How tall is Svetlana Kuznetsova?

Svetlana Kuznetsova's listed height is 174 cm (5'9"). She is taller than about 93% of women in Russian Federation.

Where 174 cm sits among women in Russian Federation

Taller than about 93% of women in Russian Federation born in 1985 (average 164.6 cm).

About Svetlana Kuznetsova

Svetlana Kuznetsova is a Russian former professional tennis player, born on 27 June 1985 in Saint Petersburg. She first took part in a WTA Tour event in 2001 and retired from the sport in 2021.

Kuznetsova was ranked as high as world No. 2 in singles and No. 3 in doubles by the WTA, and won 18 singles and 16 doubles titles. Her first major came at the 2004 US Open, where she beat her compatriot Elena Dementieva, and she added the 2009 French Open by defeating Dinara Safina in the final. She was runner-up to Justine Henin at the 2006 French Open and the 2007 US Open. In doubles she won the Australian Open in 2005 with Alicia Molik and in 2012 with Vera Zvonareva, and won her first five doubles titles with Arantxa Sanchez Vicario. She re-entered the top ten in 2016 and reached the semifinals of the WTA Finals.

Kuznetsova's listed height is 174 cm (5'9"). Her results declined after 2010 but slowly recovered through 2016, the year she returned to the top ten.

Frequently asked

How tall is Svetlana Kuznetsova in feet and inches? Svetlana Kuznetsova is 5'9" tall, which is 174 cm, 68.5 inches or 1.74 m.

Is Svetlana Kuznetsova tall for women in Russian Federation? At 174 cm she is taller than about 93% of women in Russian Federation born in 1985, whose average height is 164.6 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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