Hobart Township, Lake County, Indiana
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]
Hobart Township | |
---|---|
Coordinates: 41°32′48″N 87°15′53″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 | 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 |
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.
- "U.S. Census website". United States Census Bureau. Retrieved 2008-01-31.
- "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 2020-02-12. Retrieved 2013-05-10.
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