Suchy Dwór, Pomeranian Voivodeship
Suchy Dwór | |
|---|---|
Village | |
Suchy Dwór | |
| Coordinates: 54°34′32″N 18°27′53″E / 54.57556°N 18.46472°E | |
| Country | Poland |
| Voivodeship | Pomeranian |
| County | Puck |
| Gmina | Kosakowo |
| Population | 888 |
Suchy Dwór [ˈsuxɨ ˈdvur] is a village in the administrative district of Gmina Kosakowo, within Puck County, Pomeranian Voivodeship, in northern Poland.[1] It lies approximately 2 km (1 mi) south-west of Kosakowo, 15 km (9 mi) south of Puck, and 26 km (16 mi) north-west of the regional capital Gdańsk.
For details of the history of the region, see History of Pomerania.
References
- ^ "Główny Urząd Statystyczny" [Central Statistical Office] (in Polish). Select Miejscowości (SIMC) tab, select fragment (min. 3 znaki), enter town name in the field below, click WYSZUKAJ (Search)