Steven Seagal Looking to Sell Scottsdale Home for $3.8M

Steven Seagal Looking to Sell Scottsdale Home for $3.8M

Published: October 3, 2019 | By: American Luxury Staff

90’s action movie star Steven Seagal has adjusted the price of his desert pad once again.

According to public records, the martial artist-turned-actor bought the property in Scottsdale, Arizona, for $3.5 million in 2010, and first listed it in 2012 for $4.25 million. The price has been on a bit of a rollercoaster ride in the intervening years, dipping as low as $3.5 million, and now sits at $3.775 million.

The desert contemporary-style house features a mix of naturalistic, traditional, modern and industrial design elements. The exterior is expressed in stone, glass and copper.

Within, the home establishes a very warm contemporary living style. Warm tones, stone accents, and tile floors are the foundation of the spaces. The open-concept shared living area boasts a soaring ceiling, glass walls, clerestories, stone support pillars, and an open staircase. It’s a stylized approach that has a nice feel, albeit a slightly dated one.

There are five bedrooms and 5.5 baths in the main home; the guest house on the property brings the total number of bedrooms to six, with 7 baths throughout 9,000 square feet of interior living space. The lot is quite large, extending to just under 12 acres above the Chiricahua Golf Course.

The residence is one of several maintained by Seagal, whose film credits include ‘Above the Law’ and ‘Under Siege’, and a slew of ‘direct-to’ flicks including this year’s ‘General Commander’.

Comments Off on Steven Seagal Looking to Sell Scottsdale Home for $3.8M 5494 October 3, 2019 Real Estate October 3, 2019