Jefferson, South Dakota
city in South Dakota, United States
Jefferson is a city in the southeastern part of the U.S. state of South Dakota. It is located in Union County, and 547 people lived there at the 2010 census.[1] Jefferson became a city in 1895.[2]
Jefferson, South Dakota | |
|---|---|
 Location of Jefferson in South Dakota | |
| Coordinates: Coordinates: 42°36′16″N 96°33′46″W / 42.60444°N 96.56278°W | |
| Country | United States |
| State | South Dakota |
| Counties | Union |
| Area | |
| • Total | 0.50 sq mi (1.29 km2) |
| • Land | 0.50 sq mi (1.29 km2) |
| • Water | 0.00 sq mi (0.00 km2) |
| Elevation | 1,115 ft (340 m) |
| Population (2010)[1] | |
| • Total | 547 |
| • Density | 1,094.0/sq mi (422.4/km2) |
| Time zone | UTC-6 (CST) |
| • Summer (DST) | UTC-5 (CDT) |
References change
- ↑ 1.0 1.1 "American FactFinder". United States Census Bureau. Retrieved 2014-02-16.
- ↑ "SD Towns" (PDF). South Dakota State Historical Society. Retrieved 2014-02-16.
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