Hobart Township, Lake County, Indiana
| Hobart Township | |
|---|---|
| Township | |
 | |
| Coordinates: 41°32′48″N 87°15′53″W / 41.54667°N 87.26472°WCoordinates: 41°32′48″N 87°15′53″W / 41.54667°N 87.26472°W | |
| Country | United States |
| State | Indiana |
| County | Lake |
| Government | |
| • Type | Indiana township |
| Area | |
| • Total | 25.98 sq mi (67.3 km2) |
| • Land | 25.47 sq mi (66.0 km2) |
| • Water | 0.51 sq mi (1.3 km2) |
| Elevation[1] | 627 ft (191 m) |
| Population (2010) | |
| • Total | 39,417 |
| • Density | 1,547.4/sq mi (597.5/km2) |
| FIPS code | 18-34132[2] |
| GNIS feature ID | 453414 |
Hobart Township is one of eleven townships in Lake County, Indiana. As of the 2010 census, its population was 39,417 and it contained 16,366 housing units.[3]
Geography
According to the 2010 census, the township has a total area of 25.98 square miles (67.3 km2), of which 25.47 square miles (66.0 km2) (or 98.04%) is land and 0.51 square miles (1.3 km2) (or 1.96%) is water.[3]
References
- ↑ "US Board on Geographic Names". United States Geological Survey. 2007-10-25. Retrieved 2008-01-31.
- ↑ "American FactFinder". United States Census Bureau. Retrieved 2008-01-31.
- 1 2 "Population, Housing Units, Area, and Density: 2010 - County -- County Subdivision and Place -- 2010 Census Summary File 1". United States Census. Retrieved 2013-05-10.
External links
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