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The problem goes as follows

A swimming pool is 15 by 30 ft and a consistent 5 ft deep. You are painting he walls and floor of the pool. If a gallon can of paint covers 250 square feet how many gallons of cans would you need to buy?

How to solve this

tina
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2 Answers2

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You need to compute the total area of the five sides of the pool that you will paint, then divide by the area covered by one gallon to get the number of gallons used. Then round up if necessary because you can only buy full cans.

Ross Millikan
  • 374,822
  • There is no need to round up. – callculus42 Jul 02 '16 at 15:01
  • How to compute the total area – tina Jul 02 '16 at 15:02
  • Each of the five faces is a rectangle. Draw a picture to get the length and width of each. Can you do the bottom? Then each side is 5 feet by one side of the bottom – Ross Millikan Jul 02 '16 at 15:07
  • I am getting confused so do I multiply 5 by 5 – tina Jul 02 '16 at 15:14
  • No. I suggest you take a box and label three of the edges meeting at a point with $5,15,$ and $30$. These are the lengths of each side. All the parallel edges are the same length, so label them. Now look at each face and determine its length and width. From those you can get the area of the face. – Ross Millikan Jul 02 '16 at 15:17
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The total area of the swimming pool is the sum of 5 areas:

  • Two times a wall with width of $15$ ft and depth of $5$ ft.

The area of one wall is $15 \text{ft} \cdot 5\text{ft}=75 \ \text{ft}^2$

Therefore both walls have an area of $2\cdot 75 \ \text{ft}^2=150 \ \text{ft}^2$

  • Two times a wall with length of $30$ ft and depth of $5$ ft.

Similar calculation like above.

  • One times a wall with length of $30$ ft and width of $15$ ft.

The area of this wall is $30 \ \text{ft} \cdot 15\text{ft}=450 \ \text{ft}^2$

Sum up the all 5 areas. And finally divide this sum by $250 \ \text{ft}^2$.

enter image description here

callculus42
  • 30,550
  • I am still confused. How did you get 2? – tina Jul 02 '16 at 15:29
  • Make a sketch. I have made it for you this time. The area A and B are equal. And C and D as well. Thus you calculate the area of one area and multiply it by two. – callculus42 Jul 02 '16 at 15:42
  • @tina Does the sketch help ? Do you have a (intermediate) result ? – callculus42 Jul 02 '16 at 16:07
  • No it does not help because I am confused on why would there be a 1 and a 2 – tina Jul 02 '16 at 16:24
  • Im not sure if I undstand you right. But I try my best. You are confused about the fact that I multiply two times the areas by 2 and one time the area by 1, aren´t you ? First case: The area A is $15\cdot 5$. This area is equal to the area of B. Thus $A+B=A+A=2\cdot A$.The second case: The area C is $30\cdot 5$. This area is equal to the area of D. Thus $C+D=C+C=2\cdot C$, But the floor (E) is only one area and there is no ceiling or something like that. Thus the area of the floor is $30 \cdot 15$ – callculus42 Jul 02 '16 at 16:36
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    Ok I got it thanks – tina Jul 02 '16 at 16:51
  • That´s fine. You´re welcome. – callculus42 Jul 02 '16 at 16:52