Suchy Dwór, Pomeranian Voivodeship
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 kilometres (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.
Suchy Dwór | |
|---|---|
Village | |
  Suchy Dwór | |
| Coordinates: 54°34′32″N 18°27′53″E | |
| Country |  Poland |
| Voivodeship | Pomeranian |
| County | Puck |
| Gmina | Kosakowo |
| Population | 888 |
For details of the history of the region, see History of Pomerania.
The village has a population of 888.
References
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| Other villages | ||
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