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 13,542 at the 2010 U.S. Census.

Warrensville Heights, Ohio
Motto(s): 
"The Friendly City"
Location in Cuyahoga County and the state of Ohio
Location in Cuyahoga County and the state of Ohio
Location of Ohio in the United States
Location of Ohio in the United States
Coordinates: 41°26′19″N 81°31′24″W
CountryUnited States
StateOhio
CountyCuyahoga
Village incorporated1927 [1]
Incorporated1960 [1]
Government
  TypeMayor-council
  MayorBrad Sellers (D)
Area
  Total4.14 sq mi (10.72 km2)
  Land4.13 sq mi (10.70 km2)
  Water0.01 sq mi (0.03 km2)  0.24%
Elevation
1,037 ft (316 m)
Population
 (2010)
  Total13,542
  Estimate 
(2018[2])
13,216
  Density3,278.9/sq mi (1,266.0/km2)
 census
Time zoneUTC-5 (EST)
  Summer (DST)UTC-4 (EDT)
Zip code
44122, 44128
Area code(s)216
FIPS code39-80990[3]
GNIS feature ID1047579[4]
Websitehttp://www.cityofwarrensville.com/

References


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