Summertown, Georgia
City in Georgia, United States
From Wikipedia, the free encyclopedia
Summertown is a city in Emanuel County, Georgia, United States. The population was 121 in 2020.
Summertown, Georgia | |
|---|---|
Location in Emanuel County and the state of Georgia | |
| Coordinates: 32°44′48″N 82°16′34″W | |
| Country | United States |
| State | Georgia |
| County | Emanuel |
| Area | |
• Total | 0.79 sq mi (2.05 km2) |
| • Land | 0.79 sq mi (2.04 km2) |
| • Water | 0.0039 sq mi (0.01 km2) |
| Elevation | 253 ft (77 m) |
| Population (2020) | |
• Total | 121 |
| • Density | 154/sq mi (59.4/km2) |
| Time zone | UTC-5 (Eastern (EST)) |
| • Summer (DST) | UTC-4 (EDT) |
| ZIP code | 30401 |
| Area code | 478 |
| FIPS code | 13-74348[2] |
| GNIS feature ID | 0323743[3] |
History
Summertown was originally built up as a summer retreat, hence the name.[4] The Georgia General Assembly incorporated Summertown as a town in 1906.[5]
Geography
Summertown is located in northern Emanuel County at 32°44′48″N 82°16′34″W (32.746532, -82.276182).[6] Georgia State Route 56 passes through the east side of the city limits, leading north 5 miles (8 km) to Midville and south 11 miles (18 km) to Swainsboro, the county seat.
According to the United States Census Bureau, Summertown has a total area of 0.77 square miles (2.0 km2), of which 0.004 square miles (0.01 km2), or 0.54%, is water.[7]