Rams sign Aaron Donald to 6-year, $135M extension
After weeks of growing optimism, frequent discussions, and proximity, the Los Angeles Rams and star defensive tackle Aaron Donald have officially solidified their partnership.
The Rams have finalized a six-year contract extension with Donald worth $135 million, including $87 million guaranteed, as reported by NFL Network Insider Ian Rapoport. The deal includes a $40 million signing bonus and extends Donald’s contract to seven years, with a total value of $141 million. This agreement, later confirmed by the Rams, sets a new benchmark as the richest contract ever for a defensive player, surpassing Von Miller’s $114.5 million deal with the Denver Broncos in 2016.
Donald, previously entering the final year of his contract with a base salary of $6.89 million, will now earn over $22 million annually. His performance since being drafted in the first round in 2014 has cemented his status as one of the NFL’s elite defenders, with 39 sacks, 108 quarterback hits, and 72 tackles for loss.
A three-time All-Pro and four-time Pro Bowler, Donald comes off a stellar 2017 season where he recorded 41 tackles, 11 sacks, and five forced fumbles, earning the NFL Defensive Player of the Year award. He is also one of just four players in league history to earn Pro Bowl selections in each of his first four seasons.
With Donald locked in, the Rams can now focus on showcasing their revamped defense. Key offseason additions include defensive tackle Ndamukong Suh, a three-time All-Pro, as well as cornerbacks Aqib Talib and Marcus Peters, who together boast multiple All-Pro and Pro Bowl honors. This talented defensive lineup positions the Rams as one of the NFL’s top contenders.
Leave a Reply