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Jill has a swimming pool in her backyard. It is shaped like a rectangle and measures approximately 16.3 feet wide and 26.2 feet long. It is an average of 5 feet deep. During a few hot weeks during the summer, some water evaporates from the pool, and Jill needs to add 8 inches of water to the depth of the pool, using her garden hose. Although her water pressure varies, the water flows through Jill’s garden hose at an average rate of 10 gallons/minute.

How many liters will Jill need to add to her pool to return the water level to its original depth? How many gallons of water is this? Reminder: Volume = Length x Width x Depth

Respuesta :

Answer:

The dimensions of swimming pool are:

Length = 26.2 feet

Width = 16.3 feet

Height = 5 feet

During a few hot weeks during the summer, some water evaporates from the pool, and Jill needs to add 8 inches of water to the depth of the pool.

We will convert 8 inches o feet.

1 inch = 1/12 feet

So, 8 inch = [tex]8/12=0.66[/tex] feet

Now we have to tell how much water will be there in 0.66 feet depth.

So, volume = [tex]16.3\times26.2\times0.66=281.86[/tex] cubic feet.

1 cubic feet has 7.480 gallons

So, 281.86 cubic feet will have [tex]281.86\times7.480=2108.31[/tex] gallons of water.

The volume is the product of the area of the top of the pool and the

height of water added which is approximately 8,062 liters.

Response:

  • The volume of water Jill needs to add is 8,062 liters
  • The volume in gallons is 2,129.8 gallons

How can the required volume that Jill  needs to add be calculated?

The given dimensions of the pool are;

Width of the pool = 16.3 feet

Length of the pool = 26.2 feet

Height of water to be added = 8 inches

Required:

The volume of water to be added in liters

Solution:

First part

1 feet = 12 inches

Therefore;

[tex]8 \ inches = \dfrac{8}{12} \ feet = \mathbf{ \dfrac{2}{3} \ feet}[/tex]

Therefore;

Volume of water added is therefore;

Volume, V = Length × Width × Depth

[tex]V = 16.3 \times 26.2 \times \dfrac{2}{3} = \mathbf{284\frac{53}{75}}[/tex]

Therefore;

The volume added, V = [tex]\mathbf{284\frac{53}{75}}[/tex] ft.³

1 ft.³ = 28.31685 L

Therefore;

[tex]284\frac{53}{75} \, ft.^3 = 284\frac{53}{75} \times 28.31685 \ L \approx \mathbf{8062 \, L}[/tex]

  • The volume of water that Jill needs to add to her pool is approximately 8062 L.

Second part

1 L = 0.264172 gallons

8062 L = 8062 × 0.264172 gallons ≈ 2129.8 gallons

  • The volume of water added to the pool in gallons is approximately 2,129.8 gallons

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