Șepreuș

(Redirected from Şepreuş)

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 is located in Romania
Șepreuș
Șepreuș
Location in Romania
Coordinates: 46°34′N 21°44′E / 46.567°N 21.733°E / 46.567; 21.733
CountryRomania
CountyArad
Population
 (2021-12-01)[1]
2,752
Time zoneEET/EEST (UTC+2/+3)
Vehicle reg.AR

Population

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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

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The first documentary record of the locality Șepreuș dates back to 1407.

Economy

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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

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Șepreuș is known for its fishponds and its castle built in the 19th century.

References

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  1. ^ "Populaţia rezidentă după grupa de vârstă, pe județe și municipii, orașe, comune, la 1 decembrie 2021" (XLS). National Institute of Statistics.