geometry

∠AMB = ∠MBN = 2* ∠FBD     <==>    2 * tan^-1(2 / 8)

AM = AB / tan∠AMB         AM = 3.75

Shaded area        A = (2 * 8) - (2 * 3.75) = 8.5 u²

Let D  = (0,0)     Let   B = (2,8)

We can find the  coordinates of  F thusly  :

Construct a circle with a radius of 2 centered at   D

The equation of this circle  is   x^2 + y^2  =  4       (1)

Construct a  circle with a radius of 8  centered at  B

The eqation of this circle is  ( x - 2)^2 + (y - 8)^2  = 64  ⇒   x^2 - 4x + 4  + y^2  - 16y + 64 =  64 ⇒

x^2 - 4x +y^2  - 16y  = -4      (2)

Subtract (2)  from (1)

4x + 16y  =  8

x + 4y  = 2

x  = 2  - 4y         sub this into ( 1)  for x

(2 - 4y)^2  + y^2  =   4

16y^2 - 16y +  4  + y^2  =  4

17y^2 - 16y  =  0

y ( 17y - 16)   =  0

Solving for the  second  gives the y coordinate  of F    y  =  16/17

And    x = 2 - 4(16.17)  =  [34 - 64]/ 17  =   -30/17

F = (-30/17 , 16/17)

And the slope of BF  = [  8 - 16/17 ]  / [ 2 - -30/17 ] =   (120/17)  / (64/17)  =  120/64  =  15/8

So the equation of a line through  BF is

y = (15/8)(x - 2) + 8

y = (15/8)x  - 30/8 + 8

y = (15/8)x - 15/4 + 8

y = (15/8)x + 17/4

So   when x  = 0     this line  intersects the left side of the rectangle at  17/4 =  4.25

So  we  have a right triangle with a leg of  (8 - 4.25)  and 2  =   3.75  and  2

And the area of  two of these  triangles is      3.75 * 2  =   7.5       (3)   

So...the gray  area =  area  of rectangle  - (3)  =   8*2  -  7.5  =  16 - 7.5 =  8.5

Two 2*8 rectangles overlap, as shown below.
Find the area of the overlapping region (which is shaded)

\(\text{Let $O=(0,0)$} \\ \text{Let $A=(-1,4)$} \\ \text{Let $B=(1,4)$} \\ \\ \text{Let $E=(x_E,y_E)$} \\ \text{Let area of rectangle $=2*8$ } \\ \text{Let area of triangle $[BEG]=\dfrac{\overline{BG}*h}{2}$ } \\ \text{Let the gray area $=$ area of rectangle$-2*$ area of triangle $[BEG]$ } \\ \text{Let the gray area $= \mathbf{2*8-\overline{BG}*h}$ } \\ \text{Let $h=x_E-x_B$} \\ \text{Let $p=y_B-y_E$} \\ \text{Let $\overline{AB}=\overline{BE}=2$} \\ \text{Let $\overline{BE}^2=p*\overline{BG}$} \)

1. rotate

\(\begin{array}{|lrcll|} \hline \text{Rotation matrix } \\ & R&=&\begin{bmatrix} \cos(ß)&\sin(ß) \\ - \sin(ß)&\cos(ß) \\ \end{bmatrix} \\ & B&=&R\cdot A \\\\ & \begin{bmatrix} x_B\\y_B\end{bmatrix} &=&\begin{bmatrix} \cos(ß)&\sin(ß) \\ - \sin(ß)&\cos(ß) \\ \end{bmatrix} \begin{bmatrix} x_A\\y_A\end{bmatrix} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \begin{bmatrix} 1\\4\end{bmatrix} &=&\begin{bmatrix} \cos(ß)&\sin(ß) \\ - \sin(ß)&\cos(ß) \\ \end{bmatrix} \begin{bmatrix} -1\\4\end{bmatrix} \\\\ -\cos{ß}+4\sin(ß) &=& 1 \qquad (1) \\ \sin{ß}+4\cos(ß) &=& 4 \qquad (2) \\ \\ \mathbf{\sin(ß)} &=& \mathbf{\dfrac{8}{17}} \\\\ \mathbf{\cos(ß)} &=& \mathbf{\dfrac{15}{17}} \\ \hline \end{array}\)

2. rotate

\(\begin{array}{|rcll|} \hline \begin{bmatrix} x_E\\y_E\end{bmatrix} &=&\begin{bmatrix} \cos(ß)&\sin(ß) \\ - \sin(ß)&\cos(ß) \\ \end{bmatrix} \begin{bmatrix} 1\\4\end{bmatrix} \\\\ \cos{ß}+4\sin(ß) &=& x_E \qquad (3) \\ x_E &=& \dfrac{15}{17} + 4*\dfrac{8}{17} \\ \mathbf{x_E} &=& \mathbf{\dfrac{47}{17}} \\\\ -\sin{ß}+4\cos(ß) &=& y_E \qquad (4) \\ y_E &=& \dfrac{8}{17} + 4*\dfrac{15}{17} \\ \mathbf{y_E} &=& \mathbf{\dfrac{52}{17}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{h}&=&\mathbf{x_E-x_B} \\\\ h&=&\dfrac{47}{17}-1 \\\\ \mathbf{h}&=&\mathbf{\dfrac{30}{17}} \\\\ \hline \mathbf{p} &=& \mathbf{y_B-y_E} \\\\ p &=& 4- \dfrac{52}{17} \\\\ \mathbf{p}&=&\mathbf{\dfrac{16}{17}} \\\\ \hline \mathbf{\overline{BE}^2} &=& \mathbf{p*\overline{BG}} \quad | \quad BE=2 \\ 2^2 &=& \dfrac{16}{17}*\overline{BG} \\ \overline{BG} &=& \dfrac{4}{\dfrac{16}{17}} \\\\ \mathbf{\overline{BG}} &=& \mathbf{\dfrac{17}{4}} \\\\ \hline \mathbf{\text{Gray area }} &=& \mathbf{2*8-\overline{BG}*h} \\ &=& \mathbf{2*8-\dfrac{17}{4}*\dfrac{30}{17}} \\\\ &=& 2*8-\dfrac{30}{4} \\\\ &=& 16-\dfrac{15}{2} \\\\ &=& \dfrac{32-15}{2} \\\\ &=& \dfrac{17}{2} \\\\ \mathbf{\text{Gray area }} &=& \mathbf{8.5} \\ \hline \end{array}\)