Seward Township, Kosciusko County, Indiana
Seward Township | |
---|---|
Township | |
Coordinates: 41°06′29″N 85°56′46″W / 41.10806°N 85.94611°WCoordinates: 41°06′29″N 85°56′46″W / 41.10806°N 85.94611°W | |
Country | United States |
State | Indiana |
County | Kosciusko |
Government | |
• Type | Indiana township |
Area | |
• Total | 36.27 sq mi (93.9 km2) |
• Land | 34.98 sq mi (90.6 km2) |
• Water | 1.3 sq mi (3 km2) |
Elevation[1] | 879 ft (268 m) |
Population (2010) | |
• Total | 2,567 |
• Density | 73.4/sq mi (28.3/km2) |
FIPS code | 18-68796[2] |
GNIS feature ID | 453840 |
Seward Township is one of seventeen townships in Kosciusko County, Indiana. As of the 2010 census, its population was 2,567 and it contained 1,385 housing units.[3]
Seward Township was organized in 1859.[4]
Geography
According to the 2010 census, the township has a total area of 36.27 square miles (93.9 km2), of which 34.98 square miles (90.6 km2) (or 96.44%) is land and 1.3 square miles (3.4 km2) (or 3.58%) is water.[3]
References
- ↑ "US Board on Geographic Names". United States Geological Survey. 2007-10-25. Retrieved 2008-01-31.
- ↑ "American FactFinder". United States Census Bureau. Retrieved 2008-01-31.
- 1 2 "Population, Housing Units, Area, and Density: 2010 - County -- County Subdivision and Place -- 2010 Census Summary File 1". United States Census. Retrieved 2013-05-10.
- ↑ Biographical and Historical Record of Kosciusko County, Indiana. Lewis Publishing Company. 1887. p. 728.
External links
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