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Math Help - Surface area and volume of swimming pool

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    Surface area and volume of swimming pool

    Need help with this problem....anyone can help me??? Thanks !!!



    Find the surface area and volume of swimming pool described below:

    The pool is a rectangular shape: 70 ft x 140 ft
    The shallow end starts at 3 ft deep and continues at this depth for 30 ft before increasing over a span of 50 feet to 10 feet deep.
    Last edited by mr fantastic; November 16th 2011 at 06:22 PM. Reason: Re-titled.
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    Re: Need Help with answer...

    Quote Originally Posted by mommydearest11 View Post
    Need help with this problem....anyone can help me??? Thanks !!!



    Find the surface area and volume of swimming pool described below:

    The pool is a rectangular shape: 70 ft x 140 ft
    The shallow end starts at 3 ft deep and continues at this depth for 30 ft before increasing over a span of 50 feet to 10 feet deep.
    The cross section is a trapezium. How would you work out its area?
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    Re: Need Help with answer...

    Hello, mommydearest11!

    Find the surface area and volume of swimming pool described below:

    The pool is a rectangular shape: 70 ft x 140 ft
    The shallow end starts at 3 ft deep and continues at this depth for 30 ft
    before increasing over a span of 50 feet to 10 feet deep.

    Code:
          : - - - - - - - 140 - - - - - - - - :
          *  *  *  *  *  *  *  *  *  *  *  *  *
         3*                                   *3
          *                A                  *
          *  *  *  *  * - - - - - + - - - - - *
          :           :   *    B  :     C     *
         7:           :       *   :           *7
          *           :           *  *  *  *  *
          : - - 30  - : - - 50  - : - - 60  - :
    Surface Area

    The side is composed of rectangle A, triangle B, and rectangle C.
    . . Area of rectangle A\!:\;3\cdot140 \,=\,420
    . . Area of triangle B\!:\;\tfrac{1}{2}(50)(7) \,=\,175
    . . Area of rectangle C\!:\;7\cdot 60 \,=\,420
    . . Total area of a side: . 420 +175+420 \:=\:1015\text{ ft}^2 .[1]
    The area of two sides is: . 2 \times 1015 \:=\:2030\text{ ft}^2

    The shallow end has a wall: . 3\times 70 \:=\:210\text{ ft}^2

    The deep end has a wall: . 10\times 70 \:=\: 700\text{ ft}^2

    The "floor" is composed of three rectanges.
    . . The first has area: . 30 \times 70 \:=\:2100\text{ft}^2
    . . The second has area: . \sqrt{2549} \times 70 \:=\:70\sqrt{2549}\text{ ft}^2 .**
    . . The third has area: . 60\times 70 \:=\:4200\text{ ft}^2
    Total area of the floor: . 9240 + 70\sqrt{2549}\text{ ft}^2

    Therefore, the total surface area is: . 2030 + 210 + 700 + 6300 + 70\sqrt{2549} \:=\:9249 + 70\sqrt{2549}\text{ ft}^2


    The volume is easier to calculate.
    The volume is the area of the side times the width of the pool.
    . . And we have the area of the side at [1].

    \text{Volume} \:=\:1050 \times 70 \:=\:71,\!050\text{ ft}^3


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    ** .The second rectangle of the "floor".

    The length of this rectangle is the hypotenuse
    . . of a right triangle with sides 7 and 50.
    Hence, the hypotenuse is:. \sqrt{50^2 + 7^2} \:=\:\sqrt{2549}

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