Warrensville Heights, Ohio

From Simple English Wikipedia, the free encyclopedia
Jump to navigation Jump to search
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 / 41.43861°N 81.52333°W / 41.43861; -81.52333Coordinates: 41°26′19″N 81°31′24″W / 41.43861°N 81.52333°W / 41.43861; -81.52333
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/

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.

References[change | change source]

  1. 1.0 1.1 http://www.cityofwarrensville.com/general.htm Archived 2016-03-03 at the Wayback Machine Retrieved 30 December 2006.
  2. "Population and Housing Unit Estimates". Retrieved June 9, 2019.
  3. "U.S. Census website". United States Census Bureau. Retrieved 2008-01-31.
  4. "US Board on Geographic Names". United States Geological Survey. 2007-10-25. Retrieved 2008-01-31.