# Tetrahedral pyramid

Calculate the volume and surface of the regular tetrahedral pyramid if content area of the base is 20 cm2 and deviation angle of the side edges from the plane of the base is 60 degrees.

Correct result:

V =  36.51 cm3
S =  72.915 cm2

#### Solution:

$S_{1}=20 \ \text{cm}^2 \ \\ Φ=60 \ ^\circ \ \\ \ \\ S_{1}=a^2 \ \\ a=\sqrt{ S_{1} }=\sqrt{ 20 } \doteq 2 \ \sqrt{ 5 } \ \text{cm} \doteq 4.4721 \ \text{cm} \ \\ \ \\ u=\sqrt{ 2 } \cdot \ a=\sqrt{ 2 } \cdot \ 4.4721 \doteq 2 \ \sqrt{ 10 } \ \text{cm} \doteq 6.3246 \ \text{cm} \ \\ \ \\ h=\dfrac{ u }{ 2 } \cdot \ \tan Φ ^\circ =\dfrac{ u }{ 2 } \cdot \ \tan 60^\circ \ =\dfrac{ 6.32455532034 }{ 2 } \cdot \ \tan 60^\circ \ =\dfrac{ 6.32455532034 }{ 2 } \cdot \ 1.732051=5.47723 \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \cdot \ S_{1} \cdot \ h=\dfrac{ 1 }{ 3 } \cdot \ 20 \cdot \ 5.4772=36.51 \ \text{cm}^3$
$b=\sqrt{ h^2 + (a/2)^2 }=\sqrt{ 5.4772^2 + (4.4721/2)^2 } \doteq \sqrt{ 35 } \ \text{cm} \doteq 5.9161 \ \text{cm} \ \\ \ \\ S_{2}=a \cdot \ b/2=4.4721 \cdot \ 5.9161/2 \doteq 5 \ \sqrt{ 7 } \ \text{cm}^2 \doteq 13.2288 \ \text{cm}^2 \ \\ \ \\ S=4 \cdot \ S_{2} + S_{1}=4 \cdot \ 13.2288 + 20=72.915 \ \text{cm}^2$

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