How tall is Ivan Lendl?

Ivan Lendl's listed height is 188 cm (6'2"). He is taller than about 93% of men in the United States.

Where 188 cm sits among men in the United States

Taller than about 93% of men in the United States born in 1960 (average 177.3 cm).

About Ivan Lendl

Ivan Lendl is a Czech and American former professional tennis player and coach. He was born on 7 March 1960 in Ostrava. He is widely regarded as one of the greatest tennis players of all time.

Lendl was ranked world No. 1 in men's singles by the ATP for 270 weeks and finished as year-end No. 1 four times. He won 94 career singles titles, including eight majors, three each at the French Open and US Open and two at the Australian Open, and seven year-end championships. He was runner-up at a further eleven majors, including a record eight consecutive US Open finals. He leads the head-to-head against Jimmy Connors, 22 to 13, and John McEnroe, 21 to 15.

Lendl is listed at 188 cm (6'2"). His game was built around a hard forehand hit with heavy topspin, and his success helped popularise aggressive baseline power tennis. After retirement he coached several players, and he helped Andy Murray win three majors and reach world No. 1.

Frequently asked

How tall is Ivan Lendl in feet and inches? Ivan Lendl is 6'2" tall, which is 188 cm, 74 inches or 1.88 m.

Is Ivan Lendl tall for men in the United States? At 188 cm he is taller than about 93% of men in the United States born in 1960, whose average height is 177.3 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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