can any one help me please ..im really stuck on this..?

Question

(a) A circle has radius 2cm (in that circle there is an triangle AOB) and the chord AB has length 3cm. AO is 2cm and BO is 2cm. O is the centre of the circle.

(b) consider a paper disc whose boundary is the circle in part (a). The sector AOB is removed from the paper disc and the remaining paper is bent round to form a shallow cone by joining the edge OB to the edge OA.

(i) Which arc becomes the circle that is at the base of the cone? Find the angle subtended by this arc in the original circle (in radians), and hence find the length of the arc.

(ii) Hence find the radius of the circle that is at the base of the cone.

(iii) Find the perpendicular height of the cone.

(iv)Given that the formula for the volume of a cone is
V = 1/3 pie r^2 h
(Where r is the radius of the base and h is the perpendicular
Height), find the volume of the cone in this question.

Answer 1

Angle O = 2(arcsin(1.5)) =97.1808 degrees
Angle of remaining sector = 360 - 97.1808 = 262.8192 degrees

Change to radians:
x/(2pi) = 262.8192/360
x = 262.8192(2pi)/360 = 4.58706 radians
arc length = 4.58706(2) = 9.17412cm [this is the circumference of the base]
height of cone = √[2^2 - (1.46011)^2] = 1.36678cm
Radius of the base:
c = 2(pi)r
r = c/(2pi) = 9.17412/(2pi) = 1.46011cm
volume of the cone = (1/3)(pi)r^2h = (1/3)(pi)( 1.46011)^2(1.36678) = 3.05139cm^3