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Difficulty: MediumPerimeter and Area of Plane Shapes

A rectangular garden measuring 12 m12\text{ m} by 8 m8\text{ m} has a paved path of uniform width x mx\text{ m} constructed inside it along its perimeter. If the area of the remaining inner garden is 60 m260\text{ m}^2, what is the width xx of the path?

  1. 1 m1\text{ m}Answer
  2. B
    2 m2\text{ m}
  3. C
    1.5 m1.5\text{ m}
  4. D
    3 m3\text{ m}

Answer

The width of the path is 1 m1\text{ m}.
The total outer garden area is 12×8=96 m212 \times 8 = 96\text{ m}^2. Subtracting the path width xx from both ends gives inner dimensions (12−2x)(12 - 2x) and (8−2x)(8 - 2x). Equating the inner area (12−2x)(8−2x)=60(12 - 2x)(8 - 2x) = 60 yields x2−10x+9=0x^2 - 10x + 9 = 0. The realistic physical solution is x=1 mx = 1\text{ m}.

Step-by-Step Solution

1
Express the inner dimensions in terms of path width xx
Length = (12−2x) m(12 - 2x)\text{ m}, Width = (8−2x) m(8 - 2x)\text{ m}
The path reduces each side dimension by xx at both ends.
2
Set up the area equation for the inner rectangular garden
(12−2x)(8−2x)=60(12 - 2x)(8 - 2x) = 60
The area of a rectangle is length multiplied by width.
3
Expand and simplify the quadratic equation
96−40x+4x2=60  ⟹  4x2−40x+36=0  ⟹  x2−10x+9=096 - 40x + 4x^2 = 60 \implies 4x^2 - 40x + 36 = 0 \implies x^2 - 10x + 9 = 0
Divide the whole equation by 4 to simplify quadratic terms.
4
Solve for xx by factoring
(x−1)(x−9)=0  ⟹  x=1(x - 1)(x - 9) = 0 \implies x = 1 or x=9x = 9
Since the width xx cannot exceed half of the smaller side (x<4 mx < 4\text{ m}), x=9x = 9 is extraneous.

Key Concept

Perimeter and Area of Rectangles with Uniform Borders
Estimated Time:1m 30s
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Topics

Perimeter and Area of Plane Shapes

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