A rectangular field has a length 3 times its width. If the perimeter is 320 meters, what is the area of the field?
["How a Rectangular Field with Length Three Times Its Width Creates a Perfect Area: Insights Behind the Perimeter Puzzle", "Ever paused while solving a math riddle about a rectangular field—longer than it is wide, with exact measurements? This classic geometry question often surfaces in online learning, test prep, and curiosity-driven reading. Today, we explore one such puzzle: A rectangular field has a length three times its width, and the perimeter measures 320 meters. What is the area? Understanding this problem reveals not only a precise solution but also a broader pattern supporting design, agriculture, and property planning across the U.S. Even without technical jargon, this query reflects genuine interest in spatial reasoning and real-world applications.", "## Why This Rectangular Field Puzzle Is Trending in the U.S.", "Geometry problems like this aren’t just classroom exercises—they resonate with growing audiences curious about design efficiency, land use, and architecture. In recent years, homeowners, real estate planners, and educators increasingly engage with spatial puzzles tied to practical living spaces. The “field” metaphor blends seamlessly with discussions about farmland, backyard renovations, or even event venue layout.", "Moreover, mobile users seeking quick, informative answers favor clear, scannable content—exactly what a well-structured explanation of this field model delivers. With perimeter and area calculations transcending niche math classrooms, this topic naturally gains traction across search and Discover feeds targeting user intent: How do dimensions affect space? Can we compute area from perimeter alone?", "## How to Solve: Finding Area from Perimeter and Width Ratio", "To find the area of a rectangle where the length is three times the width and the perimeter is 320 meters: \nWe start with the perimeter formula for a rectangle:", "$$\nP = 2 \ imes (\ ext{length} + \ ext{width})\n$$", "Let the width be $ x $. Then the length is $ 3x $. Substitute into the perimeter equation:", "$$\n320 = 2 \ imes (3x + x) \n\Rightarrow 320 = 2 \ imes 4x \n\Rightarrow 320 = 8x \n\Rightarrow x = 40\n$$", "So, the width is 40 meters and the length is $ 3 \ imes 40 = 120 $ meters.", "Now calculate the area using: \n$$\n\ ext{Area} = \ ext{length} \ imes \ ext{width"]
