# The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?

Pawan Prajapati
3 years ago
. Let diameter of the sphere = d units. Radius = d . - umts. 2 Surface area = 4x(:J sq. units = rril2 sq. units. ..•(i) Diameter of new sphere = (d -t;! d ) units = d units. Radius of new sphere = 8d units. .. Surface area of new sphere = 41t(idJ = 9 n_a_i-2 sq. um•ts. ...(i'i') 16 Decrease = (M2 - :aM2 ) sq. units = 7 d 2 rrd2 sq. units. -1t 700 .. Percentage decrease = 16 x 100 = -% 1t d2 16 = 43.75%.
Dhruv Tyagi
22 Points
one year ago
Given: diameter of a sphere is decreased by 25%.

We have to find the percentage by which its Curved Surface Area decreases.

Let the radius of the sphere be r.

Then its diameter is 2r.

The Surface area of a sphere = 4πr2

The Curved surface area of the sphere = 4πr2

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 - (25/100) × (2r)

= 2r - (r/2)

= 3r/2

Radius of the new sphere = 1/2 × 3r/2 = 3r/4

Hence, curved surface area of the new sphere = 4π (3r/4)2

= 4π (9r2/16)

= (9πr2)/4

Now, decrease in the original curved surface area = 4πr - (9πr2)/4

= (16πr2- 9πr2)/4

= (7πr2)/4

So, the percentage decrease in the curved surface area is,

= [(7πr2)/4 × 1/(4πr2)] × 100%

= [7/16] × 100%

= 43.75%