Formula:
Derivation:
To find the formula of the area of the segment which is (Ag)
Use the formula which is Area of the sector (As) minus the Area of the Triangle (At) (As-At=Ag) Our formula for
Ag is R^2/2(Θ/180π-sinΘ)
So how do we derive from this? we Let As=Θ/360πr^2
Let At=1/2 r^2 sinΘ
Why is the Area of the triangle (At) 1/2 r^2 sinΘ ?
because the formula for finding the area of the triangle given 2 sides and an included angle is 1/2ab * sin C
but since the given is an isosceles triangle (both sides are equal) then a=b=r
hence, r^2.
so we start solving it
Ag= Θ/360πr^2 - 1/2r^2 sin Θ
I used commutative property to reach the formula for Ag which is R^2/2(Θ/180π-sinΘ)
- 1/2 r^2 sin Θ + Θ/360πr^2
Using Greatest Common factor..
1/2 r^2 ( -sinΘ+ Θ/180π)
Used Commutative property again
1/2 r^2 (Θ/180π-sinΘ)
= r^2/2 (Θ/180π-sinΘ)
And now we derived from the same formula.
QED
Derivation:
To find the formula of the area of the segment which is (Ag)
Use the formula which is Area of the sector (As) minus the Area of the Triangle (At) (As-At=Ag) Our formula for
Ag is R^2/2(Θ/180π-sinΘ)
So how do we derive from this? we Let As=Θ/360πr^2
Let At=1/2 r^2 sinΘ
Why is the Area of the triangle (At) 1/2 r^2 sinΘ ?
because the formula for finding the area of the triangle given 2 sides and an included angle is 1/2ab * sin C
but since the given is an isosceles triangle (both sides are equal) then a=b=r
hence, r^2.
so we start solving it
Ag= Θ/360πr^2 - 1/2r^2 sin Θ
I used commutative property to reach the formula for Ag which is R^2/2(Θ/180π-sinΘ)
- 1/2 r^2 sin Θ + Θ/360πr^2
Using Greatest Common factor..
1/2 r^2 ( -sinΘ+ Θ/180π)
Used Commutative property again
1/2 r^2 (Θ/180π-sinΘ)
= r^2/2 (Θ/180π-sinΘ)
And now we derived from the same formula.
QED