im confused can any1 help me on this question ?? or try and put me on the right track?

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

(a) no question ?

(b)
(i) It's the arc AB which becomes the circle.

Let I the middle of [AB].
sin(AOI) = AI / OA = (3/2)/2 = 3/4
AOI = arcsin(3/4)

I call now a = arcsin(3/4)

angle(AOB) = 2(AOI) = 2a radians

the length of the arc AOB is radius * angle (in radians) then equal to 2 * 2a = 4a cm

(ii) If R is the radius of the circle of the base, then
2 pi R = 4a

R = 2a / pi

(iii) we must find h.

h² + R² = OA²

h² = 4 - 4a² / pi²
= 4(pi² - a²) / pi²

h = 2sqrt(pi² - a²) / pi

then the volume is

V = 1/3 pi R² h = 1/3 * pi * 4a²/pi² * 2sqrt(pi² - a²)/pi

V = 8a² / (3pi²) sqrt(pi² - a²) cm³

a approaches 0.848 radian then
V approaches 0.588 cm³