Beulah Beach, Ohio

From Wikipedia, the free encyclopedia
Jump to: navigation, search
Beulah Beach, Ohio
Census-designated place
Beulah Beach is located in Ohio
Beulah Beach
Beulah Beach
Coordinates: 41°23′31″N 82°26′32″W / 41.39194°N 82.44222°W / 41.39194; -82.44222Coordinates: 41°23′31″N 82°26′32″W / 41.39194°N 82.44222°W / 41.39194; -82.44222
Country United States
State Ohio
County Erie
Township Vermilion
Area[1]
 • Total 0.05 sq mi (0.12 km2)
 • Land 0.05 sq mi (0.12 km2)
 • Water 0 sq mi (0 km2)
Elevation[2] 605 ft (184 m)
Population (2010)[1]
 • Total 53
 • Density 1,155/sq mi (445.8/km2)
Time zone Eastern (EST) (UTC-5)
 • Summer (DST) EDT (UTC-4)
ZIP code 44089
Area code(s) 419 / 567
FIPS code 39-06208[1]
GNIS feature ID 1048520[2]

Beulah Beach /ˈbjuːlə ˈb/[3] is an unincorporated community and census-designated place located adjacent to Lake Erie in Erie County, Ohio, United States.[4] As of the 2010 census it had a population of 53.[1] It is located within Vermilion Township.

Beulah Beach had its start in 1920 as a Christian commune.[5]

Geography[edit]

Beulah Beach is located in eastern Erie County at 41°23′31″N 82°26′32″W / 41.39194°N 82.44222°W / 41.39194; -82.44222,[6] in the western part of Vermilion Township. It is about 600 ft (180 m) above sea level.[7]

U.S. Route 6 forms the southern edge of the community, leading northeast 5 miles (8 km) to the city of Vermilion and west 6 miles (10 km) to Huron. Cleveland is 46 miles (74 km) to the east.

References[edit]