Warrensville Heights, Ohio
City in Ohio, United States / From Wikipedia, the free encyclopedia
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Warrensville Heights is a city in Cuyahoga County, Ohio, United States. It is an eastern suburb of Cleveland. The population was 13,789 at the 2020 census.
Quick Facts Country, State ...
Warrensville Heights, Ohio | |
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Motto: "The Friendly City" | |
Coordinates: 41°26′19″N 81°31′24″W | |
Country | United States |
State | Ohio |
County | Cuyahoga |
Village incorporated | 1927 [1] |
Incorporated | 1960 [1] |
Government | |
• Type | Mayor-council |
• Mayor | Brad Sellers (D) |
Area | |
• Total | 4.13 sq mi (10.69 km2) |
• Land | 4.12 sq mi (10.68 km2) |
• Water | 0.01 sq mi (0.01 km2) 0.24% |
Elevation | 1,037 ft (316 m) |
Population (2020) | |
• Total | 13,789 |
• Density | 3,344.41/sq mi (1,291.27/km2) |
census | |
Time zone | UTC-5 (EST) |
• Summer (DST) | UTC-4 (EDT) |
Zip code | 44122, 44128 |
Area code | 216 |
FIPS code | 39-80990[4] |
GNIS feature ID | 1047579[3] |
Website | www |
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