Hobart Township, Lake County, Indiana
Township in Indiana, United States / From Wikipedia, the free encyclopedia
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.[4]
More information Census, Pop. ...
Census | Pop. | Note | %± |
---|---|---|---|
1890 | 2,197 | — | |
1900 | 2,718 | 23.7% | |
1910 | 3,729 | 37.2% | |
1920 | 5,621 | 50.7% | |
1930 | 9,135 | 62.5% | |
1940 | 12,472 | 36.5% | |
1950 | 21,871 | 75.4% | |
1960 | 39,223 | 79.3% | |
1970 | 40,825 | 4.1% | |
1980 | 42,548 | 4.2% | |
1990 | 38,942 | −8.5% | |
2000 | 39,636 | 1.8% | |
2010 | 39,417 | −0.6% | |
2020 | 40,652 | 3.1% | |
Source: US Decennial Census[5] |
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Quick Facts Country, State ...
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 | |
• Total | 40,652 |
• Density | 1,547.4/sq mi (597.5/km2) |
FIPS code | 18-34132[3] |
GNIS feature ID | 453414 |
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