Warrensville Heights, Ohio
city in Cuyahoga County, Ohio, United States
Warrensville Heights is a city located in Cuyahoga County, Ohio, United States. It is an East Side suburb of Cleveland. The population was 13,789 at the 2020 U.S. Census.[4]
Warrensville Heights, Ohio | |
---|---|
Motto: "The Friendly City" | |
Coordinates: 41°26′19″N 81°31′24″W / 41.43861°N 81.52333°W | |
Country | United States |
State | Ohio |
County | Cuyahoga |
Village incorporated | 1927 [1] |
Incorporated | 1960 [1] |
Government | |
• Type | Mayor-council |
Area | |
• Total | 4.13 sq mi (10.69 km2) |
• Land | 4.12 sq mi (10.68 km2) |
• Water | 0.01 sq mi (0.01 km2) 0.24% |
Elevation | 1,037 ft (316 m) |
Population | |
• Total | 13,789 |
• Density | 3,344.41/sq mi (1,291.27/km2) |
Time zone | UTC-5 (EST) |
• Summer (DST) | UTC-4 (EDT) |
Zip code | 44122, 44128 |
Area code | 216 |
FIPS code | 39-80990[5] |
GNIS feature ID | 1047579[3] |
Website | www |
References
change- ↑ 1.0 1.1 "Welcome to Warrensville Heights! Live, explore, visit and have fun in the Friendly City". www.cityofwarrensville.com. Archived from the original on February 14, 2004. Retrieved March 9, 2023.
- ↑ "ArcGIS REST Services Directory". United States Census Bureau. Retrieved September 20, 2022.
- ↑ 3.0 3.1 U.S. Geological Survey Geographic Names Information System: Warrensville Heights, Ohio
- ↑ 4.0 4.1 "QuickFacts: Warrensville Heights city, Ohio". United States Census Bureau. Retrieved March 15, 2024.
- ↑ "U.S. Census website". United States Census Bureau. Retrieved 2008-01-31.