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).
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