Quadrilateral $A B C D$ has the following properties:

- sides $A B$ and $C D$ are parallel,
- there is a right angle at vertex $B$,
- triangle $A D B$ is isosceles with base $A B$,
- sides $B C$ and $C D$ are $10 \mathrm{~cm}$ long.

Determine the area of this quadrilateral.
We will try to determine the area of quadrilateral $ABCD$ as the sum of the areas of several triangles contained within it.

Since angle $ABC$ is a right angle and lines $AB$ and $CD$ are parallel, angle $BCD$ is also a right angle. The foot of the altitude from vertex $D$ in triangle $ABD$ is denoted as $E$.

![](https://cdn.mathpix.com/cropped/2024_04_17_05b9351128ca2eef9acfg-2.jpg?height=403&width=642&top_left_y=2097&top_left_x=730)

The isosceles triangle $ABD$ is divided into two congruent triangles by the altitude $DE$. Additionally, quadrilateral $BCDE$ is a square (it is a rectangle and $|BC| = |CD|$), and its diagonal $BD$ divides it into two congruent triangles. Therefore, triangles $BCD$, $BED$, and $AED$ are mutually congruent, and the area of each is equal to half the area of square $BCDE$, i.e.,

$$
\frac{10 \cdot 10}{2} = 50 \left(\mathrm{~cm}^{2}\right)
$$

The area of quadrilateral $ABCD$ is equal to the sum of the areas of these three triangles:

$$
S_{ABCD} = 3 \cdot 50 = 150 \left(\mathrm{~cm}^{2}\right)
$$

Evaluation. 2 points for justifying that quadrilateral $BCDE$ is a square; 2 points for dividing it into congruent triangles; 2 points for expressing the area.