A semicircular mop with a diameter of $20 \mathrm{~cm}$ is used to wipe a corner of a room's floor, such that the endpoints of the diameter continuously touch the two walls forming the right angle of the room. What area does the mop wipe in the process?
As the $AB$ diameter semicircular blade moves, if $A$ is on the $x$-axis and $B$ is on the $y$-axis, then the circle with diameter $AB$ passes through $O$. Therefore, no point in the swept area can be more than $20 \mathrm{~cm}$ away from point $O$; points outside the quarter circle centered at $O$ with a radius of $20 \mathrm{~cm}$ cannot be reached by the blade.

![](https://cdn.mathpix.com/cropped/2024_05_02_8631a47a509b2f935645g-1.jpg?height=393&width=396&top_left_y=278&top_left_x=260)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_05_02_8631a47a509b2f935645g-1.jpg?height=388&width=355&top_left_y=695&top_left_x=603)

Figure 2

![](https://cdn.mathpix.com/cropped/2024_05_02_8631a47a509b2f935645g-1.jpg?height=396&width=398&top_left_y=691&top_left_x=1427)

Figure 3

The points inside the quarter circle centered at $O$ with a radius of $10 \mathrm{~cm}$ can be covered by semicircles resting on the $x$ and $y$ axes (Figure 2).

If $P$ is a point in the ring-shaped region bounded by the two quarter circles and $OP$ intersects the outer quarter circle at point $R$, then let the projections of point $R$ on the axes be $R_{1}$ and $R_{2}$. Thus, $OR_{1}RR_{2}$ is a rectangle with diagonals $RO = R_{1}R_{2} = 20 \mathrm{~cm}$, and the center $C$ of the rectangle satisfies $OC = 10 \mathrm{~cm}$, so $C$ lies on the quarter circle centered at $O$. Point $P$ lies on segment $CR$. Therefore, the $P$ point is contained in the right triangle $R_{1}RR_{2}$, which, according to Thales' theorem, is entirely within the semicircle with diameter $R_{1}R_{2}$ passing through $R$. In this position, the blade "erases" the $P$ point.

Thus, during its path, the blade precisely erases the points of the closed quarter circle with center $O$ and radius $20 \mathrm{~cm}$, so the erased area is $\frac{1}{2} \cdot 20^{2} \pi = 100 \pi \mathrm{~cm}^{2}$.