Seward Township, Kosciusko County, Indiana

Township in Indiana, United States

41°06′29″N 85°56′46″W / 41.10806°N 85.94611°W / 41.10806; -85.94611CountryUnited StatesStateIndianaCountyKosciuskoGovernment

• TypeIndiana townshipArea

• 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

(2020)^[2]

• Total2,280 • 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 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 was organized in 1859.^[5]

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.^[4]

Cities and towns

Burket

References

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