CPhill

Since YZ  =10 sqrt 2 and XY = XZ  = 10....then XYZ is a right triangle with the right angle at  X

Call the center of the  semi-circle O  

Let the tangent  XZ  intersect the circle at P 

Draw OX.....this bisects angle YXZ

So  we have right triangle POX

Angle  OXP   =   45

Angle OPX  = 90

So angle XOP = 45

So triangle OPX  is a 45 -45- 90 right triangle  with OX =  (1/2)YZ = (1/2) 10sqrt (2)  = 5sqrt2

And OP  = XP  =  5sqrt 2 / sqrt 2 =  5 

So   OP is the radius of the semi-circle  

Area  of semi-circle  =   (1/2) pi  * (5)^2  =  12.5 pi  ≈  39.27

Here : https://web2.0calc.com/questions/square_29

[ABCD] = AD * DC  =   180

The shaded area is a trapezoid with area  =   (1/2) AD (AB + UT)  and AB = DC and UT = DC / 3  

So....the shaded area =  (1/2)(AD) (DC + DC/3)  =  (1/2)(AD)(DC)   + (1/2)(AD)(DC//3)  =

90  +  (1/6)(AD)(DC)  = 

90  +  30   =

120

If we inscribe the hexagon in a circle, the radius of the  circle  = the side length of the hexagon

The hexagon is  composed of  6 equilateral triangles  each having an area   =

(1/2)r^2 sin (60°)  = (1/2)r^2 * sqrt (3) / 2

So

A = 6 (1/2) r^2  * sqrt (3)  / 2

3/2 =  3r^2 *sqrt (3) / 2

1 = r^2 sqrt (3)

1 / sqrt (3)  = r^2

r  =  1 / 3^(1/4)  ≈  .76  = side of the hexagon

Measure of  each interior angle  =  (n-2)180  / n

Measure of each exterior angle  = 360 / n

So

(n -2) 180  /  n =  2 *360 /  n

180n - 360  =  720

180n  =  720 + 360

180 n = 1080

n =  1080 /  180  =    6

circumradius   = R =     abc   / ( 4 area)              where a,b,c are the sides

abc =  (29)(29)(42)  = 35322

area (from other answer)  =  420

4 * area =  1680

R  =    35322  /  1680    =    21.025

Inradius   =   r =    Area / semi-perimeter

semiperimeter  =  (29 + 29 + 42)  /  2 =   50  = s

Area =  sqrt ( s (s -a) (s - b) (s -c)        where a,b,c  are the side lenghts

So

sqrt [ 50 (21)(21)(8) ] =   21 * 20  =  420

So

r  =  420 / 50  =   42 / 5  =  8.4

A  zero with an even multiplicity will touch (but not go  through) the x axis     -  behavior 1

A zero with a multiplicity of 1 will  go through the x axis   - behavior 2

A  zero  with an odd multiplcity  >  1 will  "flatten out" a bit before crossing the  x axis  - behavior 3

zero                      multiplicity                       behavior

-6                              2                                      1

 0                              1                                       2

 1                              3                                       3

  4                             1                                       2                 

Corrected!!!

The triangles look to be similar  such that  Δ BCA  similar to Δ VWA

CA / BC  =  WA/ WV

28  /16  = WA  / 10

WA  = (28 * 10)   /16 =   17.5   = x

And

AB / CB  =  AV / VW

AB / 16 = 12  /10

AB =  (16 * 12)  / 10  =  19.2 =   y

x + y =   17.5  +  19.2 =   36.7 

Note     u^2/v + v^2/u =   [u^3 + v^3] / (uv)

Simplify as

x^2 - 4x +  24  =  0

uv = 24

2uv = 48

u + v =  4           square both sides

u^2  + 2uv  + v^2  = 16

u^2 +48 + v^2  =16

u^2 + v^2  = -32

u^3  + v^3  =

(u + v)(u^2  - uv + v^2)

(u+ v) ( [u^2 + v^2] - uv)

(4) ( [ -32 ] - 24 ) 

(4) ( -56)  

-224

So 

u^2/v + v^2/u  = 

[u^3 + v^3 ]  / [uv ] =

-224 /  24  =    

-28 / 3