Surface area of pool, given dimensions

Unseenbeauty28 said:
I couldnt figure out how to post the picture on here
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Note: The image is not necessary, as the exercise may be described in words:

A pool is rectangular in shape, with a half-circle at one end. The length is 80 feet; the width is 30 feet; the half-circle is appended to one of the thirty-foot sides, so that end is curved. Find the surface area of the pool.
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Note: I see no point to the "depth" notation on your graphic...?

To find the area, find the area of the rectangle, find the area of the half-circle, and add the two.

Eliz.
 
Unseenbeauty28 said:
I actually need surface area, not just area. therefore thats why i needed the depth of 4ft.
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"Area" is flat. "Surface area" is the area of the surface, rather than, say, the area of the sides (if, for instance, the re-lining of the pool were being considered).

I can conceive of no "area" that would be "surface" (being the top of the water) and that would require the depth. Are you perhaps actually supposed to find the area of the sides of the pool? Or the sides and bottom of the pool? Or are you actually supposed to find the volume of the water?

Please reply with clarification, showing what you have tried so far. Thank you.

Eliz.
 
i have to find the surface area of the whole pool, i think. I was given the information that the height was 30ft, the width was 80 ft, and the depth was 4ft. and i was given the shape of the pool, in which i showed in the picture.

what i understand to do is.

take 4*pi*r^2 = 2,826
pi= 3.14 r=15
and for half a circle it would be = 1413

and for my surface area of the rectangle part i did:

2ab+2bc+2ac= 5,680
a=4 b=3 c=80
and added them together to get = 7,093
 
Re: Surface area

Hello, Unseenbeauty28!

Code: 
                A         65            B
              * *-----------------------*
          *     :                       |
        *       :                       |
       *        :15                     |
                :                       |
      *         :                       |
    E * - - - - *O                      | 30
      *   15    :                       |
                :                       |
       *        :15                     |
        *       :                       |
          *     :                       |
              * *-----------------------*
      D - - - - - - - 80  - - - - - - - C

I will assume that there is a semicircle at the left end of the pool.

The entire length of the pool is 80 feet.
The width of the pool is 30 feet.
The radius of the semicircle is: \(\displaystyle \:OA \,=\,OE \,=\,15\)
Hence, the length of the rectangle is: \(\displaystyle \:AB \,=\,65\) ft


If they are asking for surface area of the water:
. . you know the radius of the semicircle; find its area;
. . you know the length and width of the rectangle; find the area.


Did they want the surface area of the pool
. . like how much paint it would take to cover the interior of the pool?

The curved wall has an area of: \(\displaystyle \:\frac{1}{2}\pi(30)\,\times\,4 \:=\:60\pi\) ft²

The end wall is: \(\displaystyle \:30\,\times\,4\:=\:120\) ft²

There are two side walls: \(\displaystyle \:2\,\times(65\times4) \:=\:520\) ft²

The floor has area: \(\displaystyle \:\frac{1}{2}\pi(15^2)\,+\,(65\times30)\:=\:\frac{225}{2}\pi\,+\,1950\) ft²


The total area (to be painted): \(\displaystyle \:60\pi\,+\,120\,+\,520\,+\,\frac{225}{2}\pi\,+\,1950\)

or about: \(\displaystyle \:\frac{345}{2}\pi\,+\,2590 \;\approx\;3132\) ft²

 
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