A rectangular garden is 20 meters long and 15 meters wide. If a path of uniform width is built around the garden, increasing the total area to 396 square meters, what is the width of the path? - AIKO, infinite ways to autonomy.
A rectangular garden is 20 meters long and 15 meters wide. If a path of uniform width is built around the garden, increasing the total area to 396 square meters, what is the width of the path?
A rectangular garden is 20 meters long and 15 meters wide. If a path of uniform width is built around the garden, increasing the total area to 396 square meters, what is the width of the path?
In a growing wave of interest around outdoor living spaces, homeowners and garden enthusiasts are rethinking how pathways can enhance both function and beauty. This particular inquiry—about calculating the width of a uniform path around a rectangular 300-square-meter garden that expands total area to 396 square meters—reflects rising curiosity about smart landscape design. As users increasingly seek efficient ways to improve functionality without sacrificing space, the simple math behind shared circle-illusion redesigns is gaining traction. Using precise calculations, even those new to geometry can uncover hidden dimensions that elevate garden usability.
The garden measured at 20 meters by 15 meters covers a base area of 300 square meters. By surrounding it with a uniform path and expanding the total area to 396 square meters, the math reveals exactly how much extra space a consistent walkway adds. This configuration taps into a broader trend: optimal garden layouts that balance open planting zones with accessible paths. The resulting total area rises due to the added buffer, a phenomenon that sparks both practical problem-solving and aesthetic intention.
Understanding the Context
To determine the uniform width, denote the path’s width as x meters. The outer dimensions become 20 + 2x by 15 + 2x. Multiply these:
(20 + 2x)(15 + 2x) = 396
Expanding gives:
300 + 40x + 30x + 4x² = 396
Which simplifies to:
4x² + 70x + 300 = 396
Then:
4x² + 70x – 96 = 0
Dividing through by 2:
2x² + 35x – 48 = 0
Apply the quadratic formula:
x = [–35 ± √(35² + 4·2·48)] / (2·2)
x = [–35 ± √(1225 + 384)] / 4
x = [–35 ± √1609] / 4
√1609 ≈ 40
