A rectangular plot of land measures 50 meters by 30 meters. A path of uniform width is built around it, increasing the total area to 1,980 square meters. What is the width of the path?

["Title: How to Calculate the Width of a Uniform Path Around a Rectangular Plot", "When designing a garden, backyard, or outdoor space, one common challenge is determining the dimensions and area increase caused by surrounding a rectangular plot with a uniform path. In this case, a rectangular plot measuring 50 meters by 30 meters is enhanced with a walking path of equal width all around, expanding the total area to 1,980 square meters. This article explains how to calculate the exact width of such a uniform path with a clear, step-by-step solution.", "---", "### Understanding the Problem", "We start with a rectangular plot of:\n- Length = 50 meters\n- Width = 30 meters", "A path of uniform width (let’s call it x meters) runs along the entire perimeter inside the boundary of the plot. This path reduces the center planting or usable area but increases the total area when combined with the plot. After adding the path, the overall area becomes 1,980 m².", "Our goal: Find the value of x, the width of the path.", "---", "### Step 1: Original Area of the Plot", "Calculate the area of the original plot:\n[\n\ ext{Original Area} = 50 \ imes 30 = 1,500 \ ext{ m}^2\n]", "---", "### Step 2: Dimensions After Adding the Path", "Since the path runs uniformly around all sides, it effectively subtracts x meters from both ends of both length and width. Thus, the new inner usable area (excluding path) is smaller, but the total area (plot + path) is measured from outer edge to edge.", "The new outer dimensions become:\n- New length = ( 50 + 2x ) meters (x added on each side)\n- New width = ( 30 + 2x ) meters (x added on each side)", "So, the total area is:\n[\n(50 + 2x)(30 + 2x) = 1,980\n]", "---", "### Step 3: Set Up the Equation and Expand", "Expand the expression:\n[\n(50 + 2x)(30 + 2x) = 50 \cdot 30 + 50 \cdot 2x + 30 \cdot 2x + 2x \cdot 2x = 1,500 + 100x + 60x + 4x^2\n]", "Combine like terms:\n[\n50 \ imes 2x + 30 \ imes 2x = 200x + 60x = 260x\n]", "So the equation becomes:\n[\n1,500 + 260x + 4x^2 = 1,980\n]", "---", "### Step 4: Form a Quadratic Equation", "Subtract 1,980 from both sides:\n[\n4x^2 + 260x + 1,500 - 1,980 = 0\n]", "[\n4x^2 + 260x - 480 = 0\n]", "---", "### Step 5: Simplify the Equation", "Divide every term by 4 to simplify:\n[\nx^2 + 65x - 120 = 0\n]", "---", "### Step 6: Solve the Quadratic Equation", "Use the quadratic formula:\n[\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]\nWhere ( a = 1 ), ( b = 65 ), ( c = -120 )", "[\nx = \frac{-65 \pm \sqrt{65^2 - 4(1)(-120)}}{2(1)} = \frac{-65 \pm \sqrt{4,225 + 480}}{2} = \frac{-65 \pm \sqrt{4,705}}{2}\n]", "Calculate ( \sqrt{4,705} ):\nApproximate value:\n( \sqrt{4,705} \approx 68.59 )", "Now compute:\n[\nx = \frac{-65 + 68.59}{2} = \frac{3.59}{2} \approx 1.795\n]\n[\nx = \frac{-65 - 68.59}{2} = \frac{-133.59}{2} \approx -66.8 \quad \ ext{(Discarded, negative width)}\n]", "Rounding to two decimal places:\n[\nx \approx 1.80 \ ext{ meters}\n]", "---", "### Step 7: Verification", "Compute new dimensions:\n- New length: (50 + 2(1.8) = 50 + 3.6 = 53.6) m\n- New width: (30 + 2(1.8) = 30 + 3.6 = 33.6) m\n- Area: (53.6 \ imes 33.6 = 1,797.6) m²\nBut wait—this is less than 1,980 m². There’s a slight mismatch due to rounding.", "Using more precise root ( x = \frac{-65 + \sqrt{4705}}{2} ), calculate accurately:\nActual ( \sqrt{4705} \approx 68.5816 )\n[\nx = \frac{3.5816}{2} = 1.7908 \approx 1.79 \ ext{ meters}\n]", "Now recalculate:\n- New length: (50 + 2(1.79) = 53.58)\n- New width: (30 + 2(1.79) = 33.58)\n- Area: (53.58 \ imes 33.58 \approx 1,698.7) → still off?", "Wait — recheck expansion.", "Double-check equation expansion:\n[\n(50 + 2x)(30 + 2x) = 1,500 + 100x + 60x + 4x^2 = 1,500 + 160x + 4x^2\n]", "Set equal to 1,980:\n[\n1,500 + 160x + 4x^2 = 1,980\n]", "[\n4x^2 + 160x - 480 = 0\n]", "Divide by 4:\n[\nx^2 + 40x - 120 = 0\n]", "Now solve:\n[\nx = \frac{-40 \pm \sqrt{1,600 + 480}}{2} = \frac{-40 \pm \sqrt{2,080}}{2}\n]", "[\n\sqrt{2,080} \approx 45.607\n]", "[\nx = \frac{-40 + 45.607}{2} = \frac{5.607}{2} = 2.8035 \approx 2.80 \ ext{ meters}\n]", "Try this value:\n- New length: (50 + 2(2.80) = 55.6) m\n- New width: (30 + 2(2.80) = 35.6) m\n- Area: (55.6 \ imes 35.6 = 1,978.16 \approx 1,978) — close but not exact.", "Actually, precise solving of:\n[\nx^2 + 40x - 120 = 0\n]\nGives:\n[\nx = \frac{-40 + \sqrt{1,600 + 480}}{2} = \frac{-40 + \sqrt{2,080}}{2}\n]", "[\n\sqrt{2,080} = \sqrt{16 \ imes 130} = 4\sqrt{130} \approx 4 \ imes 11.4018 = 45.607\n]", "[\nx = \frac{-40 + 45.607}{2} = \frac{5.607}{2} = 2.8035 \ ext{ m}\n]", "Now compute:\n[\n(50 + 2 \cdot 2.8035)(30 + 2 \cdot 2.8035) = (55.607)(35.607)\n]", "Calculate:\n(55.607 \ imes 35.607 \approx 1,980.000)", "Confirmed! The correct width is approximately 2.80 meters.", "---", "### Final Answer:", "The width of the uniform path around the 50 m × 30 m rectangular plot, which increases the total area to 1,980 m², is exactly ( \frac{-40 + \sqrt{2080}}{2} \approx 2.80 ) meters.", "---", "### Why This Matters", "Understanding how to calculate path dimensions around rectangular areas helps in landscaping, construction, and property development. It ensures accurate material estimates and efficient space utilization.", "For a 50m × 30m plot with a uniform path of width ~2.80 meters, the total area becomes nearly 1,980 m² — validating both formula and real-world application.", "Use this equation or calculator to quickly compute path width for any rectangular plot:\n[\n( l + 2x )( w + 2x ) = \ ext{Total Area}\n]\nSolve algebraically or numerically for precise results.", "---", "Keywords: rectangular plot area, uniform path width, perimeter expansion, garden path calculation, math for landscaping, quadratic equation area.\nMeta Description: How to calculate the width of a uniform path around a 50m × 30m plot that increases total area to 1,980 m²? Step-by-step solution with exact calculation."]