The diameter of a sphere is decreased by 25%. By what percent its curved surface area decrease?

Explanation

Let the radius of the sphere be `r`.
Then its diameter is `2r.`
The Surface area of a sphere `= 4pir^2`
The Curved surface area of the sphere`= 4pir^2`
Now it is given in the question that the diameter of the sphere is decreased by 25% hence a new sphere is formed.
Therefore, the diameter of the new sphere can be written as:`=2r-(25%) \of\ 2r`
`=>2r-25/100xx2r`
`=>2r-r/2=3r/2`
Radius of the new sphere `=1/2xx3r/2=(3r)/4`
Hence, curved surface area of the new sphere `=4pi(3r/4)^2`
`=>4pi(9r^2)/16)=(9pir^2)/4`
Now, decrease in the original curved surface area `=4pir-(9pir^2)/4=(7pir^2)/4`
So, the percentage decrease in the curved surface area is,`=((7pir^2)/4cxx1/(4pir^2))xx100=43.75%`