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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.
Seward Township | |
|---|---|
 | |
| 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 |
Seward Township was organized in 1859.[5]
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]
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