How tall is Victoria Azarenka?

Victoria Azarenka's listed height is 183 cm (6'0"). She is taller than about 99% of women in Belarus.

Where 183 cm sits among women in Belarus

Taller than about 99% of women in Belarus born in 1989 (average 166 cm).

About Victoria Azarenka

Victoria Azarenka is a Belarusian tennis player, born on 31 July 1989 in Minsk. She has been ranked as world No. 1 in singles by the WTA for a total of 51 weeks, including as year-end No. 1 in 2012.

Azarenka has won 21 WTA Tour-level singles titles, including two majors at the 2012 and 2013 Australian Opens, which made her the first Belarusian to win a major singles title. She is a three-time US Open finalist, runner-up to Serena Williams in 2012 and 2013 and to Naomi Osaka in 2020, and won the singles bronze medal at the 2012 London Olympics. In mixed doubles she won the 2007 US Open with Max Mirnyi, the 2008 French Open with Bob Bryan and the Olympic gold medal at London 2012. She also has ten WTA doubles titles.

Azarenka is listed at 183 cm (6'0"). She finished with a year-end top 10 singles ranking each year from 2009 to 2013.

Frequently asked

How tall is Victoria Azarenka in feet and inches? Victoria Azarenka is 6'0" tall, which is 183 cm, 72 inches or 1.83 m.

Is Victoria Azarenka tall for women in Belarus? At 183 cm she is taller than about 99% of women in Belarus born in 1989, whose average height is 166 cm.

Sources

Heights are as publicly listed, rounded to the nearest centimetre; listed heights for public figures are often self-reported. Biographical details adapted from Wikipedia under CC BY-SA 4.0.

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