Question

The altitude of a right circular cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which the lateral surface increases is _________.

Answer 1

Let r and h be the radius and height of the cylinder, respectively.

∴ Area of the base of the cylinder = $\mathrm{\pi }$r²

Lateral surface area of the original cylinder = 2$\mathrm{\pi }$rh .....(1)

Now,

Height of the new cylinder, H = 6h (Given) .....(2)

Let the radius of the new cylinder be R.

Area of the base of the new cylinder = $\frac{\mathrm{\pi }{r}^2}{9}$ (Given)

$\therefore \mathrm{\pi }{R}^2=\frac{\mathrm{\pi }{r}^2}{9}$

$\Rightarrow R^2=\frac{r^2}{9}$

$\Rightarrow R=\sqrt{\frac{r^2}{9}}=\frac{r}{3}$ .....(3)

Now,

Lateral surface area of the new cylinder

$=2\mathrm{\pi }RH$

$=2\mathrm{\pi }\times \frac{r}{3}\times 6h$ [From (1) and (2)]

$=4\mathrm{\pi }rh$

$=2\times 2\mathrm{\pi }rh$

= 2 × Lateral surface area of the original cylinder [From (1)]

Thus, the lateral surface area of the new cylinder increases 2 times.

The altitude of a right circular cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which the lateral surface increases is ___2___.