Șepreuș

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Șepreuș
Commune
Șepreuș is located in Romania
Șepreuș
Șepreuș
Coordinates: 46°34′N 21°44′E / 46.567°N 21.733°E / 46.567; 21.733
Country  Romania
County Arad County
Population (2002)[1] 2,472
Time zone EET (UTC+2)
 • Summer (DST) EEST (UTC+3)

Șepreuș 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.

Coordinates: 46°34′N 21°44′E / 46.567°N 21.733°E / 46.567; 21.733

References[edit]

  1. ^ Romanian census data, 2002; retrieved on March 1, 2010