Seward Township, Kosciusko County, Indiana
Township in Indiana, United States / From Wikipedia, the free encyclopedia
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.
Quick Facts Country, State ...
Seward Township | |
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Coordinates: 41°06′29″N 85°56′46″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 | 876 ft (267 m) |
Population | |
• Total | 2,280 |
• Density | 73.4/sq mi (28.3/km2) |
Time zone | UTC-5 (Eastern (EST)) |
• Summer (DST) | UTC-4 (EDT) |
FIPS code | 18-68796[3] |
GNIS feature ID | 453840 |
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Seward Township was organized in 1859.[5]