Respuesta :
Answer:
52 feet by 13 feet.
Step-by-step explanation:
Given:
The length of a rectangular pool is to be four times its width.
Sidewalk is 6 feet wide will surround the pool.
If a total area of 1600 square feet has been set aside for the area of the pool and the sidewalk.
Question asked:
What are the dimensions of the pool?
Solution:
Let width of a rectangular pool = [tex]x[/tex]
Then the length of a rectangular pool will be = [tex]4x[/tex]
[tex]Area\ of \ rectnagle= length\times breadth[/tex]
[tex]=x\times4x\\ \\ =4x^{2}[/tex]
Combined length of pool and the sidewalk = length of pool + width of sidewalk + width of sidewalk
Combined length of pool and the sidewalk = [tex]4x+6+6=4x+12[/tex]
Similarly, Combined width of pool and the sidewalk = [tex]x+6+6=x+12[/tex]
Combined area of pool and the sidewalk = 1600
[tex]Combined\ lengthh\times Combined\ r=breadth=1600\\ \\ (4x+12)(x+12)=1600\\ \\ 4x(x+12)+12(x+12)=1600\\ \\ 4x^{2} +48x+12x+144=1600\\ \\ 4x^{2} +60x+144=1600[/tex]
Subtracting both sides by 144
[tex]4x^{2} +60x=1456\\ \\ 4x^{2} +60x-1456=0\\ \\ Taking\ 4\ as\ common\\ \\ x^{2} +15x-364=0\\ \\ x^{2} +28x-13x-364=0\\ \\ x(x+28)-13(x+28)=0\\ \\ x+28=0,x-13=0\\ \\ x=-28,x=13[/tex]
Since dimensions can never be in negative, hence [tex]x=13[/tex]
Width of a rectangular pool = [tex]x[/tex] = 13 feet
The length of a rectangular pool will be = [tex]4x[/tex] = [tex]4\times13=52\ feet[/tex]
Therefore, the dimensions of the pool are 52 feet by 13 feet.
