# Surface area and volume of swimming pool

• Nov 16th 2011, 05:26 PM
mommydearest11
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.
• Nov 16th 2011, 05:33 PM
Prove It
Quote:

Originally Posted by mommydearest11
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?
• Nov 16th 2011, 06:34 PM
Soroban
Hello, mommydearest11!

Quote:

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:. $\sqrt{50^2 + 7^2} \:=\:\sqrt{2549}$