Jefferson, South Dakota

From Simple English Wikipedia, the free encyclopedia
Jump to navigation Jump to search
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
 • Total0.50 sq mi (1.29 km2)
 • Land0.50 sq mi (1.29 km2)
 • Water0.00 sq mi (0.00 km2)
1,115 ft (340 m)
 • 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.