Seward Township, Kosciusko County, Indiana

Seward Township
Township
Coordinates: 41°06′29″N 85°56′46″W / 41.10806°N 85.94611°W / 41.10806; -85.94611Coordinates: 41°06′29″N 85°56′46″W / 41.10806°N 85.94611°W / 41.10806; -85.94611
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

  1. ↑ "US Board on Geographic Names". United States Geological Survey. 2007-10-25. Retrieved 2008-01-31.
  2. ↑ "American FactFinder". United States Census Bureau. Retrieved 2008-01-31.
  3. 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.
  4. ↑ Biographical and Historical Record of Kosciusko County, Indiana. Lewis Publishing Company. 1887. p. 728.


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