A square garden is surrounded by a path of uniform width. If

swerve wrote:A square garden is surrounded by a path of uniform width. If the path and the garden both have an area of x, then what is the width of the path in terms of x?

$$\text{A. } x\sqrt{2}$$
$$\text{B. } 2\sqrt{x}-\sqrt{2}$$
$$\text{C. } \frac{\sqrt{2}}{2}-\frac{x}{4}$$
$$\text{D. } x\sqrt{2}-\frac{x}{2}$$
$$\text{E. } \frac{\sqrt{2x}}{2}-\frac{\sqrt{x}}{2}$$

The OA is E

Source: Magoosh

Say the length of the width = w

Area of the path = Area of the path incl. garden – Area of the garden

x = Area of the path incl. garden – x

Area of the path incl. garden = 2x

Area of the path incl. garden = (length of the garden + 2*width of the path)^2

Length of the garden = Square root of the area of the garden = Square root of x

Length of the garden = √x

Thus, the area of the path incl. garden = (√x + 2w)^2

=> 2x = (√x + 2w)^2

√2.√x = √x + 2w

2w = √2.√x – √x

=> w = √2x/2 – √x/2

The correct answer: E

Hope this helps!

-Jay
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swerve wrote:A square garden is surrounded by a path of uniform width. If the path and the garden both have an area of x, then what is the width of the path in terms of x?

$$\text{A. } x\sqrt{2}$$
$$\text{B. } 2\sqrt{x}-\sqrt{2}$$
$$\text{C. } \frac{\sqrt{2}}{2}-\frac{x}{4}$$
$$\text{D. } x\sqrt{2}-\frac{x}{2}$$
$$\text{E. } \frac{\sqrt{2x}}{2}-\frac{\sqrt{x}}{2}$$

The OA is E

Source: Magoosh

Since the square garden has an area of x, its side length is √x. Since the square garden is surrounded by a path of uniform width, the shape of the path and garden combined is also a square. We can let the width of the path = n, and thus the side length of the square that is the path and garden combined is √x + 2n. Since the total area of the path and garden is x + x = 2x, we have:

(√x + 2n)^2 = 2x

Taking the square root of both sides, we have:

√x + 2n = √(2x)

2n = √(2x) - √x

n = √(2x)/2 - √x/2

Answer: E