Hobart Township, Lake County, Indiana

Hobart Township
Township
Coordinates: 41°32′48″N 87°15′53″W / 41.54667°N 87.26472°W / 41.54667; -87.26472Coordinates: 41°32′48″N 87°15′53″W / 41.54667°N 87.26472°W / 41.54667; -87.26472
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

  1. "US Board on Geographic Names". United States Geological Survey. 2007-10-25. Retrieved 2008-01-31.
  2. "American FactFinder". United States Census Bureau. Retrieved 2008-01-31.
  3. 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.


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