Answer :
Answer:
Width of the walkway = 3 feet
Step-by-step explanation:
Dimensions of a pool = 15 feet × 17 feet
After the addition of a cement walkway around the pool,
New dimensions of the pool (including walkway) = (l + 2x)(w + 2x)
Area of the pool and the cement walkway = 483 square feet
(15 + 2x)(17 + 2x) = 483 [Since, l = 15 feet and w = 17 feet]
255 + 30x + 34x + 4x² = 483
4x² + 64x = 483 - 255
4x² + 64x = 228
4x² + 64x - 228 = 0
x² + 16x - 57 = 0
By using quadratic formula,
[tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex]
x = [tex]\frac{-16\pm \sqrt{(16)^2-4(1)(-57)}}{2\times 1}[/tex]
x = [tex]\frac{-16\pm 22}{2}[/tex]
x = 3.0 ft, -19 ft
Since, dimension can't be negative.
Therefore, width of the walkway = 3 ft
The width of the walkway is 3 ft
The pool measures 15 feet by 17 feet. This means the pool is rectangular.
Properties of a rectangle:
- Opposite sides are equal to each other
- Opposite sides are parallel to each other
area of the pool = lw
area = 15 × 17 = 255 ft²
A cement walkway is added around the pool, to both the width and the length.
area of the pool and the cement walkway = 483 ft²
Therefore,
area = lw
The width and length was added on both sides.
let
the side added to the width = x
the side added to the length = x
length = 2x + 17
width = 2x + 15
483 = (2x + 17)(2x + 15)
483 = 4x² + 30x + 34x + 255
4x² + 64x - 228 = 0
x² + 16x - 57
x² - 3x + 19x - 57
x(x - 3) + 19(x - 3)
(x + 19)(x - 3) = 0
x = -19 or 3
we can only use positive value of x. Therefore, x = 3
length = 2(3) + 17 = 23 ft
width = 2(3) + 15 = 21 ft
Therefore, the width of the walkway is 21 - 15 / 2 = 3 ft
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