1. **Problem statement:** A circular garbage can with diameter 48 cm is wedged into a rectangular corner formed by two perpendicular walls. We need to find: a) The distance from the corner point P to the center of the can (PA). b) The distance from the corner point P to the point of contact of the can with the wall (PB). 2. **Understanding the problem:** The can is tangent to both walls, so the radius of the can is the perpendicular distance from the center to each wall. 3. **Given data:** Diameter $d = 48$ cm, so radius $r = \frac{48}{2} = 24$ cm. 4. **Step a) Find PA (distance from corner to center):** Since the can touches both walls, the center lies at equal distances (radius) from each wall. The corner P is at the intersection of the two walls, so the center forms a right triangle with legs equal to the radius $r$. Using the Pythagorean theorem: $$PA = \sqrt{r^2 + r^2} = \sqrt{2r^2} = r\sqrt{2}$$ Substitute $r=24$: $$PA = 24\sqrt{2}$$ 5. **Step b) Find PB (distance from corner to point of contact on the wall):** The point of contact B lies on the wall, so it is directly horizontal or vertical from the corner. Since the can touches the wall at radius distance from the center, and the center is at $24\sqrt{2}$ from the corner, The distance PB is the horizontal leg of the right triangle formed by P, center, and B. Since the center is at $(24,24)$ cm from the corner (because radius is 24 on each axis), the point B is at $(24,0)$. Therefore: $$PB = 24$$ **Final answers:** - $PA = 24\sqrt{2} \approx 33.94$ cm - $PB = 24$ cm