Los Angeles Rams defensive tackle Aaron Donald is the NFL's Defensive Player of the Year for 2018.
.@RamsNFL DT @AaronDonald97 is the 2018 Defensive Player of the Year! #NFLHonors pic.twitter.com/JMVQRJmpxP
— NFL (@NFL) February 2, 2019
The Los Angeles Rams have had a dynamic offense this past season under head coach Sean McVay. However, their defense was as solid as ever. Their defensive line is one of the best in the league. Pro-Football Focus ranked them 3rd overall in the entire NFL coming into the playoffs, only behind the Jacksonville Jaguars and Philadelphia Eagles.
Donald this past season led the entire NFL in sacks with 20.5. and tackles for loss with 25. He also had 59 tackles and forced five fumbles. His presence and impact at the point of the attack is what made the Rams one of the best defenses in the entire league. It also allows Dante Fowler and Ndamukong Suh to succeed in rushing the passer as well.
Even though they did allow 24 points per game, which ranks 2oth overall in the league, the team was still able to be very effective. Donald caused many fourth down situations and helped end games early on. His dominance allowed the Rams to blow by teams all season long.
The Rams finished the season with a 13-3 record and won the NFC West Division. They've rolled through the playoffs and beat the New Orleans Saints in the NFC championship game. Without the presence of Donald in the middle, the team wouldn't nearly have an effective pass rush.
They'll now be facing off against the New England Patriots in the Super Bowl this upcoming Sunday.
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