Question 1161779
Let's draw a the circle and the right triangle to start.
The center of the circle will be called {{{O}}}, and the triangle will be {{{ABC}}} the right angle at {{{A}}}, short leg {{{AC}}}, long leg {{{AB}}},
and hypotenuse {{{red(BC)}}} .
If the length of the long leg of the right triangle had not been given, it could be calculated using the Pythagorean theorem as
{{{sqrt((120feet)^(60feet)^2)=sqrt(180)=60sqrt(3)}}}{{{feet=approximately}}}{{{103.92feet}}} when rounded to 2 decimal places.

Point {{{O}}} is the midpoint of line segment {{{red(BC)}}} . ,
because line segment {{{OB}}} is a radius of the circle and so is line segment {{{OC}}} .
The radius of the circle is {{{(1/2)120feet=60feet}}}
{{{drawing(330,330,-66,66,-66,66,
circle(0,0,60),triangle(-51.96,-30,-51.96,30,51.96,-30),
circle(0,0,0.5),rectangle(-51.96,-30,-48.96,-27),
locate(-3,1,O),locate(-15,-30,103.92feet),
locate(-3,-10,red(120feet)),red(triangle(-51.96,30,51.96,-30,0,0)),
locate(-54,-30,A),locate(52,-30,B),locate(-55,35,C),
locate(-51,10,60feet)
)}}}
Now, let's add to the drawing the circular fish tank, with a diameter of 10 feet, that is concentric with the pool.
That will be a small circle, with the same center as the large circle, because concentric means they have the same center.
{{{drawing(330,330,-66,66,-66,66,
circle(0,0,60),triangle(-51.96,-30,-51.96,30,51.96,-30),
circle(0,0,0.5),rectangle(-51.96,-30,-48.96,-27),
locate(-3,1,O),locate(-15,-30,103.92feet),
locate(-3,-10,red(120feet)),red(triangle(-51.96,30,51.96,-30,0,0)),
locate(-54,-30,A),locate(52,-30,B),locate(-55,35,C),
locate(-51,10,60feet), green(circle(0,0,5))
)}}}
I see that half of that fish tank is on the triangular central zone that contains the restaurant, and the other half is over the pool.
I would not call that "inset in the floor of the restaurant", but could be an interesting design, especially if the wall (and maybe part of the bottom of the fish tank are transparent so that the fish can be seen from the pool.
The resort owner, the architect, and the engineer will figure out how to have part of the fish tank in the restaurant and how to support the half that is over the pool.
 
1. The resort owner wants to line the largest circular edge of the pool with 2 inches of 24K gold leaf. Explain how to determine how many linear feet of gold leaf you should plan to cover.
The largest circular edge is the half circumference going from B to C.
That length is {{{(1/2)diameter*pi}}} or {{{pi*radius=pi*60feet=approximately}}}{{{188.50feet}}} .
Assuming what is covered is the top 2 inches of the pool wall, not a 2-inch ring of the horizontal surface around the pool, the length of edging of 2-inch width you should plan to use would be {{{188.50feet}}} (rounded to 2 decimal places).
 
2. What is the area of the largest section of the pool? Explain if you feel this area would be large enough to add a waterslide.
The largest section of the pool would be the half-circle section between diameter {{{red(BC)}}} and the largest circular edge of the pool (the edge part to be decorated with gold leaf).
The surface area of a circle is calculated as {{{pi*radius^2}}} , so the surface area of the largest section of the pool would be 
{{{(1/2)*pi*(60feet)^2=approximately}}}{{{highlight(5654.87)feet^2}}} (5654.87 square feet, rounded to 2 decimal places).
Waterslides come in different sizes, up to huge ones in the most popular water park that attract tourists form all over the world.
I feel that a good size for a luxury resort could be a curvy slide not taller than a playground slide covering a base area of 12 feet by 12 feet. That base area would have a surface area of {{{12feet*12feet=144feet^2}}} , which I feel the largest section of the pool is more than large enough to easily accommodate leaving plenty of space for other people to enjoy the pool without being disturbed.
 
3. Inset in the floor of the restaurant is a circular fish tank with a diameter of 10 feet. The fish tank is concentric with the pool. Describe how to find the center point for that circle.
As explained above the second drawing, the center of the circle representing the fish tank is the same as the center of the circle that represents the pool.
That is point {{{O}}} that can be located as the midpoint of hypotenuse {{{red(BC)}}} as explained above the first drawing.
 
