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 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


This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.