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 is located at 41°26′19″N 81°31′24″W (41.438653, -81.523262).
According to the United States Census Bureau, the city has a total area of 4.14 square miles (10.72 km2), of which 4.13 square miles (10.70 km2) is land and 0.01 square miles (0.03 km2) is water.
As of the 2010 United States Census of 2010, there were 13,542 people, 6,043 households, and 3,696 families living in the city. The population density was 3,278.9 inhabitants per square mile (1,266.0/km2). There were 6,743 housing units at an average density of 1,632.7 per square mile (630.4/km2). The racial makeup of the city was 3.6% White, 93.5% African American, 0.2% Native American, 0.3% Asian, 0.4% from other races and 2.0% from two or more races. Hispanic or Latino people of any race were 1.4% of the population.
There were 6,043 households, of which 31.7% had children under the age of 18 living with them, 24.5% were married couples living together, 31.8% had a female householder with no husband present, 4.9% had a male householder with no wife present, and 38.8% were non-families. 35.8% of all households were made up of individuals, and 12.8% had someone living alone who was 65 years of age or older.