CPhill

                              C

            B                             D

                           H

                          144

A  66                                                 B

This is impossible  ....the sum  of the angles in  triangle AHB  > 180°

Note that  triangle ADO  is similar to triangle AEC

AO / DO  =  AC / EC

AO  = 3 - r

DO  = r

AC =  sqrt  [ 3^2 + 3^2 ] =  3sqrt [2]

EC  =  3

(3 - r)  / r =  3sqrt (2 )  / 3

  3 - r  = r *sqrt (2)

3 = r [ 1 + sqrt 2]

r = 3 / [1 + sqrt 2] ≈ 1.24

Volume of sphere   = (4/3)(pi)(1.24)^3  ≈  2.54 pi  =  A

Volume of cone = (1/3) pi (3^2) * 3  = 9pi  =  B

A / B  ≈  2.54 / 9  =   127 / 450

Number of  diagonals  of a regular polygon  = n ( n - 3)  / 2     where n is the number of sides

So

(n) ( n - 3) / 2 =  2n

n^2 - 3n  =  4n

n^2 - 7n =  0

(n) ( n - 7)  = 0

n - 7  = 0 

n =  7

Area of a hexagon  =  6 (1/2) s^2  sin (60°)    where s is the side length

3 =  6 (1/2) s^2 *sqrt (3)  / 2

3 = 3 s^2 * sqrt (3)  / 2

s^2  = 2 / sqrt (3)

s = sqrt  [ 2  / sqrt (3) ]   ≈ 1.075 

Using the square of the distances to equate R^2

(x - 22)^2 + (y - 15)^2  = (x -22)^2  + ( y + 29)^2

y^2 - 30y + 225  =  y^2 + 58y + 841

-30y - 58y  =  841  - 225

-88y = 616

y = -616 / 88  = -7

To find x

(x -22)^2 + (-7-15)^2  = (x + 25)^2 + ( -7 -0)^2

x^2 - 44x + 484 + 484  = x^2 + 50x + 625 + 49

-44x + 968 = 50x + 674

968 - 674 =  50x + 44x

294 = 94x

x = 294/94 =  147 / 47

Center  = ( 147/47 , -7)

3.  Let  QR   =   8 + x

Let M  be the point of tangency with the radius of the large circle  and N be the same  with the smaller circle

We have similar triangles    MPR    and NQR

PR / MP =   QR / QN

(28 + x)  / 12  =  (8 + x) /  8

8(28+x)  = 12 ( 8 + x)

224 + 8x  = 96 + 12x

224  - 96 =   4x

128  = 4x

128/4  = x  =  32

QR =  8 + 32  =  40

5D =   the man hours for  the first job where D is the number of days

6(D  - 4)   = the man hrs if  one additional person is hired

Since the man hours are equal

5D = 6D - 24

D  = 24

5D  = 5 (24)  =   120 man hours

So ....let N be the number of  workers needed to complete the job 20 days earlier

N ( 24 - 20)   = 120

4N  = 120

N =  30

30  - 5  =  25   additional workers should be hired 

The smaller  gear  completes  72 revs per minute

This  is     72 * 2pi * 3  =  432pi  cm / min

They both will have the same linear velocity   

They will have different angular velocities

Exactly correct, hairyberry  .....

The shaded area always  = 25 pi   for  any R, r   such that

R^2  - r^2  =  5^2

See my response below

2.  This  problem  has no unique solution for R , r

But it does have a definite solution for the shaded area

R = radius of  large circle

r = radius of  small circle

R^2  - r^2   = (chord length / 2)^2

R^2 - r^2  =  5^2

One obvious  integer solution is  that R  = 13  and r =  12  since R^2 - r^2 =  5^2

Shaded area =  pi ( 13^2 - 12^2)  =  pi [ 169 -144 ]  =  25 pi

But

R^2  - 5^2  =  r^2

If we let  R = 20

And r  =  sqrt (375)

The shaded area =  pi (20^2 - (sqrt 375)^2 )  =  pi [ 400 - 375] =  25 pi

In general, any  R^2 - r^2   producing 5^2 will work