CPhill

Largest angle =  4 / ( 2 + 3 + 4) * 180   =  (4/9) *180   = 80°

The exterior angle is supplemental  =  180 - 80   =  100°

Area =  12sqrt (46) =  (1/2) (15) * altitude

12sqrt (46)   / 7.5  =  altitude ≈  10.85

Call the legs a, b

a + b + 10  = 24

a + b  =14      square both sides

a^2 + 2ab +b^2  =  196         { a^2 + b^2    = 10^2 = 100}

a^2 + b^2   + 2ab =  196

100 + 2ab = 196

2ab = 96

ab = 48

Area  =  (1/2)ab  =  24

x  + 2x + 10  + 3(2x + 10) - 10   = 180

9x  + 30  =  180

9x  =  150

x = (50/3)°    smallest angle

(130/3)°  =  next  smallest angle

(360/3)° = 120°  =   largest angle

AB =  18

AD = 4

Area =  AB * AD   =  18 * 4   =   72

Side of square 1  = 1/2

Radius of circle 1    =    (1/2)sqrt(2) / 2  =  sqrt (2)/4

Side of square 2  =  1

Radius of circle 2 =  (1)sqrt (2) / 2

Area circle 2   /Area of circle 1   =  [ sqrt (2) / 2 ] ^2   / [ sqrt (2) / 4]^2   = 16 / 4  =  4 /  1

Last one

Let the center of the largest semi-circle = E

Let the center of the next largest semi-circle = F

Let the center of the smallest semi-circle =  D

Let  the center of the circle  = O.....let its radius = r

OE  = 8 - r

OF = 5 + r

OD  = 3 + r

DE = 5

FE   = 3

 cos (OEF)  = -cos OED

Law of Cosines

OF^2  = FE^2 + OE^2  - 2 [ FE * OE]cos (OEF)

OD^2  =  DE^2  +OE^2 - 2[ DE * OE]cos (OED)

(5 + r)^2  = 3^2  + (8- r)^2 + 2[3 * (8-r)]cos (OED) 

(3 + r)^2  = 5^2 +  (8-r)^2  - 2[5 *(8-r)] cos (OED)

cos (OED)  =  [ (5+ r)^2  - 9 - (8 -r)^2 ] / [6 (8-r)]

cos (OED) = [  (3 + r)^2  - 25 - (8-r)^2 ] / [-10 (8 -r)]

[(5 + r)^2  - 9 - (8-r)^2] / 6   =  [(3 +r)^2 -25 - (8-r)^2 ] / (-10)

[ r^2 + 10r + 25  - 9 - r^2 + 16r -64 ] /6  =  [ r^2 + 6r + 9  -25 - r^2 + 16r - 64 ] / -10

[ 26r - 48 ]  / 6 =  [ 22r  - 80 ] / -10

-260r + 480  =  132r - 480

960  =  392r

r = 960  / 392    = 120  / 49

Second one

Law of Cosines

BC^2  = 4^2 + 5^2  -  2 (4*5)cos 60°

BC^2  = 41  - 20

BC = sqrt (21)

Law of Cosines again

4^2  = 5^2  + 21  - 2(5 sqrt 21) cos BCD

[16 - 25 - 21] / [-2 * 5 *sqrt 21]   =cos BCD

[ -30] / [ -10 sqrt 21 ] = cos BCD

3/sqrt21 =  cos BCD   =  sin ACE  .......cos ACE  = sqrt 12 /  sqrt 21 

Angle ACE + Angle  CAE + Angle AEC  =180

Angle ACE + Angle CAE + 60  = 180

Angle ACE + Angle CAE   = 120

Angle CAE  =  120 - Angle ACE

sin (angle CAE)  =  sin (120  - angle ACE)  = 

sin (120)cos ACE -  cos (120)sin ACE  =

sqrt 3 / 2  * sqrt 12 /sqrt 21  + 1 / 2  *  3/sqrt 21  =

3/sqrt 21  + (3/2)/sqrt 21 =

4.5  /sqrt 21

Law of Sines

CE   / sin CAE  =   AE  / sin  ACE

4 / (4.5/sqrt21)   = AE / (3/sqrt 21)

[3 * 4] / 4.5  =  AE  =  8/3 

Simplify

10a - 3b  = 2

22a - (16+ m)b = 17+ k

Multiply  the first by  22  and the second by 10

220a  - 66b  =  44

220a - 10(16+m)b = 10(17+k)

To  have infinite solutions

10 (16 + m)  = 66   →  160 + 10m = 66 → 10m = -94  → m = -94/10  = -47/5

and

10(17 + k)  = 44   →  170 + 10k = 44 → 10k = -126 →  k = -126/10  = -63/5

Trivial solution (a,b)  = (0,0)

a + ab^2  = 250ab + ab^3    

a - ab^2  =  240ab  - 50b^2    add these    and then subtract them

2a   =   490ab + ab^3  - 50b^2

2ab^2 = 10ab + ab^3 + 50b^2    add these 

2a + 2ab^2  = 500ab + 2ab^3       divide through by  a

2  + 2b^2  = 500b + 2b^3

1 + b^2  = 250b + b^3

b^3 - b^2 + 250b - 1  =  0 

b ≈ .004

a ≈  -.02