Calculate the volume (V) and the surface (S) of a regular quadrilateral prism whose height is 28.6 cm and the deviation of the body diagonal from the base plane is 50°.

Result

V =  8235.598 cm3
S =  2517.207 cm2

#### Solution:

$c=28.6 \ \text{cm} \ \\ A=50 \ ^\circ \ \\ \tan A=c:u \ \\ \ \\ u=c/\tan( A ^\circ \rightarrow\ \text{rad})=c/\tan( A ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=28.6/\tan( 50 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=23.99825 \ \\ \ \\ a=u /\sqrt{ 2 }=23.9982 /\sqrt{ 2 } \doteq 16.9693 \ \text{cm} \ \\ S_{1}=a^2=16.9693^2 \doteq 287.958 \ \text{cm}^2 \ \\ \ \\ V=S_{1} \cdot \ c=287.958 \cdot \ 28.6 \doteq 8235.5985 \doteq 8235.598 \ \text{cm}^3$
$S=2 \cdot \ S_{1} + 4 \cdot \ a \cdot \ c=2 \cdot \ 287.958 + 4 \cdot \ 16.9693 \cdot \ 28.6 \doteq 2517.2067 \doteq 2517.207 \ \text{cm}^2$

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