A rectangular field has a length 4 meters more than twice its width. If the area is 104 square meters, what is the width?
["Title: How to Find the Width of a Rectangular Field with Given Area and Length Relationship", "When faced with a real-world geometry problem, such as determining the dimensions of a rectangular field, math becomes not only practical but also essential. One common scenario involves relating the length and width through a mathematical expression and using the area to find unknown dimensions. In this article, we’ll solve the problem step-by-step: a rectangular field where the length is 4 meters more than twice its width, and the total area is 104 square meters. What is the width?", "---", "### Understanding the Problem", "We are given:", "- The length (L) is 4 meters more than twice the width (W) → ( L = 2W + 4 )\n- The area (A) of the rectangle is 104 square meters → ( A = L \ imes W = 104 )", "Our goal is to find the width W.", "---", "### Step-by-Step Solution", "Start by substituting the expression for length into the area formula.", "[\nA = L \ imes W\n]\n[\n104 = (2W + 4) \ imes W\n]", "Now expand and rewrite the equation in standard quadratic form:", "[\n2W^2 + 4W = 104\n]", "Subtract 104 from both sides to set the equation to zero:", "[\n2W^2 + 4W - 104 = 0\n]", "To simplify, divide every term by 2:", "[\nW^2 + 2W - 52 = 0\n]", "---", "### Solving the Quadratic Equation", "We now solve the quadratic equation:", "[\nW^2 + 2W - 52 = 0\n]", "Since this equation doesn’t factor easily, we apply the quadratic formula:", "[\nW = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n]", "Here, ( a = 1 ), ( b = 2 ), and ( c = -52 ). Plug in the values:", "[\nW = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-52)}}{2(1)}\n]\n[\nW = \frac{-2 \pm \sqrt{4 + 208}}{2}\n]\n[\nW = \frac{-2 \pm \sqrt{212}}{2}\n]", "Simplify ( \sqrt{212} ). Note that ( 212 = 4 \ imes 53 ), so:", "[\n\sqrt{212} = \sqrt{4 \ imes 53} = 2\sqrt{53}\n]", "Thus:", "[\nW = \frac{-2 \pm 2\sqrt{53}}{2} = -1 \pm \sqrt{53}\n]", "We discard the negative root since width cannot be negative:", "[\nW = -1 + \sqrt{53}\n]", "---", "### Calculating the Width Numerically", "Estimate ( \sqrt{53} ). Since ( 7^2 = 49 ) and ( 8^2 = 64 ), ( \sqrt{53} \approx 7.28 ).", "Then:", "[\nW \approx -1 + 7.28 = 6.28 \ ext{ meters}\n]", "But let’s confirm the exact solution to maintain accuracy.", "---", "### Verifying the exact solution", "We found:", "[\nW = -1 + \sqrt{53}\n]", "Check if this satisfies the area condition:", "[\nW \approx 6.28,\quad L = 2W + 4 \approx 2(6.28) + 4 = 12.56 + 4 = 16.56\n]\n[\nA \approx 6.28 \ imes 16.56 \approx 104\n]", "The solution checks out.", "---", "### Why This Matters (Practical Insight)", "Understanding how to derive and solve equations from real-life geometry problems—like the rectangular field—builds critical thinking and problem-solving skills. Whether you’re designing a farm layout, planning a garden, or constructing a space, knowing how to translate word problems into algebra is invaluable.", "The exact width is:", "[\n\boxed{W = -1 + \sqrt{53} \ ext{ meters}}\n]", "But for practical use, this approximates to 6.28 meters, though the precise mathematical answer remains in the radical form.", "---", "### Final Answer", "The width of the rectangular field is ( -1 + \sqrt{53} ) meters, approximately 6.28 meters. This solution comes directly from solving the quadratic equation derived from the area and length–width relationship.", "---", "Keywords: rectangular field dimensions, solving quadratic equations, area = length × width, width calculation, geometry problem solution, real-world math examples, algebra practice", "Meta Description: Find the exact and approximate width of a rectangular field where length is 4 meters more than twice the width and the area is 104 m². Solve step-by-step with algebra."]
