A 6,426 sq. ft. modern in Beverly Hills changed hands this month. Its owner, until recently, was Saturday Night Live alum David Spade, who acquired it about two decades ago for $4 million. After Spade listed it in January for $19.995 million, the 1978-built house found a buyer in very short order; only a month elapsed from listing to closing. The quick sale and the final price of $19.5 million are as much indicators of a hot market as the home’s desirability.

The four-bedroom property is located in Trousdale Estates. Its common area is dominated by a large central living area, which is broken into levels — the design mirrors the exterior architecture, which breaks the roofline into a series of monolithic stacked planes. Upper and lower sections are separated by a partial staircase, with a gallery section above. On the ground level, a freestanding fireplace accent wall brings some room-to-room definition to bear, while a bit of interior landscaping rises up from the floor and connects the first and second stories. A thoughtful design, and one which generally hasn’t aged poorly.

Specialty rooms include a lounge with a wet bar; this spot also features a solarium-like extension with floor-to-ceiling glass on three sides. Sliding glass opens the home to its terraces and patios, with the pool area positioned to take in the views of downtown offered by the location; but the .71-acre lot has more space to fill than that, and below this outdoor living area is a tennis court.

Last October, Spade picked up a new residence in Los Angeles, tossing down $13.85 million to close on a 2020-built property — another modern.