Hobart Township, Lake County, Indiana

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Hobart Township
Township
Map highlighting Hobart Township, Lake County, Indiana.svg
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[edit]

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[edit]

  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. ^ a b "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[edit]