The total surface area of a hollow metal cylinder, open at both ends, of external radius 8 cm and height 10 cm, is $338 \pi \ cm^2$. Taking r to be the internal radius, write down an equation in r and use it to determine the thickness of the metal in the cylinder.
For the hollow cylinder,
outer radius $=R=8 \ cm$
inner radius $=r$
height $=h=10 \ cm$
total surface area (TSA) $=338 \pi \ cm^2$
$\begin{aligned}
\text{TSA} & = \text{Lateral surface area} + \text{Base area} \\
& = 2 \pi h \left(R+r\right) + 2 \pi \left(R^2 - r^2\right)
\end{aligned}$
$\begin{aligned}
\text{i.e. } & 2 \times \pi \times 10 \times \left(8+r\right) + 2 \times \pi \left(8^2 - r^2\right) = 338 \pi \\
\text{i.e. } & 20 \left(8+r\right) + 2 \left(8+r\right) \left(8-r\right) = 338 \\
\text{i.e. } & \left(8+r\right) \left(10+8-r\right) = 169 \\
\text{i.e. } & \left(8+r\right) \left(18-r\right) = 169 \\
\text{i.e. } & 144-8r+18r-r^2=169 \\
\text{i.e. } & r^2 - 10r+25 = 0 \\
\text{i.e. } & \left(r-5\right)^2 = 0 \\
\implies & r = 5
\end{aligned}$
$\therefore$ Inner radius $= 5 \ cm$
$\therefore$ Thickness of metal in the cylinder $= R-r=8-5=3 \ cm$