Seward Township, Kosciusko County, Indiana

Seward Township is one of seventeen townships in Kosciusko County, Indiana, United States. As of the 2020 census, its population was 2,280 (down from 2,567 at 2010[4]) and it contained 1,357 housing units.

Seward Township
Coordinates: 41°06′29″N 85°56′46″W / 41.10806°N 85.94611°W / 41.10806; -85.94611
CountryUnited States
StateIndiana
CountyKosciusko
Government
 • TypeIndiana township
Area
 • Total
36.27 sq mi (93.9 km2)
 • Land34.98 sq mi (90.6 km2)
 • Water1.3 sq mi (3 km2)
Elevation876 ft (267 m)
Population
 • Total
2,280
 • Density73.4/sq mi (28.3/km2)
Time zoneUTC-5 (Eastern (EST))
 • Summer (DST)UTC-4 (EDT)
FIPS code18-68796[3]
GNIS feature ID453840

Seward Township was organized in 1859.[5]

Geography

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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.[4]

Cities and towns

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References

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  1. ^ "US Board on Geographic Names". United States Geological Survey. October 25, 2007. Retrieved January 31, 2008.
  2. ^ "Census Bureau profile: Seward Township, Kosciusko County, Indiana". United States Census Bureau. May 2023. Retrieved April 2, 2024.
  3. ^ "U.S. Census website". United States Census Bureau. Retrieved January 31, 2008.
  4. ^ a b "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 February 12, 2020. Retrieved May 10, 2013.
  5. ^ Biographical and Historical Record of Kosciusko County, Indiana. Lewis Publishing Company. 1887. p. 728.
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