C D M$ corresponding to the sides $A B$ and $C D$. Determining the height corresponding to a side from the area, we get that the distance $m$ between the parallel sides of the trapezoid is:

$$
m=\frac{2 \cdot T_{A B M}}{A B}+\frac{2 \cdot T_{C D M}}{C D}=\frac{4}{A B}+\frac{16}{C D}=\frac{4}{A B}+\frac{8}{A B}=\frac{12}{A B}
$$

From this, the area $T$ of the trapezoid is:

$$
T=\frac{1}{2}(A B+C D) \cdot m=\frac{1}{2}(A B+2 \cdot A B) \cdot \frac{12}{A B}=18
$$

This solves the problem.

II. solution. We will show that the areas of the triangles $A B M, B C M, C D M$ form a geometric sequence.

The heights of the triangles $A B M$ and $B C M$ corresponding to vertex $B$ are the same, so the ratio of their areas is equal to the ratio of the sides opposite to $B$:

$$
\frac{T_{A B M}}{T_{B C M}}=\frac{A M}{M C}
$$

The heights of the triangles $B C M$ and $C D M$ corresponding to vertex $C$ are the same, so:

$$
\frac{T_{B C M}}{T_{C D M}}=\frac{B M}{M D}
$$

Since the triangles $A B M$ and $C D M$ are similar, the right-hand sides of (1) and (2) are equal, so the areas indeed form a geometric sequence. Therefore, if $T_{A B M}=T, \quad T_{C D M}=t$, then $T_{B C M}=\sqrt{T t}$. We know that the area of the triangle $D A M$ is equal to the area of the triangle $B C M$ (this follows from the above proof, but we can also get it more simply by adding the triangle $C D M$ to both, and thus arriving at the triangles $B C D$ and $A C D$ of equal area), so the area of the trapezoid is:

$$
T+t+2 \sqrt{T t}
$$

Substituting our data, the desired area is: $2+8+2 \cdot \sqrt{2 \cdot 8}=18$.