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Seek shadows in math problems
As shown in the figure, let the intersection of the sector and the circle be E, connecting Ce, OE and OC.

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.