# A man built a walk of uniform width around a rectangular pool. If the area of the walk is 253...

## Question:

A man built a walk of uniform width around a rectangular pool. If the area of the walk is 253 {eq}ft^2 {/eq} and the dimensions of the pool are 9 feet by 3 feet, how wide is the walk?

## The Area of a Rectangle:

A rectangle is a quadrilateral that has two pairs of parallel sides. The parallel sides of a rectangle are equal, but the adjacent sides are not. The area of a rectangle is the amount of space covered by the figure, and it is calculated as {eq}A = l\times w {/eq}.

The area of a rectangle is given by:

• {eq}A = l\times w {/eq}, where {eq}l {/eq} is the length and {eq}w {/eq} is the width of the rectangle.

Let the width of the walk be {eq}x {/eq}. If the length of the pool is {eq}9\; \rm ft {/eq} and the width is {eq}3\; \rm ft {/eq}, then the length of the pool plus the walk is:

• {eq}9 + 2x {/eq}

And the width of the pool plus the walk is:

• {eq}3 + 2x {/eq}

The area of the walk will be equal to the area of the pool plus the walk minus the area of the pool.

• {eq}(9 + 2x)(3 + 2x) - (9\times 3) {/eq}
• {eq}27 + 24x + 4x^2 - 27 {/eq}
• {eq}24x + 4x^2 {/eq}

If the area of the walk is 253 square foot, then:

• {eq}24x + 4x^2 = 253 {/eq}
• {eq}4x^2+ 24x - 253 = 0 {/eq}

• {eq}x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} {/eq}
• {eq}a = 4, \; b = 24, \; c = -253 {/eq}. Therefore:
• {eq}x = \dfrac{ -24 \pm \sqrt{24^2 - 4(4)(-253)}}{2\times 4} {/eq}
• {eq}x = \dfrac{ -24 \pm 68}{8} {/eq}
• {eq}x = -3\pm 8.5 {/eq}
• {eq}x = 5.5 \; \rm ft, \quad x = -11.5 \; \rm ft {/eq}

Considering the positive value of x:

• {eq}\boxed{x = 5.5\; \rm ft} {/eq}

The width of the walk is 5.5 feet.