As can be seen from the figure, there is the following relationship:

S shadow =S sector CAD+S circle -2S blank

And s sector CAD =1/2 * π/2 * ac2 = 4π; S circle = π * OA 2 = 2.5 2 π = 6.25 π.

S blank =S△ABC+S sector OAE+S sector OBE

It is known that OC=OE=OA=OB=2.5 and CE=AC=4.

From cosine theorem OE 2 = OC 2+Ce 2-2OC * Ce * COS ∠ OCE

Available cos∠OCE=CE/(2OC)=4/(2*2.5)=4/5.

We can get ∠ OCE ≈ 37, ∴ BCE = 90-37 * 2 = 16.

And ∠ ACE = 2 ∠ OCE = 37 * 2 = 74, ∠ BOE = 2 ∠ BCE = 16 * 2 = 32.

s△ABC = 1/2 * AC * BC = 1/2 * 4 * 3 = 6，

S quadrilateral OA CEO = 2s △ oce = 2 *1/2 * oc * ce * sin ∠ oce.

=2* 1/2*2.5*4*3/5=6

S department OAE=S department CAE-S quadrilateral OACEO

= 1/2 * 74/ 180 *π*ac^2-6

= 1/2*0.4 1 1π*4^2-6

=3.3π-6

S sector OBE =1/2 * 32/180 * π * ob2.

= 1/2*0. 178π*2.5^2

=0.56π

∴S blank =S△ABC+S sector OAE+S sector OBE

=6+3.3π-6+0.56π

=3.86π

∴S shadow =S sector CAD+S circle -2S blank.

=4π+6.25π-2*3.86π

=2.53π

≈7.95

Is the area of the shaded area.