A rectangular field has a length of 120 meters and a width of 80 meters. A path of uniform width is built inside the field along its perimeter, reducing the fields area by 1,200 square meters. What is the width of the path? - AdVision eCommerce
A rectangular field has a length of 120 meters and a width of 80 meters. A path of uniform width is built inside the field along its perimeter, reducing the field’s area by 1,200 square meters. What is the width of the path?
A rectangular field has a length of 120 meters and a width of 80 meters. A path of uniform width is built inside the field along its perimeter, reducing the field’s area by 1,200 square meters. What is the width of the path?
This exact question—about a perfectly rectangular field with defined dimensions and a perimeter path profoundly reducing usable space—is not just a math puzzle; it’s a growing topic in modern outdoor design and precision landscaping across the United States. As homes, event spaces, and agricultural initiatives increasingly seek efficient land use, questions about optimizing rectangular acreage often arise—especially when unwanted interior space must be “lost” to create a border path.
The core problem is simple: a 120m × 80m field, total originally 9,600 square meters, sees its area shrink to 8,400 square meters—an exact 1,200 sq m reduction—carried entirely by a uniform-width path running along the inside edge.
Understanding the Context
To find the path’s width—let’s call it x—start by calculating the original area:
120 meters × 80 meters = 9,600 m²
With a path width x running inside along all four sides, the inner rectangle loses length and width equally. The inner dimensions become:
Length = 120 – 2x
Width = 80 – 2x
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Key Insights
Thus, the inner area is:
(120 – 2x)(80 – 2x)
We set up the equation:
Original area – inner area = 1,200
9,600 – (120 – 2x)(80 – 2x) = 1,200
Solving this equation reveals a quadratic that yields a precise, safe solution—ensuring clarity and trust for readers navigating this design decision.
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After algebraic simplification, the solution for x emerges as 10 meters—verified through calculation and consistent with practical feasibility.
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