J. D. Drew Net Worth
J.D. Drew is a former American professional baseball player who has a net worth of $40 million. During his time in Major League Baseball, which lasted from 1999 to 2011, J.D. earned $108 million in salary. During his final five seasons in the league alone, he earned $14 million per year from the Red Sox.
David "Jonathan" Drew was born on November 20, 1975 in Valdosta, Georgia., Drew attended college at Florida State University where he received both the Dick Howser trophy and the Golden Spikes Award in 1997. He made his Major League Baseball debut on September 8, 1998 with the St. Louis Cardinals with whom he acted as right fielder. Following the Cardinals, Drew played for the Atlanta Braves in 2004, the Los Angeles Dodgers from 2005 to 2006, and finished out his career with the Boston red Sox from 2007 to 2011. He has a batting average of .278, a record of 242 home runs, and a total of 795 runs batted in. In 2008, Drew became an All-star as well as the MLB All-star Game MVP. He won the World Series with the Boston Red Sox in 2007. Drew holds strong to his Christian beliefs and married his girlfriend, Sheigh, in 2001. Drew made his last MLB appearance on September 28, 2011 with the Boston Red Sox after his statistics slowly began to deteriorate. He has two brothers by the names of Tim and Stephen who have also played in Major League Baseball, making them Tim and J.D. the first brothers to be drafted in the same year.
J. D. Drew
|Net Worth:||$40 Million|
|Date of Birth:||Nov 20, 1975 (46 years old)|
|Height:||6 ft (1.8542 m)|
|Nationality:||United States of America|