## Task 5 - 050935

In the parallelogram $A B C D$, let $A B = C D = a, B C = A D = b, (a > b)$, and $A E = h_{a}$, where $E$ is the foot of the perpendicular dropped from point $A$ to side $C D$ or its extension. Furthermore, a circular disk with a radius of length $r$ is given. The center of the circular disk traverses all sides of the parallelogram.

Calculate the area $F$ covered by the circular disk!
There are two cases to distinguish:

1. For $2 r \geq h_{a}$, the entire parallelogram area is covered.
2. For $2 r < h_{a}$, assuming $h_{a} < h_{b}$, and therefore also $2 r < h_{b}$.

In Case 1, the covered area $F$ can be decomposed into the partial areas $F_{1}$ to $F_{9}$.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0901.jpg?height=314&width=717&top_left_y=1219&top_left_x=675)

Here,

- $F_{1}$ and $F_{5}$ are rectangles with side lengths $a$ and $r$,
- $F_{3}$ and $F_{7}$ are rectangles with side lengths $b$ and $r$,
- $F_{2}, F_{4}, F_{6}$, and $F_{7}$ are sectors of a circle with radius $r$ whose angles sum up to an angle of $360^{\circ}$. Therefore, the four sectors can be combined into a full circle.

Thus, for the area $I(F)$ of the covered area,

$$
I(F)=2\left(I\left(F_{1}\right)+I\left(F_{3}\right)\right)+I\left(F_{9}\right)+\pi r^{2}=2(a r+b r)+\pi r^{2}+a h_{a}
$$

In Case 2, the same partial areas as in Case 1 are formed outside the parallelogram. The area of the parallelogram is not completely covered. The area $F_{10}$ remains uncovered.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0901.jpg?height=208&width=642&top_left_y=2169&top_left_x=707)

$F_{10}$ is a parallelogram with side length $(a-2 x)$ and the corresponding height $\left(h_{a}-2 r\right)$, where $x$ is the side length of one of the corner rhombi.

According to the second theorem of rays, $x: r = b: h_{a}$, and thus $x = \frac{b r}{h_{a}}$.

Therefore, for the area $I\left(F^{\prime}\right)$ of the covered area in Case 2,

$$
I\left(F^{\prime}\right)=I(F)-I\left(F_{10}\right)=2 a r+2 b r+\pi r^{2}+a h_{a}-\left(a h_{a}-2 b r-2 a r+\frac{4 b r^{2}}{h_{a}}\right)=
$$

$$
=4\left(a r+b r-\frac{b r^{2}}{h_{a}}\right)+\pi r^{2}
$$

## Adopted from $[5]$