Sabina, Ohio

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Sabina is a village in Clinton County, Ohio, United States. As of the 2000 census, the village had a total population of 2,780.

Contents

History

The town of Sabina was laid out by Warren Sabin, after whom it was named, in 1830, on land originally entered by P. Neville. The original plat of the town was recorded on the 6th of December, 1830, and contained thirty-seven lots. The corporation of the town was extended to include additional territory August 9, 1813. In 1859, the town was incorporated, and M. Morris appointed Mayor.

Geography

Sabina is located at 39°29′21″N 83°38′6″W / 39.48917°N 83.635°W / 39.48917; -83.635 (39.489243, -83.635079)[4], along Routes 22 and 3 about ten miles east of Wilmington, the county seat. It is also located within an hour's drive from the Columbus, Dayton, and Cincinnati metro areas.

According to the United States Census Bureau, the village has a total area of 1.3 square miles (3.3 km²), all of it land.

Sabina is located 31 miles south of Springfield and 47 miles southwest of Columbus.

Demographics

As of the census[3] of 2000, there were 2,780 people, 1,075 households, and 762 families residing in the village. The population density was 2,149.6 people per square mile (832.1/km²). There were 1,173 housing units at an average density of 907.0/sq mi (351.1/km²). The racial makeup of the village was 97.48% White, 0.61% African American, 0.29% Native American, 0.43% Asian, 0.07% from other races, and 1.12% from two or more races. Hispanic or Latino of any race were 1.15% of the population.

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