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.

Re: Need Help with answer...

Quote:

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**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?

Re: Need Help with answer...

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 $\displaystyle A$, triangle $\displaystyle B$, and rectangle $\displaystyle C.$

. . Area of rectangle $\displaystyle A\!:\;3\cdot140 \,=\,420$

. . Area of triangle $\displaystyle B\!:\;\tfrac{1}{2}(50)(7) \,=\,175$

. . Area of rectangle $\displaystyle C\!:\;7\cdot 60 \,=\,420$

. . Total area of a side: .$\displaystyle 420 +175+420 \:=\:1015\text{ ft}^2$ .[1]

The area of *two* sides is: .$\displaystyle 2 \times 1015 \:=\:2030\text{ ft}^2$

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

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

The "floor" is composed of three rectanges.

. . The first has area: .$\displaystyle 30 \times 70 \:=\:2100\text{ft}^2$

. . The second has area: .$\displaystyle \sqrt{2549} \times 70 \:=\:70\sqrt{2549}\text{ ft}^2$ .**

. . The third has area: .$\displaystyle 60\times 70 \:=\:4200\text{ ft}^2$

Total area of the floor: .$\displaystyle 9240 + 70\sqrt{2549}\text{ ft}^2$

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

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].

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

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

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