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 | |
| 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]
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]