4. In an isosceles trapezoid $A B C D$ the diagonals are perpendicular to each other and the height is $28 \mathrm{~mm}$. Calculate the area of trapezoid $A B C D$.
4. Let's introduce the labels as in the figure:

![](https://cdn.mathpix.com/cropped/2024_05_30_e7bdd2a259b40b39e2f1g-2.jpg?height=411&width=536&top_left_y=474&top_left_x=820)

$1 \mathrm{BOD}$

Since trapezoid $A B C D$ is isosceles, it follows that $|B C|=|A D|$ and $\angle A B C=\angle D A B$. Furthermore, $\overline{A B}$ is a common side of triangles $A B C$ and $A B D$, so they are congruent by the $S-K-S$ theorem.

From this congruence, it follows that $\angle C A B=\angle A B D$, i.e., $\angle S A B=\angle A B S$. Therefore, triangle $A B S$ is an isosceles right triangle, so $\angle S A B=45^{\circ}$.

Now, let $\overline{E F}$ be the height of trapezoid $A B C D$ that contains point $S$, and let $v_{1}=|E S|$ and $v_{2}=|S F|$. Thus, $\angle S A E=45^{\circ}, \angle A E S=90^{\circ}$, so triangle $A E S$ is an isosceles right triangle, which means $|A E|=|E S|$. Since $\overline{E S}$ is the height of the isosceles triangle $A B S$, it follows that $|A E|=|E B|=\frac{a}{2}$. This means $\frac{a}{2}=v_{1}$.

Similarly, $\frac{c}{2}=v_{2}$, so $v=v_{1}+v_{2}=\frac{a}{2}+\frac{c}{2}=\frac{a+c}{2}$.

Therefore, the area of trapezoid $P(A B C D)=\frac{a+c}{2} \cdot v=v \cdot v=28 \cdot 28=784 \mathrm{~mm}^{2}$.