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Șepreuș (Hungarian: Seprős) is a commune in Arad County, Romania, is situated on the northern part of the Teuz Plateau, it stretches over 5768 ha. It is composed of a single village, Șepreuș, situated at 63 km from Arad.
Șepreuș | |
|---|---|
  Șepreuș Location in Romania | |
| Coordinates: 46°34′N 21°44′E / 46.567°N 21.733°E | |
| Country | Romania |
| County | Arad |
| Population (2021-12-01)[1] | 2,752 |
| Time zone | EET/EEST (UTC+2/+3) |
| Vehicle reg. | AR |
Population edit
According to the last census, the population of the commune counts 2472 inhabitants, out of which 89.6% are Romanians, 0.8% Hungarians, 9.3% Roma and 0.3% are of other or undeclared nationalities.
History edit
The first documentary record of the locality Șepreuș dates back to 1407.
Economy edit
The economy of the commune is based on agriculture, mainly on growing of grain, maize, sunflower, barley, sugar-beet, vegetable, oil plants, fodder-crop and technical crops.
Tourism edit
Șepreuș is known for its fishponds and its castle built in the 19th century.
References edit
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