Question

The length of a rectangular pool is to be four times its width and a sidewalk 6ft 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, what are the dimensions of the pool?

Answer 1

To find the dimensions of the rectangular pool, we can set up an equation based on the given information. The width of the pool is 14 feet and the length is 56 feet.

Step-by-step explanation:

To find the dimensions of the pool, we can set up an equation using the given information. Let's assume the width of the pool is x feet. According to the problem, the length of the pool is four times its width, so the length would be 4x feet. We also know that a sidewalk 6ft wide will surround the pool. This means the dimensions of the pool plus the sidewalk are (x+12)ft by (4x+12)ft.

To find the area, we need to multiply the length by the width: (x+12)(4x+12) = 1600. Expanding this equation gives 4x^2 + 48x + 144 = 1600. Rearranging the equation gives 4x^2 + 48x - 1456 = 0.

Now we can solve this quadratic equation. We can either factor it or use the quadratic formula. In this case, factoring is not easy, so we will use the quadratic formula which states that for an equation ax^2 + bx + c = 0, the solutions are given by x = (-b ± sqrt(b^2 - 4ac))/(2a).

For our equation, a = 4, b = 48, and c = -1456. Plugging in these values in the formula, we can find the value of x. After solving, we get x = 14 or x = -26. Since the width cannot be negative, we take x = 14. Therefore, the width of the pool is 14 feet and the length is 4 times the width, which is 56 feet.

Answer 2

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 =
`x`

Then the length of a rectangular pool will be =
`4x`

`Area\ of \ rectnagle= length* breadth`

`=x*4x\\ \\ =4x^(2)`

Combined length of pool and the sidewalk = length of pool + width of sidewalk + width of sidewalk

Combined length of pool and the sidewalk =
`4x+6+6=4x+12`

Similarly, Combined width of pool and the sidewalk =
`x+6+6=x+12`

Combined area of pool and the sidewalk = 1600

`Combined\ lengthh* 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`

Subtracting both sides by 144

`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`

Since dimensions can never be in negative, hence
`x=13`

Width of a rectangular pool =
`x` = 13 feet

The length of a rectangular pool will be =
`4x` =
`4*13=52\ feet`

Therefore, the dimensions of the pool are 52 feet by 13 feet.