Tennis: Halep defeats Pliskova and Romania’s Fed Cup team leads Czech team 2-1
Feb 10, 2019
Tennis: Halep defeats Pliskova and Romania’s Fed Cup team leads Czech team 2-1.
Romanian tennis player Simona Halep, WTA’s 3rd, defeated Karolina Pliskova (WTA’s 5th), in three sets, 6-4, 5-7, 6-4, on Sunday, and Romania’s Fed Cup team leads the Czech Republic team with 2-1 after the first three matches of the meeting taking place in Ostrava, in the World Group quarterfinals.
Halep won after 2 hours and 37 minutes.
In the matches on Saturday, Karolina Pliskova defeated Mihaela Buzarnescu, 6-1, 6-4, and Simona Halep overcame Katerina Siniakova, 6-4, 6-0.
In the last singles match at Ostrava, Mihaela Buzarnescu meets Katerina Kiniakova. The doubles match will oppose the pairings Irina Begu/Monica Niculescu - Barbora Krejcikova/Katerina Siniakova.
Romania’s team at Fed Cup for the meeting at the weekend is made up of players Simona Halep (WTA’s 3rd), Mihaela Buzarnescu (WTA’s 29th), Irina Camelia Begu (WTA’s 75th), Ana Bogdan (WTA’s 105th) and Monica Niculescu (WTA’s 106th).
The Czech team includes: Karolina Pliskova (WTA’s 5th), Katerina Siniakova (NO 1 in WTA’s doubles ranking), Barbora Krejcikova (No 2 in WTA’s doubles ranking) and Marketa Vondrousova (WTA’s 72nd).
The Czech Republic won both matches with Romania so far in Fed Cup, in 1980 in Western Berlin, 2-1, in the World Group quarterfinals, and in 2016, in Cluj-Napoca, 3-2, in the first round of the World Group. AGERPRES (RO - editor: Marius Tone; EN - editor: Adina Panaitescu)
[Read the article in Agerpres]