Marie Bouzková is a tennis pro from the Czech Republic.
She stands at a height of 5 feet 11 inches which is equivalent to 180 cm
How tall is Marie Bouzková?
On August 31, 2020, she attained her highest WTA singles ranking of 46. On May 9, 2022, she attained her highest WTA doubles ranking of No. 24.
Marie Bouzková has won three WTA Tour doubles titles so far.
At ITF Women’s Circuit tournaments, Bouzková has won 12 singles titles and three doubles titles.
She won the 2014 US Open girls’ singles title by defeating Anhelina Kalinina in the final. Sesil Karatantcheva defeated her in the opening round of the 2015 Abierto Mexicano Telcel, where she made her WTA Tour debut.
Bouzková became well-known after making it to the 2019 Rogers Cup semifinals.
She defeated Simona Halep, Jeena Ostapenko, and Sloane Stephens before falling to Serena Williams.
She forced the former world No. 1 to three sets despite losing, winning the first set 6-1.
How much does Marie Bouzková weigh?
Marie also weighs 143 In which is equivalent to 65 kg in body.
