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

Let the diameter of a sphere be d.

After decreasing, diameter of the sphere
= d – \(\frac{25}{100} \) x d
= d –\(\frac{1}{4} d = \frac{3}{4} d \)

Since, surface area of a sphere
= \( 4\pi r^2 \)
= \( \pi (2r)^2 \)
= \( \pi d^2 \)

Surface area of a sphere, when diameter of the sphere is

= \(\pi (\frac{3}{4} d)^2 \)
= \(\pi \frac{9}{16} d^2 \)

Now, decrease percentage in curved surface area
= \(\frac{\pi d^2 - \pi \frac{9}{16} d^2 }{\pi d^2} × 100 \)
= \(\frac{16-9}{16} × 100 \)
= \( \frac{7}{16} × 100 \)
= 43.75%