If a cylinder’s radius and height are each shrunk down to a third of the original size, what would be the formula to find the modified surface area? .

H, r

Surface Area = 2πrh 

If the h, is now h/3 and the r, now r/3

Surface Area = 2π*r/3*h/3
 
                       = 2πrh/9

The modified surface area is reduced by 9 times.

The surface area of the cylinder is equal to the sum of the lateral area and the areas of the base,
                              SA = LA + 2BA
                             SA = 2πrh + 2πr²
With the given shrinkage of the dimensions, the equation becomes
                             SA = (2π)(r/3)(h/3) + 2π(r/3)²
This is simplified into,
                                SA = 2πrh/9 + 2πr²/9

Ive seen this before! the answer is b