pls come help urgent?

Question

8. A cake with the shape of a right cylinder is cut into 16 parts of the same
size. The following diagram shows a piece of the cake.

(the diagram i send later)
The cross-section of the cake is a sector OCD with centre O. Given thatthe length of CD = 12 cm. Find
(a) ÐDOC in rad,
(b) the length of OD,
(c) the total surface area of the cake if the thickness of the
cake = 0.2 OD.
( d ) the volume of the piece of cake above

can explain why the answer for C is tat?
what formulae should use to solve the total surface area?

Answer 1

(a) ÐDOC = π/8
(b) OD = 12*16/(2π) = 30.55775 cm
(c) S = π(96/π)^2 + 0.2(192)(96/π)
S = 4,106.961 cm^2
( d ) V = 0.2(1/2)(π/8)(96/π)^3 = 1,120.531 cm^3

Answer 2

Well, the angle of each arc is 360/16 = 22.5

The length of OD (the radius) is half the diameter 12/2 = 6

Surface area is the same as the area of a circle, which is pi * r ^2

3.14(6^2) = 3.14(36) = 113.04

Volume of the cake = surface area * height or 113.04 * 6 = 678.24

Volume of 1/16 piece = 678.24 / 16 = 42.39

Answer 3

I assume CD is the line from C to D, not the arc from C to D.

Let theta be the angle DOC. So theta=360/16 degrees. Say the perpendicular from O to CD meets CD at E. Then OED is a right triangle with angle at O = theta/2. Say r=OD. Then sin(theta/2)=6/r. So r=6*csc(360/32).

The arc CD is r*theta (measured in radians) = r*pi/8. The thickness is 1.2*csc(360/32). The area of the top (and bottom) is r^2 * theta/2 (in radians). The area of the side is CD* its thickness. The total area is their sum

The volume is pi*r^2*thickness/16.

Answer 4

c