A rectangular garden measuring 20 meters by 15 meters is surrounded by a uniform path of width \( x \) meters. If the total area including the path is 504 square meters, what is \( x \)?

["Title: How to Calculate the Width of a Path Surrounding a Rectangular Garden (Garden & Path Area Problem)", "---", "Introduction\nIf you’re designing a rectangular garden surrounded by a uniform path, one of the most common challenges is determining the path width based on total area. In this article, we’ll solve a practical problem: a garden measuring 20 meters by 15 meters with a uniform path of width ( x ) meters around it, resulting in a total area of 504 square meters. We’ll walk through the step-by-step math to find the value of ( x ), making it easy to plan your garden layout and maximize space.", "---", "The Problem Explained\nWe start with a rectangular garden of dimensions:", "- Length = 20 meters\n- Width = 15 meters", "A uniform path surrounds the garden with width ( x ) meters on all sides. This means the total dimensions including the path become:", "- Total Length = 20 + 2x (add ( x ) on both the left and right)\n- Total Width = 15 + 2x (add ( x ) on both the top and bottom)", "The total area including the path is given as 504 square meters.", "---", "Step-by-Step Calculation", "1. Write the expression for total area including the path:\n [\n \ ext{Total Area} = (\ ext{Total Length}) \ imes (\ ext{Total Width}) = (20 + 2x)(15 + 2x)\n ]", "2. Set up the equation using the given total area:\n [\n (20 + 2x)(15 + 2x) = 504\n ]", "3. Expand the left-hand side:\n [\n (20)(15) + 20(2x) + 2x(15) + 2x(2x) = 504\n ]\n [\n 300 + 40x + 30x + 4x^2 = 504\n ]\n [\n 4x^2 + 70x + 300 = 504\n ]", "4. Bring all terms to one side:\n [\n 4x^2 + 70x + 300 - 504 = 0\n ]\n [\n 4x^2 + 70x - 204 = 0\n ]", "5. Simplify the quadratic equation:\n Divide all terms by 2:\n [\n 2x^2 + 35x - 102 = 0\n ]", "6. Solve using the quadratic formula:\n The quadratic equation ( ax^2 + bx + c = 0 ) has solutions:\n [\n x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n ]\n Here, ( a = 2 ), ( b = 35 ), ( c = -102 )", "Compute the discriminant:\n [\n \Delta = 35^2 - 4(2)(-102) = 1225 + 816 = 2041\n ]\n [\n x = \frac{-35 \pm \sqrt{2041}}{4}\n ]", "Approximate ( \sqrt{2041} \approx 45.18 )\n [\n x = \frac{-35 \pm 45.18}{4}\n ]", "Taking the positive root (since ( x ) must be positive):\n [\n x = \frac{-35 + 45.18}{4} = \frac{10.18}{4} \approx 2.545\n ]", "However, let’s check if the equation can be solved exactly.", "Wait — let’s re-verify calculations:", "Actually, double-check expansion:", "[\n (20 + 2x)(15 + 2x) = 300 + 40x + 30x + 4x^2 = 4x^2 + 70x + 300\n ]\n Set equal to 504:\n [\n 4x^2 + 70x + 300 = 504 \Rightarrow 4x^2 + 70x - 204 = 0\n ]", "Try factoring:\n Divide by 2:\n [\n 2x^2 + 35x - 102 = 0\n ]", "Use quadratic formula again:\n [\n x = \frac{-35 \pm \sqrt{35^2 - 4(2)(-102)}}{2(2)} = \frac{-35 \pm \sqrt{1225 + 816}}{4} = \frac{-35 \pm \sqrt{2041}}{4}\n ]", "Since ( \sqrt{2041} ) is irrational, we keep it exact or approximate:", "[\n x \approx \frac{-35 + 45.18}{4} \approx \frac{10.18}{4} \approx 2.545\n ]", "But let’s test ( x = 3 ):", "Total length = 20 + 6 = 26\n Total width = 15 + 6 = 21\n Area = 26 × 21 = 546 → too big.", "Try ( x = 2 ):\n Total length = 24, width = 19 → 24 × 19 = 456 → too small.", "Try ( x = 2.5 ):\n Length = 25, width = 20 → 25 × 20 = 500 → very close.", "Try ( x = 2.545 ):\n 20 + 2(2.545) = 25.09\n 15 + 2(2.545) = 20.09\n Area ≈ 25.09 × 20.09 ≈ 504.0 (matches)", "Therefore, the exact width of the path is:\n [\n x = \frac{-35 + \sqrt{2041}}{4} \approx 2.545 \ ext{ meters}\n ]", "But can we simplify? Let's factor the quadratic:\n Try finding rational root. Try ( x = 3 ): 2(9)+35(3)-102 = 18+105-102=21≠0\n ( x = 2 ): 8 + 70 - 102 = -24\n ( x = 2.5 ): 2(6.25)=12.5, 35×2.5=87.5 → 12.5+87.5-102= -2 → close.\n ( x = 2.54 ): 2(6.4516)=12.903, 35×2.54=88.9 → sum=101.803 – 102 ≈ -0.197\n ( x = 2.55 ): 2(6.5025)=13.005, 35×2.55=89.25 → 102.255 – 102 = 0.255", "So root between 2.54 and 2.55.", "However, in practical landscaping, keeping decimal precision is acceptable.", "---", "Final Answer:\nThe uniform path width ( x ) that results in a total area of 504 square meters is approximately:\n[\n\boxed{x = \frac{-35 + \sqrt{2041}}{4}} \approx 2.545 \ ext{ meters}\n]\nOr rounded to two decimal places:\n[\n\boxed{x \approx 2.55 \ ext{ meters}}\n]", "---", "Why This Matters\nAccurately calculating path dimensions prevents overspending on materials and ensures uniform planting zones. Use this method whenever expanding garden spaces with surrounding walkways.", "---", "Bonus Tips:\n- Always double-check units—here, meters applied consistently.\n- Use the quadratic formula when area involves product of binomials.\n- Round height/width only when designing construction (use exact value in calculations).", "---", "Keywords:\nrectangular garden path width, area of garden with path, 20m by 15m garden, solve path width equation, quadratic garden path problem, rectangular area with uniform border", "---", "Related Reads:\n- How to Determine Pathwidth in Rectangular Gardens\n- Area and Perimeter Calculations for Garden Designs\n- Solving Quadratic Equations in Real Estate and Landscaping", "---", "Unlock your landscaping potential — precise measurements lead to beautiful results!"]