How tall is Sofia Kenin?

Sofia Kenin's listed height is 170 cm (5'7"). She is taller than about 77% of women in Russian Federation.

Where 170 cm sits among women in Russian Federation

Taller than about 77% of women in Russian Federation born in 1998 (average 165.3 cm).

About Sofia Kenin

Sofia Kenin, known as Sonya, is an American professional tennis player, born on 14 November 1998 in Moscow. Coached primarily by her father, she developed into a promising junior, reaching No. 2 in the world after winning the Orange Bowl at age 16 and finishing runner-up in the 2015 US Open girls' singles.

Kenin entered the top 100 of the WTA rankings in 2018 as a teenager. She won her first three titles in 2019 and was named WTA Most Improved Player of the Year. By winning the 2020 Australian Open she became American No. 1 and the youngest American to win a Grand Slam women's singles title since Serena Williams in 1999. She finished runner-up at the French Open that year and was named the 2020 WTA Player of the Year. She has won five singles and four doubles titles on the WTA Tour.

Kenin has a listed height of 170 cm (5'7"). She has been ranked as high as world No. 4 in singles and No. 21 in doubles, and her doubles titles include the 2024 Miami Open with Bethanie Mattek-Sands.

Frequently asked

How tall is Sofia Kenin in feet and inches? Sofia Kenin is 5'7" tall, which is 170 cm, 66.9 inches or 1.70 m.

Is Sofia Kenin tall for women in Russian Federation? At 170 cm she is taller than about 77% of women in Russian Federation born in 1998, whose average height is 165.3 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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