Question

the circumference of the base of a cylinder is 24\pi mm. a similar cylinder has a base with circumference of 60\pi mm. the lateral area of the larger cylinder is 210\pi mm². what is the lateral area of the smaller cylinder? 17 1\pi mm² 33 6\pi mm² 60\pi mm² 84\pi mm²

Explanation:

Step1: Find the ratio of the circumferences

The ratio of the circumferences of two similar - cylinders is equal to the ratio of their corresponding linear dimensions (radius or height). Let $C_1 = 24\pi$ and $C_2=60\pi$. The ratio of the circumferences is $\frac{C_1}{C_2}=\frac{24\pi}{60\pi}=\frac{2}{5}$.

Step2: Find the ratio of the lateral - areas

For two similar solids, if the ratio of their corresponding linear dimensions is $a:b$, the ratio of their surface areas (including lateral - areas) is $a^{2}:b^{2}$. Since the ratio of the linear dimensions is $\frac{2}{5}$, the ratio of the lateral - areas of the two similar cylinders, $A_1$ (smaller) and $A_2$ (larger), is $(\frac{2}{5})^2=\frac{4}{25}$.

Step3: Set up a proportion to find the lateral - area of the smaller cylinder

We know that $\frac{A_1}{A_2}=\frac{4}{25}$, and $A_2 = 210\pi$. Substituting the values into the proportion, we get $A_1=\frac{4}{25}\times A_2$.
$A_1=\frac{4}{25}\times210\pi=\frac{4\times210\pi}{25}=\frac{840\pi}{25}=33.6\pi$ $mm^{2}$.

Answer:

$33.6\pi\ mm^{2}$