As can be seen from the figure, there is the following relationship:
S shadow =S sector CAD+S circle -2S blank
And then what? 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.
Through cosine theorem? OE^2=OC^2+CE^2-2OC*CE*cos∠OCE
Available? cos∠OCE=CE/(2OC)=4/(2*2.5)=4/5
Available? ∠OCE≈37 ,∴∠BCE=90 -37 *2= 16
And then what? ∠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.