Șepreuș
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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 |
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.
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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