In an isosceles trapezoid $A B C D$ with bases $B C$ and $A D$, the diagonals intersect at point $O$. Find the perimeter of the trapezoid if $B O=\frac{7}{8}, O D=\frac{25}{8}, \angle A B D=90^{\circ}$.
Denote $B C=7 x, A D=25 x$ and consider the right triangle $A B D$.

## Solution

From the similarity of triangles $B O C$ and $D O A$, it follows that

$$
\frac{B C}{A D}=\frac{B O}{O D}=\frac{7}{25}
$$

Let $B C=7 x, A D=25 x$. Let $P$ be the projection of vertex $B$ onto the base $A D$. Then $A P=\frac{1}{2}(A D-B C)=9 x$.

From the right triangle $A B D$, we find that

$$
A D \cdot D P=B D^{2} \text {, or } 25 \cdot 16 x^{2}=16 .
$$

Thus, $x=\frac{1}{5}$. Therefore,

$$
\begin{gathered}
B C=\frac{7}{5}, A D=5, B P=\sqrt{A P \cdot P D}=12 x=\frac{12}{5} \\
C D=A B=\sqrt{B P^{2}+A P^{2}}=\sqrt{\frac{144}{25}+\frac{81}{25}}=\frac{15}{5}=3
\end{gathered}
$$

Therefore, the perimeter of trapezoid $A B C D$ is

$$
\frac{7}{5}+5+6=\frac{62}{5}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_633abcb7b0e2bd041761g-07.jpg?height=294&width=598&top_left_y=1019&top_left_x=730)

## Answer

$\frac{62}{5}$.