Answer:
The surface area of the second cylinder is equal to
![9.33\ m^(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/5ibypsp1ekq1uosba0b8tvaah6vr4sdmbn.png)
Explanation:
we know that
If two figures are similar then
the ratio of their surfaces areas is equal to the scale factor squared
Let
z-------> the scale factor
x------> the surface area of the smaller cylinder (second cylinder)
y-------> the surface area of the original cylinder (first cylinder)
so
![z^(2)=(x)/(y)](https://img.qammunity.org/2020/formulas/mathematics/high-school/bsr5zpx86e0gikgp398wuhrw2lup269tnz.png)
Step 1
Find the scale factor
![z=(2)/(6)=(1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/high-school/i6rcacpho5k6ns8hza8vikn42s7eeukyza.png)
Step 2
Find the surface area of the second cylinder
we have
![z=(1)/(3)](https://img.qammunity.org/2020/formulas/mathematics/high-school/x33dhux51cze1l6epoodiu821e15u4m9up.png)
![y=84\ m^(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/4skgsnlh0v036q2rhijqtbfdzkvidy2mf1.png)
substitute and solve for x
![((1)/(3))^(2)=(x)/(84)](https://img.qammunity.org/2020/formulas/mathematics/high-school/5pxly616eav69l08o3c70oxljdkmhy9jkc.png)
![x=(1)/(9)*84=9.33\ m^(2)](https://img.qammunity.org/2020/formulas/mathematics/high-school/2qlzdoeknqoclp3j6y5s54en2qhcyu4iti.png)