Warrensville Heights, Ohio

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
City
Motto: 
"The Friendly City"
Location in Cuyahoga County and the state of Ohio
Location in Cuyahoga County and the state of Ohio
Coordinates: 41°26′19″N 81°31′24″W
CountryUnited States
StateOhio
CountyCuyahoga
Village incorporated1927 [1]
Incorporated1960 [1]
Government
 • TypeMayor-council
Area
[2]
 • Total4.13 sq mi (10.69 km2)
 • Land4.12 sq mi (10.68 km2)
 • Water0.01 sq mi (0.01 km2)  0.24%
Elevation
[3]
1,037 ft (316 m)
Population
 (2020)[4]
 • Total13,789
 • Density3,344.41/sq mi (1,291.27/km2)
Time zoneUTC-5 (EST)
 • Summer (DST)UTC-4 (EDT)
Zip code
44122, 44128
Area code216
FIPS code39-80990[5]
GNIS feature ID1047579[3]
Websitewww.cityofwarrensville.com

References
  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.
  2. "ArcGIS REST Services Directory". United States Census Bureau. Retrieved September 20, 2022.
  3. U.S. Geological Survey Geographic Names Information System: Warrensville Heights, Ohio
  4. "QuickFacts: Warrensville Heights city, Ohio". United States Census Bureau. Retrieved March 15, 2024.
  5. "U.S. Census website". United States Census Bureau. Retrieved 2008-01-31.


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