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Hayesville is a city in Keokuk County, Iowa, United States. The population was 50 at the 2010 census.
Hayesville is located at 41°15′52″N 92°14′56″W (41.264570, -92.248880).
According to the United States Census Bureau, the city has a total area of 0.25 square miles (0.65 km2), all of it land.
As of the census of 2010, there were 50 people, 24 households, and 16 families living in the city. The population density was 200.0 inhabitants per square mile (77.2/km2). There were 25 housing units at an average density of 100.0 per square mile (38.6/km2). The racial makeup of the city was 100.0% White.
There were 24 households, of which 16.7% had children under the age of 18 living with them, 58.3% were married couples living together, 4.2% had a female householder with no husband present, 4.2% had a male householder with no wife present, and 33.3% were non-families. 20.8% of all households were made up of individuals, and 12.5% had someone living alone who was 65 years of age or older. The average household size was 2.08 and the average family size was 2.44.
The median age in the city was 55.3 years.