Question

1)The diameter of a spere si decreased by 20%. By what percentage does its curved surface area decrease?

Answer 1

Let the diameter of sphere be "d"

surface area of sphere , S = 4πr²

= π (2r)²

= π(d)²

Diameter of sphere decreases by 20%

New diameter = d - (d/5) = [5d- d] / 5 = 4d/5

New surface area , S' = π(4d/5)²

= (16/25) πd²

Change in surface area of sphere = S - S'

= πd² – (16/25) πd²

= (9/25) πd²

 

Decrease in surface area = [(9/25πd²) / πd²] x 100 = 9 / 25 × 100 = 36 %

 

Therefore the decrease in S.A of sphere is 36%

 

An alternate method:

 

Let the diameter of sphere be "d"
curved surface area of sphere = 4πr2
= π (2r)2
= π(d)2
Given diameter of sphere decreases by 20%
New diameter = d - (d/5) = 4d/5
New curved surface area = π(4d/5)2
                    = (16/25) πd2
Change in surface area of sphere = πd2 – (16/25) πd2
= (9/25) πd2

decrease in curved surface area = [9/25) πd2/ πd2] x 100 = 36%
Therefore surface area decreases by 36%

Hope This Helps :)