Surface area of the top

A cylinder is three times as high as it is wide. The length of the cylinder’s diagonal is 20 cm. Find the surface area of the top of the cylinder.

Correct result:

S =  31.4159 cm²

Solution:

$$\begin{gathered} u=20\text{ cm } \\ h=3D \\ {u}^2=h^2+D^2 \\ {u}^2=9D^2+D^2 \\ D={\displaystyle \frac{u}{\sqrt{9+1}}}={\displaystyle \frac{20}{\sqrt{9+1}}}=2\sqrt{10}\text{ cm}\doteq 6.3246\text{ cm } \\ r=D\mathrm{/}2=6.3246\mathrm{/}2=\sqrt{10}\text{ cm}\doteq 3.1623\text{ cm } \\ h=3\cdot D=3\cdot 6.3246=6\sqrt{10}\text{ cm}\doteq 18.9737\text{ cm } \\ {u}_2=\sqrt{h^2+D^2}=\sqrt{18.9737^2+6.3246^2}=20\text{ cm } \\ u=u_2(check-ok) \\ S=\pi \cdot r^2=3.1416\cdot 3.1623^2\doteq 31.4159 \\ S=10\pi =31.4159{\text{cm}}^2 \end{gathered}$$

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