A rectangular garden has a length that is 3 meters more than twice its width. If the area of the garden is 165 square meters, what is the width of the garden? - Treasure Valley Movers

A rectangular garden has a length that is 3 meters more than twice its width. If the area of the garden is 165 square meters, what is the width of the garden?

Curious homeowners and gardeners across the U.S. are increasingly turning to precise, math-driven landscaping solutions—especially when planning spaces that balance beauty, function, and budget. This particular garden shape—where the length stretches 3 meters beyond twice the width—has become a favorite for those seeking structured, easy-to-design rectangular plots. When paired with a known area of 165 square meters, curious seekers ask: What’s the true width? Solving this involves a foundation in algebra, but the results reveal more than just numbers—insights that mirror real-world planning challenges.

Why This Garden Shape Is Trending in the U.S.

Modern backyard design increasingly emphasizes functionality without sacrificing aesthetics. A rectangular garden with proportional, calculated dimensions reflects smart space planning common in DIY landscaping communities. Platforms like Pinterest and gardening blogs show growing interest in rectangular layouts because they fit neatly into urban and suburban yards, simplify irrigation and planting beds, and support sustainable planting guides. The data shows that readers researching garden dimensions—especially with exact ratios—often combine practical construction needs with style preferences, making this type of problem highly relevant.

How to Calcate the Width Using Math (A Step-by-Step Breakdown)

To find the width when a rectangular garden’s length is 3 meters more than twice its width and the area is 165 square meters, begin by defining variables:

Let w = width in meters
Then, length = 2w + 3

Area of a rectangle is:
Area = width × length
So:
w × (2w + 3) = 165

Expanding:
2w² + 3w = 165

Rewriting in standard quadratic form:
2w² + 3w – 165 = 0

This equation can be solved using the quadratic formula, a reliable method always relevant in mathematical and real-life problem-solving.
w = [–b ± √(b² – 4ac)] / (2a), where a = 2, b = 3, c = –165

Calculate discriminant:
b² – 4ac = 9 + 1320 = 1329

Width values:
w = [–3 ± √1329] / 4

√1329 ≈ 36.46 (approximate, mobile-friendly decimal precision)
w = (–3 + 36.46)/4 ≈ 33.46 / 4 ≈ 8.37 meters

Only the positive result makes sense physically. This precise width reflects both the constraints