The area of rectangle $A B C D$ is $100 \sqrt{5}$. Let $P$ be the point on side $A B$ that is closer to $A$ and divides $A B$ into a ratio of $1:4$. We know that segment $P D$ is perpendicular to diagonal $A C$. Calculate the perimeter of the rectangle.
I. solution. The angles $A D P \varangle$ and $C A B \varangle$ are right angles. The triangles $A C B$ and $D P A$ have two equal angles, so the two triangles are similar. Thus, the ratios of their corresponding sides are equal, that is:

$$
\frac{\frac{a}{5}}{b}=\frac{b}{a}, \quad \text { that is } \quad a^{2}=5 b^{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_e401403720c778d82f0cg-1.jpg?height=309&width=501&top_left_y=412&top_left_x=801)

From the relationship $T=a b=100 \sqrt{5}$, we get $a=\frac{100 \sqrt{5}}{b}$. Substituting this into the above equation, we get $\frac{50000}{b^{2}}=5 b^{2}$, that is, $b^{4}=10000$. Only $b=10$ is valid, as it represents a length. Then $a=10 \sqrt{5}$. The perimeter of the rectangle is: $K=2 a+2 b=20 \sqrt{5}+20$

II. solution. Let's set up a coordinate system where point $A$ is the origin, side $A B$ lies on the $x$-axis, and side $A D$ lies on the $y$-axis. Thus, the coordinates of the points are: $A(0 ; 0), B(b ; 0), D(0 ; d), C(b ; d) ;(b, d>0)$, since $A B C D$ is a rectangle.

Point $P$ is the point on side $A B$ closest to $A$, dividing it into fifths, so $P\left(\frac{b}{5} ; 0\right)$. Since $\overrightarrow{P D} \perp \overrightarrow{A C}$, their dot product is 0. The coordinates of these vectors are: $\overrightarrow{P D}\left(-\frac{b}{5} ; d\right), \overrightarrow{A C}(b ; d)$.

Thus

$$
\overrightarrow{P D} \cdot \overrightarrow{A C}=-\frac{b^{2}}{5}+d^{2}=0, \quad \text { that is } \quad b^{2}=5 d^{2}
$$

Since the area of $A B C D$ is $T=b d=100 \sqrt{5}$, we have $b=\frac{100 \sqrt{5}}{d}$. Substituting this, we get $\left(\frac{100 \sqrt{5}}{d}\right)^{2}=5 d^{2}$, from which only $d=10(d>0)$ is possible. The perimeter of $A B C D$ is:

$$
K=2(b+d)=2\left(\frac{100 \sqrt{5}}{d}+d\right)=\frac{200 \sqrt{5}}{d}+2 d=20 \sqrt{5}+20
$$

Thus, the perimeter of the rectangle is $20 \sqrt{5}+20 \approx 64.72$.