Respuesta :
For this case we have that the pool area without including the walkway isA = (16) * (32) = 512ft ^ 2
Then, we have that the entire area of the pool including the walkway is
Aw = 924 ft2.
(16 + 2w) * (32 + 2w) = 924
512 + 32w + 64w + 4w ^ 2 = 924
512 + 96w + 4w ^ 2 = 924
Solving for w> 0
-412 + 96w + 4w ^ 2 = 0
w = 3.72 ft
answer
the width of the walkway is w = 3.72 ft
Then, we have that the entire area of the pool including the walkway is
Aw = 924 ft2.
(16 + 2w) * (32 + 2w) = 924
512 + 32w + 64w + 4w ^ 2 = 924
512 + 96w + 4w ^ 2 = 924
Solving for w> 0
-412 + 96w + 4w ^ 2 = 0
w = 3.72 ft
answer
the width of the walkway is w = 3.72 ft
Answer:
The width of walkway=3.72 feet
Step-by-step explanation:
We are given that a rectangular pool has a concrete pathway surrounding the pool.
Let x be the width of walkway.
Length of outer rectangle =[tex]32+2x[/tex]
Breadth of outer rectangle =[tex]16+2x[/tex]
Length of pool=32 feet
Breadth of rectangle =16 feet
Area of pool= [tex]32\times 16[/tex]=512 square feet .
The antire area of the pool including the walkway =924 square feet
Area of entire pool including walkway=[tex](32+2x)(16+2x)[/tex]
[tex](32+2x)(16+2x)=924[/tex]
[tex]512+64x+32x+4x^2=924[/tex]
[tex]4x^2+96x+512-924=0[/tex]
[tex]4x^2+96x-412=0[/tex]
Divide the equation by 4 then we get
[tex]x^2+24x-103=0[/tex]
Quadratic formula for quadratic equation
[tex]ax^2+bx+c=0[/tex]
[tex]D=b^2-4ac[/tex]
[tex]x=\frac{-b\pm\sqrtD}{2a}[/tex]
We have a=1, b=24,c=-103
By using quadratic formula we solve quadratic equation
[tex]D=(24)^2-4\times 1\times (-103)[/tex]
[tex]D=576+412=988[/tex]
[tex] x=\frac{-24\pm \sqrt{988}}{2}[/tex]
[tex]x=\frac{-24\pm 31.43}{2}[/tex]
[tex] x=\frac{-24+31.43}{2}[/tex] and [tex]x=\frac{-24-31.43}{2}[/tex]
[tex]x=\frac{7.43}{2} [/tex] and [tex]x={-55.43}{2}[/tex]
x=3.72 and x=-27.7
The value of x=-27.7 is not possible because the width is always a natural number.
Hence, the width of walkway =x=3.72 feet.
