8,9

The diagonal of a parallelogram, equal to $b$, is perpendicular to the side of the parallelogram, equal to $a$. Find the second diagonal of the parallelogram.
Apply the Pythagorean theorem and the theorem on the sum of the squares of the diagonals of a parallelogram

## Answer

Let the diagonal $B D$ of the parallelogram $A B C D$ be perpendicular to its side $A B$, and $A B=C D=a, B D=b$. By the Pythagorean theorem from the right triangle $A B D$, we find that

$$
A D^{2}=A B^{2}+B D^{2}=a^{2}+b^{2}
$$

By the theorem on the sum of the squares of the diagonals of a parallelogram,

$$
A C^{2}+B D^{2}=2 \cdot A B^{2}+2 \cdot A D^{2}, \text { or } A C^{2}+b^{2}=2 a^{2}+2\left(a^{2}+b^{2}\right),
$$

from which $A C^{2}=4 a^{2}+b^{2}$.