1. Prove the Pythagorean Theorem by a method not used in class.. § 12.1

There are over 260 of them. You should not have had too much trouble finding another one.

2. On the three sides of a right triangle construct semicircles with centers at the midpoints of the sides. Calculate the area of each of the three semicircles. Do you see a relationship?

Do you think it works for other geometric figures?

a

b

c2 2

a

1 a aA

2 2 8

2 2

b

1 b bA

2 2 8

2 2

c

1 c cA

2 2 8

2 2 2SO a b c8

3. Find the ratio of the volume to the surface area of a cube.

It is good to have an easy one once in a while!

V 1

A 6

4. A sphere is circumscribed by a cylinder. Find the ratio of the two surface areas. Find the ratio of the two volumes.

Use unit radius.

Sphere –

Area – 4πr 2 Volume -

Cylinder

Area - 6πr 2 Volume - 2 πr 3

The ratios are the same for Sphere/Cylinder = 2/3

3r3

4

5. Find the volume of a unit regular octagon.

Dissect it into two pyramids. The trick is to find the altitude of the pyramid.

1

√2

h 2 = 1 2 – (√2/2) 2 = √2/2

h

V = (2) (1/3) (1) (√2/2) = √2/3

6. What is the volume of the Great Pyramid of Giza that had a side measure of 756 ft and an altitude of 481 feet? If it took 30 years of 6 day weeks working 10 hours a day, how many cubic feet were put in place each hour?

V = bh/3 = 100,017,216 cubic feet.

756

481

Time = 93600 hours

V/time = 29,370 cubic feet per hour.

That is a volume about the size of over two classrooms per hour!!

7. An auto tunnel through a mountain is being planned. It will be semicircular cylinder with a radius of 30 feet and a length of 5000 feet. How many cubic feet of dirt will have to be removed? If a dump truck has a bed of dimensions 7 feet by 10 feed by 6 feet, how many loads will be required to carry away the dirt?

Volume - (1/2)(5000)(π 302) = 7068583 cubic feet

7068583/420 = 16,830 dump truck loads

8. Investigate the Archimedean solids. What characteristics do they have in common?

Faces are regular polygons.

6. What is the shape of the cylinder with minimum surface area for a given volume?

V = πhr 2 is fixed. Solve for h

h = V/ πr 2

r

h

SA = 2πr 2 + 2πrh and substitute for h.

SA = 2πr 2 + (2πr)(V/ πr 2)

SA = 2πr 2 + 2V/r and take the derivative

0 = 4r - 2V/r 2 and take the derivative