Warrensville Heights, Ohio

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 / 41.43861°N 81.52333°W / 41.43861; -81.52333
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

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]

References

  1. 1 2 "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. 1 2 U.S. Geological Survey Geographic Names Information System: Warrensville Heights, Ohio
  4. 1 2 "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.


This article is issued from Wikipedia. The text is available under Creative Commons Attribution-Share Alike 4.0 unless otherwise noted. Additional terms may apply for the media files.