CPhill

We can  consider  just a 1 x 1 square and  two  circles  with  equations  x^2 + y^2 =1  and x^2 + y^2 = 1/4

The area of the  smal square= 1

The area between  the  circles is   (1/4) pi ( 1^2  - (1/2)^2)  =  (3/4)pi / 4  = (3/16) pi

The probability  is  (3/16)pI / 1  ≈ .589 ≈  58.9%

Triangles EAC , ABF and BCD  are all 45 - 45 - 90 right triangles

Their legs are    6/sqrt 2 ,  8/sqrt 2  and 10/sqrt 2   respectively   

R =  (1/2) ( 6/sqrt 2) ^2  =   9

S =  (1/2) (8/sqrt 2)^2  =    16

T =  (1/2) (10/sqrt 2)^2  =  25

(R + S)  / T  =   (9 + 16)  / 25  =    1

It looks like  RS should be 28  and PQ = 7

Area of  trapezoid =  (1/2)(height of triangle QSR) ( sum of  base lengths)

12 =  (1/2)(height of triangle QSR) ( 28 + 7)

12 = (35/2)  ( height of triangle QSR)

12 ( 2 / 35)  = height of  triangle QSR

24/35 =  height of  triangle QSR

Area of triangle QSR = (1/2)RS (height of QSR) =  (1/2)(28)(24/35) =  (14/ 35)*24 = (2/5) (24) =  48/5  = 9.6

[PQS ] =  area of trapezoid - area of triangle QSR =  12 - 9.6 =  2.4

Let the side of the  square = S

QT = S/2

Triangle QUT is a  45 -45 -90  right triangle

Each leg is   (S/2) (1/sqrt 2) =  S / [ 2sqrt 2 ]

Its area =  (1/2)(product of the  legs) =  (1/2) (S/[2sqrt 2 ] )^2  =  (1/16)S^2

And  triangle QSR =  1/2 area of the square=  S^2/2

So  [ RTUS ]  =  S^2/2 - S^2/16 =  (7/16)S^2

So   [ QUT ] / [ RTUS ]   =   [ (1/16)S^2 ] / [ (7/16)S^2 ]  =  1 / 7

Let G be the total  number of  games   you  play

Let .75 G be the number of  games  you  have played

Let . 45 (.75)G be the number of  games  you  have won

Let .25G be the games  left

Let  .50G = the number of  wins  you  must have

And number of wins you  need - Number of  wins  you  have =  No. of  wins you need in  your remaining  schedule

So

[ No. of wins you  need - No, of  wins you have  ] / [ No. of Games Left ]

[ .50G - .45 (.75)G ] /  .25G  =

(.1625 G )  / ( .25G) =  .65    (you need to  win  65%  of your remaining  games )

\( \frac{-3 \pm 3 \sqrt{29}}{14} \)

The discriminant =  3^2 * 29  =   9*29 =  261

So

3^2  - 4 (7) (k)  =  261

9 - 28k = 261

-28k =  252

k = -252 / 28   =    -9 

Inradius =   Area /  Semi-perimeter

Semi-perimeter =  ( 5 + 6 + 8)  / 2 =   19 / 2 =   9.5

Area = sqrt [ 9.5 ( 9.5 - 5) (9.5 -6) (9.5 - 8) ]   ≈ 14.9812

Inradius =  14.9812 /  9.5   ≈ 1.58

Thx....since corrected  !!!

The denominator  cannot  = 0 

Simplifying the  denominator  we get

4x^2 -8x + c    →   this is an upward turning parabola

To have a domain of all  real numbers, this parabola must lie above the  x  axis

So.....the discrimiant  must  be <  0  ....so....

(-8)^2 -4(4) ( c) <  0

64  - 16c <  0

64 < 16c

4 < c

So

c >  4

The smallest integer  is  5

Area of semi-circle =  (1/2) pi (diameter / 2)^2  = (1/2) pi (10)^2  =  50pi      (1)

BC =   sqrt [ 20^2  -10^2 ] =  sqrt [ 300 ]  = 10sqrt 3 

Area of right triangle ABC  = (1/2) AB * BC =   (1/2) (10) (10sqrt 3)  = 50 sqrt 3       (2) 

Shaded area =   (1) - (2)   =    50 pi  - 50sqrt 3   

a + b =     50  + 50sqrt 3  ≈   136.6