Question

If the radius of a sphere is doubled, then the percentage increase in the surface area is _________.

Answer 1

Let the radius of the original sphere be r.

∴ Surface area of the original sphere, S₁ = 4$\mathrm{\pi }$r²

If the radius of the sphere is doubled, then the radius of the new sphere is 2r.

∴ Surface area of the new sphere, S₂ = 4$\mathrm{\pi }$(2r)² = 16$\mathrm{\pi }$r²

Now,

Percentage increase in the surface area of sphere

$=\frac{\mathrm{Increase}\mathrm{in}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{sphere}}{\mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{original}\mathrm{sphere}}\times 100\%$

$=\frac{S_2-S_1}{S_1}\times 100\%$

$=\frac{16\mathrm{\pi }{r}^2-4\mathrm{\pi }{r}^2}{4\mathrm{\pi }{r}^2}\times 100\%$

$=\frac{12\mathrm{\pi }{r}^2}{4\mathrm{\pi }{r}^2}\times 100\%$

$=300\%$

Thus, the percentage increase in the surface area of the sphere is 300%.

If the radius of a sphere is doubled, then the percentage increase in the surface area is ___300%___.