Mensuration Practice Test 2

Lateral surface area of the cylinder = 2πrh
Area of the base = $\mathit{\pi }{\mathit{r}}^2$ = $9\left(\mathit{\pi }{\mathit{r}}^2\right)$
${\mathit{r}}^2$ = $9{\mathit{R}}^2$
$\frac{{\mathit{r}}^2}{9}$ = ${\mathit{R}}^2$
R = $\frac{\mathit{r}}{3}$
Height of the cylinder H = 6h
New lateral surface area = 2π RH
= $2\mathit{\pi }\left(\frac{\mathit{r}}{3}\right)6\mathit{h}$ = 4πrh
New lateral surface area is two times.