Seward Township, Kosciusko County, Indiana

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Seward Township
Map highlighting Seward Township, Kosciusko County, Indiana.svg
Coordinates: 41°06′29″N85°56′46″W / 41.10806°N 85.94611°W / 41.10806; -85.94611 Coordinates: 41°06′29″N85°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
  Total36.27 sq mi (93.9 km2)
  Land34.98 sq mi (90.6 km2)
  Water1.3 sq mi (3 km2)
Elevation
[1]
876 ft (267 m)
Population
 (2010)
  Total2,567
  Density73.4/sq mi (28.3/km2)
Time zone UTC-5 (Eastern (EST))
  Summer (DST) UTC-4 (EDT)
FIPS code 18-68796 [2]
GNIS feature ID 453840

Seward Township is one of seventeen townships in Kosciusko County, Indiana, United States. As of the 2010 census, its population was 2,567 and it contained 1,385 housing units. [3]

Contents

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]

Cities and towns

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References

  1. "US Board on Geographic Names". United States Geological Survey. 2007-10-25. Retrieved 2008-01-31.
  2. "U.S. Census website". 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. Archived from the original on 2020-02-12. Retrieved 2013-05-10.
  4. Biographical and Historical Record of Kosciusko County, Indiana. Lewis Publishing Company. 1887. p. 728.