Ferdinand, Indiana | |
|---|---|
 | |
 Flag  Logo | |
 Location of Ferdinand in Dubois County, Indiana. | |
| Coordinates: 38°13′37″N 86°51′47″W / 38.22694°N 86.86306°W | |
| Country | United States |
| State | Indiana |
| County | Dubois |
| Township | Ferdinand |
| Area | |
| • Total | 2.33 sq mi (6.04 km2) |
| • Land | 2.30 sq mi (5.95 km2) |
| • Water | 0.04 sq mi (0.10 km2) |
| Elevation | 535 ft (163 m) |
| Population (2020) | |
| • Total | 2,157 |
| • Density | 939.05/sq mi (362.64/km2) |
| Time zone | UTC-5 (Eastern (EST)) |
| • Summer (DST) | UTC-5 (EDT) |
| ZIP code | 47532 |
| Area code | 812 |
| FIPS code | 18-22990[3] |
| GNIS feature ID | 2396936[2] |
| Website | http://www.ferdinandindiana.org |
Ferdinand is a town in Ferdinand Township, Dubois County, in the U.S. state of Indiana.[2] The population was 2,157 at the 2020 census.
- ^ "2020 U.S. Gazetteer Files". United States Census Bureau. Retrieved March 16, 2022.
- ^ a b c U.S. Geological Survey Geographic Names Information System: Ferdinand, Indiana
- ^ "U.S. Census website". United States Census Bureau. Retrieved January 31, 2008.