Six-years, $134M with $87M guaranteed for star DT Aaron Donald.
Days after the New York Giants made Odell Beckham Jr. the NFL's highest-paid wide receiver of all-time and the Green Bay Packers made Aaron Rodgers the highest-paid player in league history, the Los Angeles Rams have reached an agreement with Aaron Donald that will make him the highest-paid defensive player ever.
According to multiple reports, Donald and the Rams agreed to a six-year deal worth $134 million, including $87 million guaranteed. That contract surpasses the six-year, $114.5 million (with $70M in guarantees) deal that Von Miller signed with the Denver Broncos in 2016. Donald, the Rams' 13th overall pick in the 2014 NFL Draft, was entering the final year of his rookie contract and scheduled to earn $6.9 million this season. As a result of his new deal, that number will bump up to $8.892, according to ESPN's Adam Schefter.
In his first four seasons, Donald is a four-time Pro Bowl selection and three-time All-Pro, in addition to being named 2014 Defensive Rookie of the Year. He boasts career numbers of 148 tackles, 39 sacks and nine forced fumbles.
The 27-year old defensive tackle has held out of training camp as negotiations were carried out, but he will be ready for the team's Monday night opener against the Oakland Raiders on September 10.
The Rams finished atop the NFC West last season with an 11-5 record and have added defensive tackle Ndamukong Suh as well as cornerbacks Aqib Talib, Marcus Peters and Sam Shield, to replace Robert Quinn, Alec Ogletree and Trumaine Johnson. They also acquired veteran wide receiver Brandin Cooks in a trade with the New England Patriots.