Hexaprism container

Calculate the volume and surface of a container in the shape of a regular hexagonal prism with a height of 1.4 m with a base edge of 3dm and a corresponding height of 2.6 dm.

Correct result:

V =  327.6 dm³
S =  298.8 dm²

Solution:

$$\begin{gathered} a=3\text{ dm } \\ {v}_1=2.6\text{ dm } \\ h=1.4\text{ m}\to \text{ dm}=1.4\cdot 10\text{ dm}=14\text{ dm } \\ n=6 \\ {S}_1={\displaystyle \frac{a\cdot v_1}{2}}={\displaystyle \frac{3\cdot 2.6}{2}}={\displaystyle \frac{39}{10}}=3.9{\text{dm}}^2 \\ {S}_6=n\cdot S_1=6\cdot 3.9={\displaystyle \frac{117}{5}}=23.4{\text{dm}}^2 \\ V=S_6\cdot h=23.4\cdot 14={\displaystyle \frac{1638}{5}}={\displaystyle \frac{1638}{5}}{\text{dm}}^3=327.6{\text{dm}}^3 \end{gathered}$$

$S=2\cdot S_6+n\cdot a\cdot h=2\cdot 23.4+6\cdot 3\cdot 14={\displaystyle \frac{1494}{5}}={\displaystyle \frac{1494}{5}}{\text{dm}}^2=298.8{\text{dm}}^2$

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