4A. The lengths of the bases of a trapezoid are $a=25, b=15$, and the length of one of the legs is $c=8$. Determine the perimeter and area of the trapezoid, if the sum of the angles on the larger base is $90^{\circ}$.
Solution. We continue the legs of the trapezoid $ABCD$ to their intersection at point $E$. From the condition of the problem, the sum of the angles of the larger base in the trapezoid, which are the base angles of triangle $ABE$, is $90^{\circ}$, meaning that this triangle is a right triangle with a right angle at vertex $E$.

Triangles $ABE$ and $DCE$ are similar due to the parallelism of the bases in the trapezoid, so we can set up a proportion (see the diagram for the given data): 25:15 $=(8+x): x$, from which

![](https://cdn.mathpix.com/cropped/2024_06_05_e56627a2e774dea33268g-03.jpg?height=586&width=560&top_left_y=596&top_left_x=919)
we get $x=12$.

From the right triangle $DCE$, we have: $\overline{CE}=\sqrt{15^{2}-12^{2}}=9$, and from the right triangle $ABE$ we have

$$
\overline{BE}=\sqrt{25^{2}-20^{2}}=15, \text { i.e. } \overline{BC}=\overline{BE}-\overline{CE}=15-9=6
$$

Accordingly, for the perimeter of the trapezoid, we get

$$
L=25+8+15+6=54
$$

and for the area

$$
P=P_{ABE}-P_{DCE}=\frac{20 \cdot 15}{2}-\frac{12 \cdot 9}{2}=150-54=96
$$