It is known that P is a point outside ⊙O, PA cuts ⊙O to A, AB is the diameter of ⊙O, PB crosses ⊙O to C, PA = 2 cm, and Pb = 4 cm.

get

Triangle PAB is a right triangle,

In the right triangle PAB:

∠ABC=30，

AB = 2ì3。

Therefore, the area PAB of a right triangle

=PAxAB/2

=2√3.

From < AOC = 2 < ABC = 60

And OA=OC=AB/2=√3,

get

Triangle AOC is an equilateral triangle.

Area of equilateral triangle

=3√3/4.

Because AB is a diameter ≧O,

Triangle ABC is a right triangle.

In the right triangle ABC:

∠ABC=30，

AB = 2ì3。

get

AC=AB/2=√3，

BC=3。

Area of right triangle ABC

=ACxBC/2

=3√3/2.

Sector OAC area

=60 π(√3)? /360

=π/2.

but

Area s of the shaded part in the figure

= area of right triangle PAB- area of fan-shaped OAC-area of triangle OBC.

In ...

Triangle area OBC = right triangle area ABC- equilateral triangle area AOC,

get

Area s of the shaded part in the figure

= area of right triangle PAB- area of fan-shaped OAC-area of triangle OBC.

= right triangle area PAB- sector OAC area-(right triangle area ABC- equilateral triangle area AOC)

= right triangle area PAB- sector OAC area-right triangle area ABC+ equilateral triangle area AOC)

The ∴ area s of the shaded part in the figure

= right triangle area PAB- sector OAC area-right triangle area ABC+ equilateral triangle area AOC)

=2√3-π-(3√3/2)+(3√3/4)

=(5√3/4)-(π /2).