CPhill

radius =  8*15*12  /  4*Area

radius = 1440  /  [ 4 *   sqrt  [ 17.5 (17.5 - 8)(17.5 - 15)(17.5 -12) ]  ]  ≈  7.53 

a^3 - b^3    = 1

a      b         ab      a + b

1      0          0           1

0     -1          0          -1

Slope of line =  (-7-2) / ( 4 - -8)  = -9 / 12  =  -3/4

Using the slope and ( 4, -7)   the equation of the line is

y = (-3/4) ( x - 4)  - 7

y = -3/4x + 3  - 7         mulriply through by 4

4y = -3x - 16

3x + 4y = -16

Simplify L1

-3y = -11x - 14

y = (11/3)x - 14

The slope of a perpendicular line is -3/11

Using (5, -6), the equation of the perpendicular line L2  is

y = (-3/11)(x - 5)   -6

y = (-3/11)x + 15/11 - 66/11

y = (-3/11)x - 51/11

The radius of the circle can be  found as

25 pi = pi *r^2

25 = r^2

r = 5

Ler R be the radius of the sphere

R^2 =   [ 5^2 + r^2 ] =  [ 5^2 + 5^2 ]  =  50

Surface area of sphere  =

4 pi R^2  = 

4 pi * 50  =

200 pi

If one side is  12, the sum of the lengths of the other two sides must be > 12

The smallest possible perimeter  is  > 24

RT / TS =  2/5

So  TS  = ( 5) / (2 + 5) SR=  (5/7)SR = (5/7)QP

Triangle SUT similar to triangle QUP

So height of  SUP  = (5/7) height of QUP

So VR  = (5/7)QV

So  QR  =  QV + VR  =  QV + (5/7)QV

So  QR  =  (12/7)QV

So (7/12)QR = QV

So  QV / QR  = 7/12 

And triangles QVU and QRS are similar

So 

UV / QV  = SR / QR

UV  =  QV * SR / QR  =   

UV  = (QV / QR) * SR  =  (7/12) (20) =  140  / 12 =  35 / 3 =  11 + 2/3 

Radius of semi-circle  =  (9 + 16 + 9) / 2  = 34 / 2  =  17

Height of trap  =   sqrt [    17^2 - (QP/2)^2 =  sqrt [ 289  - (16/2)^2] = sqrt [ 289 - 64 ] = 25

Area of  trap  =   (1/2) (25) ( 34 + 16)  =  25 * 25  =  625

P is not defined.....

[ LOM ] =  (1/2) r^2 sin (90°)  =   (1/2)(1)^2 * (1)  =  1/2