## Task A-2.6.

In a dining room measuring $6 \mathrm{~m} \times 6 \mathrm{~m}$, one corner is occupied by a rectangular card table with dimensions $2 \mathrm{~m} \times$ $1.5 \mathrm{~m}$. What is the radius of the largest circular rug that can be spread on the dining room floor?
## Solution.

Let's mark the floor plan of the bathroom as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_30_4cdaff2107ef283b1d94g-11.jpg?height=685&width=694&top_left_y=614&top_left_x=612)

Consider the carpet (circle) that can be placed on the bathroom floor. If the carpet does not touch the wall $\overline{B C}$, there is no obstacle to moving it towards that wall. Similarly, if it does not touch the wall $\overline{C D}$, we can move it so that it touches. In other words, if a carpet of a certain radius can fit in the bathroom, it can be placed so that it touches two opposite walls.

This means that the carpet of the largest radius touches the walls $\overline{B C}$ and $\overline{C D}$. Therefore, the center of the carpet lies on the diagonal $\overline{A C}$.

Of course, the carpet of the largest radius also touches the bathtub at point $F$, because otherwise we can increase the size of the carpet.

Let $N$ be the foot of the perpendicular from point $S$ to the line $F G$. Consider the triangle $F N S$ and apply the Pythagorean theorem: $|F N|^{2} + |S N|^{2} = |F S|^{2}$.

Let the radius of the circle/carpet be $r$. Then $|F N| = |F H| - |N H| = 4.5 - r$. Similarly, $|S N| = 4 - r$.

Thus, we have

$$
(4.5 - r)^{2} + (4 - r)^{2} = r^{2}
$$

Simplifying, we get the quadratic equation $r^{2} - 17r + 36.25 = 0$, whose solutions are $r = 2.5$ and $r = 14.5$.