4. If the pool’s depth averages 4 feet, how much water will it take to fill the pool? (Hint: the volume of a cylinder is the area of the base times the depth.)
The volume of a cylinder is the surface area of the base times the height of the cylinder.
the surface area of a circle of {{{radius=60feet}}} is
{{{pi*(60feet)^2=11309.7335}}} square feet (rounded to 4 decimal places.
A cylinder with that base surface area filled to a depth of 4 feet has a volume of
{{{(11309.7335feet^2)(4feet)=highlight(45239.93)}}} cubic feet (rounded to 2 decimal places).
IF the triangular section containing the restaurant and, the entire fish tank are somehow suspended above the pool water, it would take about {{{highlight(45240)}}} cubic feet of water to fill the pool to an average depth of 4 feet.
HOWEVER, IF the triangular section containing the restaurant is firmly supported on solid ground or other solid materials,
with no pool water under that section, it would take less water to fill the pool to an average depth of 4 feet.
In that case the 4-foot high prism with a triangular base under the section of the pool area containing the restaurant will not need to be filled with water.
The volume of water saved would be 4 feet times the area of that right triangle that has legs measuring 60 feet and approximately 103.92 feet.
As those legs are perpendicular to each other one can be considered the base and the other the height of the triangle.
The surface area of the triangle can be calculated as
{{{(1/2)*base*height=(1/2)(60feet)(103.92feet)=3117.60feet^2}}}
A triangular prism with that base and a height of {{{4feet}}} has a volume of
{{{(4feet)(3117.60feet^2)=12470.40feet^3}}} .
That design would require 12470.40 cubic feet of water less that the one with the triangular section suspended above the pool water.
It would require {{{45239.93feet^3-12470.40feet^3=32769.53feet^3}}} or approximately {{{highlight(32770)}}} cubic feet of water to fill the pool to an average depth of 4 feet.
 
5. City ordinances only allow the resort to use 40,000 cubic feet of water in the pool without the fish tank. Show how to calculate the maximum depth the pool can be to stay under the 40,000-cubic-foot threshold.
City engineers may do a quick calculation of the amount of water in a circular pool  with a diameter of 120 feet,
and decide that the maximum depth of the water over such a pool,
with its base surface area of approximately {{{pi*(60feet)^2=11309.73}}} would be 
{{{40000feet^3/11309.73feet^2=approximately}}}{{{(3.54feet)}}} .
They probably would round down to {{{3.5feet}}} to be safe.
 
6. How much water do you need to fill just the pool without the fish tank, if the average depth is 4 feet? Explain the difference in your calculations for questions 5 and 6. Does it change the maximum average depth of your water?
The volume of water in the fish tank can be calculated as the volume of a cylinder with a diameter of {{{10feet}}} , multiplying the base surface area by the fill depth.
The surface area of a circle can be calculated as {{{pi*radius^2}}} and {{{radius=(1/2)*diameter}}} .
For the fish tank, {{{radius=(1/2)*(10feet)=5feet}}} 
The base surface area of the fish tank is {{{pi*(5feet)^2=approximately}}}{{{78.5398feet^2}}}
If filled to a depth of 4 feet the fish tank would hold a water volume of approximately
{{{(78.5398feet^2)(4feet)=314.16ft^3}}} (rounded to 2 decimal places.
That volume of about 314 cubic feet is very small compared to the volumes of the  pool in the two possible designs described in the response to part 4:
about {{{45240}}} cubic feet if the triangular section containing the restaurant and the fish tank is suspended over the pool water, and
about {{{32770}}} cubic feet if the triangular section containing the restaurant and the fish tank is on a solid foundation without any pool water below it.
The water in the fish tank, being less than 1% of the pool water for either design would make an insignificant difference in the total water needed to fill pool with or without fish tank.
The most important difference in the calculations done so far for question 6, and the calculation for questions 5,
is that they ask for different kinds of measurements for different items, calculated for different purposes.
In answering question 5, we calculated a maximum water depth in feet allowed for a cylindrical pool in a certain community.
In answering question 6, so far we have only been asked for and calculated the volume in cubic feet needed to fill a fish tank to a depth of 4 feet.
We can calculate more realistically, the maximum depth that we can fill pool only and pool plus fish tank using only 40,000 cubic feet of water. 
Or we can fill the fish tank to a depth of 4 feet with 314 cubic feet of water and figure out how deeply we can fill the pool, with the rest of the 40,000 cubic water allowance.
Either way, the water used for the fish tank will make little difference. What makes a big difference is the actual design of the pool, as described in the answer to part 4.