How do you solve these math problems and what are the answers?

North High School will have its first graduating class in the year 2001. To celebrate, it has designed watches with a gold leaf sector between the 12 and the 1 as shown.

Next year's graduating class will have gold between the 12 and the 2 as shown. Each of the first twelve graduating classes will have a similar design. This way, when graduates meet other alumni many years from now, they will instantly be able to determine which year they graduated.

As student body president, Susan is concerned with the cost of the watches and needs to tell the jeweler the amount of gold each watch will require. If the radius of the face of the watch is 2 cm, help Susan out by making a table with the areas of gold leaf for each of the first 12 graduating classes.

If telephone poles (18 feet tall) suspend a telephone wire off the ground along the equator, how much longer than the equator would the wire be? (Ignore the sag between poles.)

If telephone poles (18 feet tall) suspend a telephone wire off the ground along the equator, how much longer than the equator would the wire be? (Ignore the sag between poles.) the equator is 7926 miles long.


North High School will have its first graduating class in the year 2001. To celebrate, it has designed watches with a gold leaf sector between the 12 and the 1 as shown.

Next year's graduating class will have gold between the 12 and the 2 as shown. Each of the first twelve graduating classes will have a similar design. This way, when graduates meet other alumni many years from now, they will instantly be able to determine which year they graduated.

As student body president, Susan is concerned with the cost of the watches and needs to tell the jeweler the amount of gold each watch will require. If the radius of the face of the watch is 2 cm, help Susan out by making a table with the areas of gold leaf for each of the first 12 graduating classes.

For the watch question, find the area of the face of the watch.
It should be (pi)r^2= (pi)* (2cm)^2= 12.5664 cm^2, then divide that by 12, for the 12 equal portions the face is divided into and you get 1.0472 cm^2 for each section.
So the area of gold leaf for the 1st class is 1.0472 cm^2, 2nd class is 2*1.0472 cm^2, 3rd class is 3* 1.0472 cm^2, etc.


For the telephone pole question, work backwards using
2(pi)r = circumference to find the radius of the earth (you should get 1261.46 mi). Then and the height of the poles to the radius and use the same equation to find the circumference (length ) of the wire, and subtract the two. I think the difference is 113.097 mi.

Your watch problem:

Calculate the area of the watch face - pi * r^2
1/12th of that is the area of the first sector
Add sectors until you've completed the table

Your wire problem:

The circumference of the earth at the equator is 2 * pi * r
The length of the wire circumscribing the equator is calculated by adding the height of the poles to the radius.

Since the telephones are 18 feet tall and there would be one on each end of the diameter of the sphere, the diameter would be increased by 36 feet. If you slice the earth along the equator, you have a circle.

The diameter of the earth is 7926 mi. = 41,849,280 ft.
The circumference of the earth is 3.14(41849280) = 131,406,739.2 ft.

The diameter of the earth plus the telephone poles is 7,926mi. + 36 ft. = 41849280ft. + 36ft. = 41849316ft.

Now find the circumference of your circle using C = pi * d

C = 3.14(41849316) = 131,406,852.2 ft. for the circumference with the telephone poles.

Now subtract your circumference with the telephone poles - original circumference.

You'll find that the wire would be 113 ft. longer than the equator.

I hope this is correct.