Drew Hankinson - Height in feet/cm
How Tall Is Drew Hankinson?Drew Hankinson height is 6 feet 7 inches. Or in metric units - 201 centimetres. |
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Name | Drew Hankinson |
Date of birth | December 22, 1983 |
Zodiac sign | Capricorn |
Height | 6ft 7in (201 cm) |
Gender | Male |
Birthplace | Maryland, United States |
Country of nationality | United States of America |
Profession | Wrestler |
Also known as | Festus, Andrew William Hankinson, Imposter Kane |
Lists | Wrestler from United States, Wrestler who are 6ft 7in (201 cm) |
People of the same height
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