Funky Circle Areas

Here's my best effort  ....

If the perimeter of the rectangle is 26....then 

2(W + L )  = 26

W + L  = 13

Then  L  =  13 - W

Call point Q  =   ( W , L )  =  (W , 13 - W)

So.....using the equation of a circle with a center O located at  (0,0)  and a radius of 10, we have

W^2  + (13 - W)^2  =  100

W^2 + W^2 - 26W + 169  = 100

2W^2 - 26W + 69  =  0

Solving this for W  we get that W  =  [ 13 + √31] / 2  = OP

And L  =   [ 13  - √31 ] / 2  = PQ 

So  PR  =  (1/2) √ [  ( 13 + √31)^2 + (13 - √13)^2 ] =  (1/2) √[ 382 + 26√31 - 26√13 ]

So.... the perimeter  =

arc BQA  + AP  + PR  + BR  =

5pi + (10 - OP) + PR + (10 - PQ)  =

5pi  + (10 -  [ 13 + √31] / 2 )  + (1/2) √[ 382 + 26√31 - 26√13 ]  + (10 -  [ 13  - √31 ] / 2 )  ≈

33.1125  units

CPhill: I will give it a go. Please check my reasoning! Thanks.

The diagonal OQ of the rectangle = radius of the circle = 10

Cos(45) =OP / 10 =~ 7.07 =10/sqrt(2)

OR = 26 / 2 - 7.07=~ 5.93 =13 - 5sqrt(2)

10 - OR = BR =~4.07=5sqrt(2) - 3

AP = 10 - 7.07 =~2.93=10 - 5sqrt(2)

Arc AQB =[10 x 2 x Pi] / 4 =~15.71 =5pi

So, the perimeter of the shaded area =10 + [5sqrt(2) - 3] + [10 - 5sqrt(2)] + 5pi =17 + 5pi

Noice guys! Thanks to CPhill and guest!