Question

A rectangular swimming pool measures 16 m by 20 m. A path of uniform width is built around the pool. If the area of the path is 100 m², find the width of the path, giving your answer correct to 2 decimal places. 16 20​

Answer 1

To find the width of the path around the rectangular swimming pool, you can set up an equation. Let's assume the width of the path is "w" meters.

The total area of the pool and the path together will be the area of the larger rectangle minus the area of the pool itself:

Total Area = (20 + 2w) meters (length) × (16 + 2w) meters (width)

And you know that the area of the path is 100 m²:

Area of the Path = Total Area - Area of the Pool

100 m² = (20 + 2w)(16 + 2w) - (20)(16)

Now, you can solve for "w":

100 = (20 + 2w)(16 + 2w) - 320

100 = (320 + 32w + 40w + 4w^2) - 320

Simplify:

4w^2 + 72w = 100

4w^2 + 72w - 100 = 0

Now, we have a quadratic equation. You can solve it using the quadratic formula:

w = (-b ± √(b² - 4ac)) / 2a

In this case, a = 4, b = 72, and c = -100. Plug these values into the formula:

w = (-72 ± √(72² - 4 × 4 × -100)) / (2 × 4)

w = (-72 ± √(5184 + 1600)) / 8

w = (-72 ± √(6784)) / 8

w = (-72 ± 82.42) / 8

Now, calculate both possible solutions:

w = (-72 + 82.42) / 8

w ≈ 10.42 / 8

w ≈ 1.3025

w = (-72 - 82.42) / 8

w ≈ -154.42 / 8

w ≈ -19.3025

Since the width of the path cannot be negative, the width of the path around the pool is approximately 1.30 meters (correct to 2 decimal places).

Answer 2

Width = 1.30 meters

Explanation:

To find the width of the path, we can set up an equation using the given information.

Let's denote the width of the path as "x" meters.

The area of the pool with the path around it can be represented as the total area (including the pool and the path) minus the area of just the pool.

This is because the path surrounds the pool.

Total area with path = Area of pool + Area of path

The area of the pool is 16 m × 20 m, or 320 square meters.

So, we have:

Total area with path = 320 m² + 100 m²

= 420 m²

Now, the total area with the path is equal to the overall dimensions (length and width) of the pool and path. The pool's length and width each increase by 2x due to the path (one width added to each side).

Total area with path = (16 + 2x)(20 + 2x)

Now, set up an equation:

420 = (16 + 2x)(20 + 2x)

We can solve this quadratic equation for "x." Start by expanding the right side:

420 = (16 + 2x)(20 + 2x) = 320 + 32x + 40x + 4x²

Combine like terms:

420 = 320 + 72x + 4x²

Subtract 320 from both sides:

4x² + 72x + 320 = 420

Now, simplify:

4x² + 72x - 100 = 0

Divide the entire equation by 4 to make it easier to work with:

x² + 18x - 25 = 0

Now, we can solve this quadratic equation. Wecan use the quadratic formula:

In this case, a = 1, b = 18, and c = -25.

Substitute these values into the formula:

`\sf x = (-18 \pm√( 18^2 - 4(1)(-25)))/(2(1))`

Simplify further:

`\sf x = (-18 \pm √(324 + 100))/(2)`

`\sf x = (-18 \pm √(424 ))/(2)`

Now, we have two possible solutions for "x":

`\sf x = (-18 + √(424 ))/(2) \approx (2.59)/(2) \approx 1.295`

`\sf x = (-18 + √(424 ))/(2) \approx (-38.59)/(2) \approx -19.295`

Since the width of the path cannot be negative, we take the positive solution:

The width of the path is approximately 1.30 meters (correct to 2 decimal places).