CPhill

It's actually easier to answer (b) first

Let A  =  the amount that Adam started with

Let B  = the amount that Bill started with

Let C = the amount that Carl started with

We know that, since each paid an  equal amount, then

(1/2) A  = (6/7)B = (3/4) C

Which implies that

A =  (2/1)(6/7)B =  (12/7)B

C =  (4/3)(6/7)B = (24/21)B  =  (8/7)B

And we know that  

(1/7)B + (1/4)C   =  44820

So....subbing for C ,we have

(1/7)B   + (1/4)(8/7)B  =  44820

( 1/7)B + (8/28)B = 44820

( 1/7)B + ( 2/7)B =  44820

(3/7)B =  44820      mult both sides by 7/3

B = 44820 (7/3)   = $104580     what  Bill started with

A= (12/7)(104580) = $179280     what Adam started with

C = (8/7)(104580)  = $119520   what  Carl started with

And the flat cost

(6/7)(104580)  + (1/2)(179280)  +  (3/4)(119520)  =

$89640   + $89640   + $89640   = 

$268920

Let A be the total area  of the largest square

The smallest  square occupies [ (2 ) / (2 +9)]A  =   2/11  of the total area  =  (2/11)A

The shaded area occupies  [ (3) / (3 + 7)]A =   3/10 of the total area  =  (3/10)A

Shaded area                                  (3/10)A      (3/10)

____________________   =        ______  =    ____   =   (3/10) (11/2)  =   33/20 =  33 : 20

Area of smallest square                (2/11)A       (2/11)

108 (1/3) =   $36 on  earphones

108 (1/4) =  $27  on the flash drive

He has

108  - 36 - 27   =   $45  left

Call the total number of people at the concert  = N

(2/3)N  wear hats

Then (1/3)N don't wear hats

And of these, (2/5) don't wear T-shirts, either =  (1/3)(2/5)N = (2/15)N

So

(2/15)N =  840

N = (15/2) 840   =  6300 people at the concert

One rev =  2pi * r

So..

2 * pi * (311 mm)   =   ???   mm  

I'm assuming that, as in  your drawing,  AD = 5   (instead of 6)

We can let AM  and BN be the equal height (s) of the tapezoid 

So triangles  AMD  and  BNC are right

And triangles PXD and PYC are right

And   angle AMD = angle  PXD      ( since both = 90° )

And angle BNC = angle PYC         ( since both = 90° )

And angle ADM  = angle XDP

And angle BCN = angle YCP   

So, by AA congruency....triangle AMD is similar to triangle PXD

And triangle BNC is similar to triangle PYC

So  

AM / AD =  PX / PD       (1)

And

BN / BC = PY / PC        (2)

Let   PD = (5/12)DC   and   PC = (7/12)DC

So, by (1)

h / 5 = PX / [(5/12)DC ]    ⇒    h/ 5  = (12/5) PX / DC    ⇒  h = 12PX / DC      

And by (2)

h/ 7 =  PY / [ (7/12)DC ]   ⇒  h / 7  = ( 12/7) PY / DC   ⇒  h =  12PY/DC

Which implies that

PX / DC  =  PY /DC

Which implies that

PX  = PY

-27   = -3 * x

-27   /  -3   =  x

9  =  x

This is impossible

Look  at triangle   BDI

BD =  3    DI  =  3      and  BI   = 6

By the triangle inequality theorem,  BD + DI   must be greater than BI

But

BD + DI  =  BI

So......this inner triangle cannot exist

We  have that

AQ * BQ   =  CQ * DQ

Let CQ = x   and  DQ =  38 - x

So

6 * 14  =  x * (38 - x)

84  = 38x - x^2                  reararrange   as

x^2 - 38x + 84  = 0            

Using the Quadratic Formula   the  minimum for x = CQ  ≈  2.3567

Volume of a square pyramid  =   (1/3) base^2 * height  =

(1/3) ( 5)^2 * 15   =

(1/3) ( 25) * 15  =

125 in^3