Question

What is the surface area of a cylindrical ring where the diameter of the cross section is 6.3 in and the centerline has a length of 48in

Answer 1

The correct answer is 950.02 in^2

Answer 2

Answer:  The surface area of a cylindrical ring  is 950.4 square in.

Step-by-step explanation:

We know that the surface area of an open cylinder is given by :-

$S.A.=2\pi rh$ , where r= radius and h= height

As per given , diameter of cylindrical ring = 6.3 in

then , $r=\dfrac{6.3\text{ in}}{2}=3.15\text{ in}$

h= 48 in

Now , the surface area of the cylindrical ring = $2(\dfrac{22}{7})(3.15)(48)$

$=950.4\text{ square in.}$

Hence, the surface area of a cylindrical ring  is 950.4 square in.