Jefferson, South Dakota

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Jefferson, South Dakota
Location of Jefferson in South Dakota
Location of Jefferson in South Dakota
Coordinates: Coordinates: 42°36′16″N 96°33′46″W / 42.60444°N 96.56278°W / 42.60444; -96.56278
CountryUnited States
StateSouth Dakota
CountiesUnion
Area
 • Total0.50 sq mi (1.29 km2)
 • Land0.50 sq mi (1.29 km2)
 • Water0.00 sq mi (0.00 km2)
Elevation
1,115 ft (340 m)
Population
 (2010)[1]
 • Total547
 • Density1,094.0/sq mi (422.4/km2)
Time zoneUTC-6 (CST)
 • Summer (DST)UTC-5 (CDT)

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]

References[change | change source]

  1. 1.0 1.1 "American FactFinder". United States Census Bureau. Retrieved 2014-02-16.
  2. "SD Towns" (PDF). South Dakota State Historical Society. Retrieved 2014-02-16.